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MrB

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Posts posted by MrB

  1. 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

     

    1403396701241.png

     

  2. First you get value from your db.

    The query is

    $q=mysql_query("select * from tbl_name where id=$id") or die(mysql_error());$res=mysql_fetch_assoc($q);

    Then in your text box

    <input type="text" name="name" value="<?php echo $res['username']">

    You need to add

    error_reporting('E_ALL ^ E_NOTICE');

    at first of your file to avoid errors.

     

     

    thank u 4 ur reply :-)

    i solve it

  3. my first login system code was

    <!DOCTYPE html><html><body bgcolor="#A3FFD1"><?php//error handlingfunction customError($errno, $errstr) {  echo "<b>Error:</b>[$errno] $errstr";}//custom error handlingset_error_handler("customError");$username = $_POST['username'];$password = $_POST['password'];if (username == '123' AND password =='1234') {	if(file_exists('exam.php'))	 	require('exam.php');	else{		echo"login error";	}}else{echo "<font size=40 ><b>ِAccess Denied </b></font>" ;}?></body></html>

    and after that i found some security login code

    <!DOCTYPE html><html><body bgcolor="#A3FFD1"><?php//uses local host just for testing$connect =mysql_connect("localhost","root","") or die ("error with connection");//my database called 'members' and table called 'users'mysql_select_db("members",$connect) or die ("error with data connection");error_reporting(0);if ($_POST['login']){   if($_POST['username'] && $_POST['password']) {      $username=mysql_real_escape_string($_POST['username']);     $password=mysql_real_escape_string(hash("sha512",$_POST['password']));     $user=mysql_fetch_array(mysql_query("SELECT * FROM 'users' WHERE 'username' = '$username' "));     if($user='0'){         die ("u r not registered with our data base ,, U CANNOT LOGIN , u might <a href='contactme.php' </a> ");                }if ($username['password'] != $password){       die("Incorrect password,..try again or <a href='contactme.php' ");                                     }$salt = hash("sha512",rand().rand().rand());setcookie ("c_user",hash("sha512",$username),ttime()+24*60*60,"/"));setcookie ("c_salt",$salt,time()+24*60*60,"/");$userID = $user['ID'];mysql_query("UPDATE 'users' SET 'salt' = '$salt'  WHERE 'ID' = '$userID' ");if(file_exists('exam.php'))require('exam.php');else{echo"Error ,.. No such page please try to <a href='login.php>log in </a>'";}}else{echo "<font size=40 ><b>Access Denied </b></font>" ;}    }}?></body></html>
  4. now where can i called the user value and how i put them in the input text

     

    <input type="text" name="LofT" disabled="" size="5" value='<?php $_POST[val1]; ