Search the Community

When I add a new column in MYSQL in phpmyadmin a new row is created (i guess its called a row) for example: if i first did this query with a php script. $sql = "INSERT INTO MyGuests (firstname, lastname, email)VALUES ('John', 'Doe', 'john@example.com')"; And later I go to phpmyadmin and add in MyGuests (telephonenumber) and VALUE this with 12345678 then this is done in a new row. I would rather have this all in one row. Can you maximize the number of rows and still add columns.

I am trying to start this online ordering system for a business. Basically identical to how the pizza-hut and dominos online ordering! How do I go about that? Do I learn how to write code? Buy software for my business? And if I do pursue software from another company, if the business becomes bigger, will that limit the rights that's I have for it to be mine? Please help. I'm new to this and really want to learn the ropes here!

--

hello there!!i m building an app for the elections..in this app i ve build a rest service that takes data(votes,etc) from a .js file of another website,and presents them to some fields.the problem is that.on that js file,the creators provide only "party_id" and not "party_name" which is contained in another .js file.i copied the id's and the names to my database.and now i want that:when the first js file gives the "Party_id",to search the database and provide me the "party_name"..what i want to do is to compare the values given from the rest service and the database collection and to return the correct name. i tried this code: var objectId = value; //Debug information that you can see in browser console.console.log("objectId = " + objectId);var whereObject = {"kwdikos": objectId};return JSON.stringify(whereObject); as shown here,it returns all the database array!! http://imageshack.com/a/img673/9877/EkpHcb.png but as it seems,it returns it without comparing..i mean,i want it when the id sent is 2,to return ND,and when id_sent is 1,to return PASOK.it just displays all the database array without criteria match!i tried to do it by using IF statement,but it didnt let me do it!! the correct results should be these:http://imagizer.imageshack.us/a/img537/7291/d2mTx9.png thanks for your time guys!!!

How would you go about making a hyperlink that can only be visited once before changing to another one? (Like tracking the amount of times the link was visited, and when it reaches a certain number it switches.)

i have a form in which i have a field date of birth , and a textbox to fill date of birth , when a person fills his date of birth like 01/01/1990 , then it will be saved into database , and when i retrieve this date form database then this comes 01/01/1990 12:00 AM , i want only date part from it. This is my form to fillup date of birth form.aspx <tr> <td width="30%"> Date of Birth <br /> </td> <td style="width: 1px"> :<asp:TextBox ID="txtDateofBirth" runat="server" placeholder="dd/mm/yyyy"></asp:TextBox> </td> <td class="td2"> </td> </tr> this is code for form.aspx.cs SqlCommand cmd = new SqlCommand("insert into ApplicantForm(NameofApplicant, dateofbirth)" + "values( @NameofApplicant, convert(Datetime, @dateofBirth, 103)", con); cmd.Parameters.AddWithValue("@dateofBirth",txtDateofBirth.Text); this is code for print of this date of birth print.aspx <asp:Label ID="txtDOB" runat="server" ></asp:Label> this is code print.aspx.cs SqlCommand cmd = new SqlCommand("Select * from ApplicantForm Where ApplicantId=" + ((Request.QueryString["ApplicantId"])) + "", con); dr = cmd.ExecuteReader(); txtDOB.Text = dr["dob"].ToString();

http://stackoverflow.com/q/26809213/4080419 GOAL: To have multiple people edit a Google Spreadsheet and have all edits display on a webpage (not Google site) where my css takes over the formatting and display of the content coming from the Spreadsheet. PROBLEM: I am brand new to PHP, as in I've been doing tutorials all week and this is my first time working with it. Likewise, I'm brand new to Google API's and don't fully understand how they work. I don't know what code to use or really how to modify it. WHAT I'VE TRIED: Google API obviously has the code for how to fetch the data to put into a page, so I've copied this, though I don't know if I put it in correctly/Dreamweaver says "There is a syntax error on line 32. Code hinting may not work until you fix this error". Additionally, I've read that perhaps Google Fusion Tables might be better because I can download it as a CSV version and then have that be what used in my actual script (though I don't fully understand how/if that would continually update itelf?): <html xmlns="http://www.w3.org/1999/xhtml"><head><title>Untitled Document</title><!--#include file="php-5.6.2.tar.bz2" --><?php include("php-5.6.2.tar.bz2"); ?><style type="text/css">#page {margin-left:auto;margin-right: auto;top: 15%;bottom: 15%;width: 1000px;}.menue {border-color:rgb(0,0,0);border-width:thick;height:300px;width: 300px;position: abosolute;margin-left: 10%;margin-right: 10%;}</style></head><body><div id="page"><div id="Skillet_Product_List_and_Price" class="menue"><?phpusing Google.GData.Client;using Google.GData.Spreadsheets;namespace MySpreadsheetIntegration{class Program{static void Main(string[] args){SpreadsheetsService service = new SpreadsheetsService("https://docs.google.com/a/emich.edu/spreadsheets/d/1OAFhMrayek9BKvO6ZI9EIPmCeY61oLL7-wfZOIg2xV8/edit#gid=724477237");// TODO: Authorize the service object for a specific user (see other sections)// Instantiate a SpreadsheetQuery object to retrieve spreadsheets.SpreadsheetQuery query = new SpreadsheetQuery();// Make a request to the API and get all spreadsheets.SpreadsheetFeed feed = service.Query(query);if (feed.Entries.Count == 0){// TODO: There were no spreadsheets, act accordingly.}// TODO: Choose a spreadsheet more intelligently based on your// app's needs.SpreadsheetEntry spreadsheet = (SpreadsheetEntry)feed.Entries[0];Console.WriteLine(spreadsheet.Title.Text);// Get the first worksheet of the first spreadsheet.// TODO: Choose a worksheet more intelligently based on your// app's needs.WorksheetFeed wsFeed = spreadsheet.Worksheets;WorksheetEntry worksheet = (WorksheetEntry)wsFeed.Entries[0];// Fetch the cell feed of the worksheet.CellQuery cellQuery = new CellQuery(worksheet.CellFeedLink);cellQuery.MinimumRow = 2;cellQuery.MinimumColumn = 1;cellQuery.MaximumColumn = 2;CellFeed cellFeed = service.Query(cellQuery);// Iterate through each cell, printing its value.foreach (CellEntry cell in cellFeed.Entries){// Print the cell's address in A1 notationConsole.WriteLine(cell.Title.Text);// Print the cell's address in R1C1 notationConsole.WriteLine(cell.Id.Uri.Content.Substring(cell.Id.Uri.Content.LastIndexOf("/") + 1));// Print the cell's formula or text valueConsole.WriteLine(cell.InputValue);// Print the cell's calculated value if the cell's value is numeric// Prints empty string if cell's value is not numericConsole.WriteLine(cell.NumericValue);// Print the cell's displayed value (useful if the cell has a formula)Console.WriteLine(cell.Value);}}}}?></div></div></body></html> But I've also tried this, which I was playing with : $file = "https://docs.google.com/a/emich.edu/spreadsheets/d/1OAFhMrayek9BKvO6ZI9EIPmCeY61oLL7-wfZOIg2xV8/edit#gid=724477237"$data = file_get_contents($file) or die('Could not read file!');echo $datafunction readRows() {var sheet = SpreadsheetApp.getActiveSheet(Eateries Menu Pricing.xlsx);var rows = sheet.getDataRange(A2:22);var numRows = rows.getNumRows(22);var values = rows.getValues();for (var i = 0; i <= numRows - 1; i++) {var row = values;Logger.log(row);}};var ss = SpreadsheetApp.getActiveSpreadsheet();var sheet = ss.getSheets()[0];// This represents ALL the datavar range = sheet.getDataRange();var values = range.getValues();//This logs the spreadsheet in CSV format with a trailing commafor (var i = 0; i < values.length; i++) {var row = "";for (var j = 0; j < values.length; j++) {if (values[j]) {row = row + values[j];}row = row + ",";}Logger.log(row);} And finally I've tried this just to try displaying a website thinking it'll work: <?php$myvar = "https://docs.google.com/a/emich.edu/spreadsheets/d/1O8-h-xmGE49K4x9sHrRL-c 97bHPXhma2eKkqrYAvog/edit#gid=1568863245";$var = fopen($myvar,"rb");echo stream_get_contents($var);?> DISCLAIMER: I understand that somewhat similar-ish questions have been asked before, but I am so new to PHP that I really need to be talked through this process. I've been looking everywhere online to try to understand this all, and I'm really trying to understand the process and how this all works

hello w3school mmbrs i have a q. thats makes me confused and its how can i insert data into WEB FORM from a DATABASE Assume : i have a database called 'home' and my table called 'vals' and the table 'vals' have 3 Fields 'username','val1','val2' and i create a web form that's have 2 inputs (val1,val2) so i want to put the values FROM THE DATABASE INTO MY WEB FORM note* , i want to insert data into web form (not from webform) any idea please ?!

I am making packet type system when in my **Packet Table** , parent packet ( `id` which is primary key ) and its N -Sub Packet is under (`parent_id`) is stored , below is my table structure :**Packet_table** id | packet_name | parent_id | 1 | 01 | 0 2 | 02 | 0 3 | 03 | 1 4 | 04 | 1 5 | 05 | 1 6 | 06 | 4 7 | 07 | 4 8 | 08 | 3 9 | 09 | 5 10 | 010 | 2 ........................so on and on with N packets in same table Below is what i have tried *but its not getting* `id` N sub packet detail properly: SELECT p.`packet_name` AS MAIN, s.`packet_name` AS SUB FROM packet_table s LEFT JOIN packet_table p ON s.`parent_id` = p.`id`thats as per above table : **id** ( which is primary / auto increment )id = 1 -> main packet (01) , its sub and N sub packets are :01 - > 03,04,0504 -> 06,0703 -> 0805 -> 09 **in short** 01 -> 03 -> 08 04 -> 06 , 07 05 -> 09 its not necessary above design format mysql code .. just simple N sub packet query will doabove is just few but in my case there will be N number of sub packet for each (id).So how it be can achieve .Note : it can be same as Category and N sub Category type

Hello W3S! It's been a while since i've been online here... sorry about that Anyways. I have some trouble with a dynamic menu i'm trying to make with MySQLI... not sure if that is the problem anyhow... Here is the code as a start, i will explain under the code below what i'm trying to do: // File that we are on (viewing / watching)$tab = pathinfo( $_SERVER['SCRIPT_NAME'], PATHINFO_FILENAME );$menu_res = query("SELECT * FROM menu WHERE menu_file_url = ".$tab);$menu_row = mysqli_fetch_row($menu_res);if( $tab == $menu_row['menu_file_url'] || $menu_row['menu_accessible'] == "no" ) stderr("Page Error", "We are currently working on this page! Go to another page to keep browsing! Thanks for your patience! :)");if(isset($CURUSER)){ $menu_while_res = query(" SELECT * FROM menu WHERE menu_accessible = 'yes' AND menu_view = 'user' OR menu_view = 'both' ORDER BY menu_order_id ASC");}else{ $menu_while_res = query(" SELECT * FROM menu WHERE menu_accessible = 'yes' AND menu_view = 'guest' OR menu_view = 'both' ORDER BY menu_order_id ASC");}$HTMLOUT .= "<ul class='nav_first'>"; while ($menu_while_row = mysqli_fetch_array($menu_while_res, MYSQLI_ASSOC)) { // Menu Items Loaded Here $tabarray = array( $menu_while_row['menu_array_id_name'] => "<li><a href='".$menu_while_row['menu_file_url']."'>".$menu_while_row['menu_name']."</a></li>", ); // K = Key // V = Value foreach($tabarray as $k => $v) { if( $tab == $k ) $HTMLOUT .= str_replace("<li>", "<li class='nav_active'>", $v); else $HTMLOUT .= $v; } // Unset Menu For re-load again unset($tabarray); }$HTMLOUT .= "</ul>"; Currently i'm trying to make a dynamic menu with MySQLI! It's working perfectly... but when i tried to "expand" the project a bit longer and try to make a dynamic menu with errors on pages if the users are not allowed to view a specific file, then i get nothing... What i'm trying to do is to controle in the database with "Enum" as my DB setup that is "no" is has been set on one of the menu items (menu_accessible).. then the item will NOT show on the menu! AND if the user still tries to enter that specific page by URL, he will then get an error message saying that we are working on the website page... The code i'm trying to insert into this project is this little peace of code here: // File that we are on (viewing / watching)$tab = pathinfo( $_SERVER['SCRIPT_NAME'], PATHINFO_FILENAME );$menu_res = query("SELECT * FROM menu WHERE menu_file_url = ".$tab);$menu_row = mysqli_fetch_row($menu_res);if( $tab == $menu_row['menu_file_url'] || $menu_row['menu_accessible'] == "no" ) stderr("Page Error", "We are currently working on this page! Go to another page to keep browsing! Thanks for your patience! :)"); However, i get no respond on the code! Even when i have checked if the "$tab" variable is real and related to the name inside the DB (which it is!)... so if possible, can anyone help and tell me what i'm doing wrong here Oo? Thanks alot by the way! And sorry for the long goodbye hehe ...studies and all, killing me! -.-' Anyways, hope some answers or good tips... really need this one Thanks in advance! Mr rootKID

i want to insert student result that he calculate some equation and send it to his cell in users database my problem is i cant send the result of some student to his own cell in the users table when i try to do it ,,its always (result) go to the first user that i logged in with ,, i think the issues with my SESSION can any 1 help please ??? my index.php code was <html><head><title>Home Work Page</title></head><body><form action="login.php?login=yes" method="POST"><table border=1> <tr> <td> username : <input type="text" name="username"/></td><br /><td> password :<input type="password" name="password"/> </td> </tr> </table><input type="submit" name="resultbtn" value='login'/></form></body></html> and my login.php code is <?phperror_reporting(0);$username = $_POST['username'];$password= $_POST['password'];$login = $_GET['login'];setcookie("username","$username",time()+15);if($login=='yes') {$con = mysql_connect("localhost","root","");mysql_select_db("members");$get = mysql_query("SELECT count(id) FROM users WHERE username='$username' and password='$password' ");$resultt = mysql_result($get,0);if($resultt !=1){ echo "error with login"; }else { $_SESSION['username'] = $username; $sql = "SELECT * FROM users WHERE username='$username' "; $myval = mysql_query($sql,$con);echo "welcome back " . $_COOKIE['username'];echo"<br/><br/><br/><br/><html dir='rtl'><meta charset ='Windows-1256'></html><table> <tr> <th> </th></tr>";if ($val = mysql_fetch_array($myval)) {echo "<tr>";echo " <td>" . "the question : calculate the triangle area if H = " . "</td>";echo " <td> <input type='text' value= $val[val1] disabled='disabled' size=5/> ". "</td>";echo " <td>and L = <input type='text' value= $val[val2] disabled='disabled' size=5/> ". "</td>";echo "</tr>"; }echo "</table>";}}echo"<form action='result.php' method='POST'><br/><br/><br/><br/><table><tr><td> Hello student the result will be : <input type='text' name='add' ></td><td><input type='submit' name='submit' value='send results'></td></tr></table></form>";?> and the result.php (the page with my problems) code is <?phperror_reporting(0);$val3 = $_POST["add"];//connectmysql_connect("localhost", "root", "");mysql_select_db("members");//insert$insert_query = mysql_query("UPDATE users SET results='$val3' WHERE username ='$_SESSION[username]' ");$query = mysqli_query($insert_query) or die(mysqli_error());//check whether the data insertion was successfulif(!$insert_query)echo "<p>Sorry! Something went wrong.</p>";elseecho "<p>Thanks! Your Results has been Sent.</p>"; my logout.php code is <?phpsession_start();unset($_SESSION['username']);session_destroy();header ("location: index.php");?> my table looks like

Hi, I'm trying to let JavaScript check if a givin user exist in the database. It seems that the _check-user.php always returns 0, but if I fill in a name that doesn't exist, and echo out the result variable in JS, the echo will return 1. Is there someone who could help me? JavaScript part: function checkUser() { $.post("_check_user.php", { 'username': document.getElementById("username").value }, function(result) { if (result == 1) { document.getElementById("checkUser").className = "succes"; document.getElementById("checkUser").innerHTML = "Name is available"; }else{ document.getElementById("checkUser").className = "errormsg"; document.getElementById("checkUser").innerHTML = "Name is not available!"; } });} _check-user.php: <?phpinclude("config.php");$result = mysql_query("SELECT login FROM users WHERE login='".clean_string($_POST["username"])."'");if(mysql_num_rows($result)>0){ echo 0;}else{ echo 1;}?>

Ok so I have been looking for how to do this all day(Very long time) and I can't find it anywhere. What I am trying to do is power my website with what I call URL variables the things that YouTube uses E.G. Page.php?video=1 but when it comes to selecting data from my database I have an issue. I have a table for ID and some others I will call them EX1 to EX5, what I am trying to do is the URL "?" thing says a number and that number is turned into a variable(don't know how) but that variable indicates the ID, but I don't want to put the ID on the page, I want the ID to tell my PHP code where in the tables the EX1 - EX5 are, so they can become a variable for later use on the page. Sum up. PHP takes the URL?name=1 turns into variable(1) - variable indicates where in the columns to look LOOKS UNDER ID finds 1 - now it needs to make the EX1 - EX5 in the same column into variables. ID EX1 EX2..... 1 I don't know how to do this can anyone help me?

Okay I'm biulding a website for a client which is required to do the following: Financial Advisor's from the company i am working for needs to be able to login to the website (by creating a account using a code unique to the company). Then be able to add client infomation to their own client list on a database (seperate from other advisors). From there the Advisors need to be able to go back and see a simplified list of their clients (i.e. just the clients name for example with a search option to find clients easily). Then when the click on the client's name they will be taken to a different page which displays the client's information in full, with the ability to edit this information. I understand the difficulty of this task and my skills are more directed towards the design aspect of websites, as you can probably make out I have little experince with databases and PHP, however I have taught myself other computer languages and not worried about the need to learn. All i'm asking for is for someone to outline what needs to be done and point me in the right direction as to where I can aquire the knowledge needed to complete this project. Thanks a million Ieuan.

i'm trying to make a install script. for that i wish to CREATE a DATABASE and at the same time CREATE some TABLEs for it with a single form submit? for this can DATABASE be selected within the "CREATE TABLE" query? if yes how? or there is any other idea?

can anyone explain how to use fpdf to coonvert fetched html format into pdf

In Chromium 32.0.1700.107 Ubuntu 12.04, I can't locate the directory in sqlite3 database? Any one to help me.

Good day Every one, I have checked the w3Schools web site W3Schools JavaScript Tutorial and W3Schools JavaScript Reference I can't seem to find anything. Any one can help me. Thanks,

I have many select insert update query , i have written transaction query but want to know if its safe or some need to be corrected. Also if all is ok i should get successful "Transaction completed successfully!" , but if something get wrong then error message showed be seen and ROLLBACK should be done , but its NOT .... not rollback is done Below is what I have tried. Please let me know if something is wrong: function autocommitfalse(){global $db;return mysqli_autocommit($db, FALSE);}function autocommittrue(){global $db;return mysqli_autocommit($db, TRUE);}function commitquery(){global $db;return mysqli_commit($db);}function rollbackedquery(){global $db;return mysqli_rollback($db);} =============================================================================== try {global $db;autocommitfalse();-----------------$SQLSELECT = "SELECT * FROM BC ";$result_as = mysqli_query($db,$SQLSELECT);$rsfetch = mysql_fetch_array($result_as);if(!$result_as ){throw new Exception('Wrong SQL SELECT: ' . $SQLSELECT. ' Error: '.db_error_msg($db) . db_error_no());}-----------------$sqla ="INSERT INTO A (sd) VALUES ('df') where id=".$rsfetch['dh'];$result_a = mysqli_query($db,$sqla);if(!$result_a){throw new Exception('Wrong SQL SELECT: ' . $result_a. ' Error: '.db_error_msg($db) . db_error_no());}-----------------$sqlb ="INSERT INTO B (sdds) VALUES ('trtrt') where id=".$rsfetch['dh'];$result_b =mysqli_query($db,$sqlb);if(!$result_{throw new Exception('Wrong SQL SELECT: ' . $result_a. ' Error: '.db_error_msg($db) . db_error_no());}-----------------$a = mysqli_insert_id($db);-----------------$sqlCb ="UPDATE CB SET A ='NM' where id=".$a ;$result_Cb = mysqli_query($db,$sqlCb);if($result_Cb === false || mysqli_affected_rows($db) == 0 ){throw new Exception('Wrong SQL SELECT: ' . $result_a. ' Error: '.db_error_msg($db) . db_error_no());}-----------------commitquery();echo 'Transaction completed successfully!';} catch (Exception $e) {echo"<br >";echo "<table align=center><tr><td>";echo $e->getMessage();echo "</td></tr></table>";echo"<br >";echo display_error(_("Transaction failed: transaction rolled back"));rollbackedquery();}autocommittrue(); Please let me know its correct or not , if need to correct then what , as still not rollbacked with above code..

're stuck with what field type in the list below. I will make 5 ferently databases that will eventually be built into the apps up a tablet. has tried me on something not quite get me to the way I think to have Her is one of databases everything is written in Norwegian.... Kjøretøy idKommune Nr Kommune FylkeStasjons Navn Pre-hospitaltjenesteAlarmsentral 113Aktive Kallesignal BilummerTypebil Reg Nr Fabrikkmerke Modell Påbygg Betegnelse Års Modell HistorikkStatus

Hi all,Im am trying to use a web form to send data to a table called "people" in my database using php.The script store the first time you submit the for and the data goes to the correct table too. The problem is it only allows you to do it only once meaning only 1 entry is registrered. If I try again, I get error message. Then i delete the entry from the database, I refresh the form, send again and it registers. I just need to know how to make it register more than one person's details. Below is the code <?phpif (isset($_POST['submitted'])) {include('w-hill-reg-form-database-connect.php');$fname = $_POST['fname'];$lname = $_POST['lname'];$sqlinsert = "INSERT INTO people (firstname, lastname) VALUES ('$fname', '$lname')";if (!mysqli_query($dbcon, $sqlinsert)) { die ('Error inserting new record');}$newrecord = "New record added";}?><html><head><title>CALLING ALL TENNIS PLAYERS!</title></head><body><form method="post" action="w-hill-registration-form-php.php"><input type="hidden" name="submitted" value="true" /><fieldset><legend>New People</legend> <label>First Name: <input type="text" name="fname" /></label> <label>Last Name: <input type="text" name="lname" /></label></fieldset><br/><input type="submit" value="Add new person" /></form><?phpecho $newrecord // New record added statement added at the top?></body></html>

I have made a form where some radio buttons are contained.The selection made by the user is stored in the db tables as a number, corresponding to the different values of each of the radio buttons. Here are the radio buttons with corresponding values: <td><input type="radio" class="formBuzType" name="buztype" value="1" > <label>test1</label><br></td> <td> <input type="radio" class="formBuzType" name="buztype" value="3"> <label>test3</label> <br></td> <td> <input type="radio" class="formBuzType" name="buztype" value="4"> <label>test4</label> <br></td> <td><input type="radio" class="formBuzType" name="buztype" value="5"> <label>test5</label> <br></td> <td><input type="radio" class="formBuzType" name="buztype" value="2"><label>test2</label> <br></td> If the user has chosen test3 then "3" will be stored in the database. Getting the value 3 from the db is not difficult(a simple SELECT query)...what I cannot do is making the appropriate preselection in the form depending on the fetched from the db value. To say it simpler...since 3 is in the db, I want when the radio buttons appear to the user(HTML) the value 3 option is selected so that the user can know what he has chosen the last time...how I will do that?

I am creating an admission form which will ask for the information such as name, class, level, and such other information, so, I don't know how to connect to the database and how to make the database and how the database will store all that information. Can anybody explain it with real example with the code?

Hello programmers! I have been given a project to make an online school admission system, but I don't have much ideas how to do that. So, I need your help. What things I should consider and how the look of the website should be and how to maintain the database? What to understand first? etcSo, please help me as much as you can. By suggesting some samples that you have etc.Thanks.

I'm working on a website that I want to be accessible to speakers of many languages but I'm wondering how to architect such a website. I seem to have two options and I hope you guys can help me choose: 1) Database I could make a table that lists all the messages and then have individual language tables with translations of those messages. Each page would then make a dozen or more queries to get all the messages/content to populate each page. I'm a bit worried about what the effect this may have on my databases and whether it is too much for them to handle in addition to all the other queries such as registration, login etc. 2) PHP array Using PHP arrays is also a popular solution. I think both MediaWiki (which drives Wikipedia etc) and phpMyAdmin go with this solution. It's easier to crowd source the translation this way since most people are familiar with pure text files. I'm worried about all the memory that is consumed by this approach and at what point it becomes inviable since every translation gets loaded into memory for every page request.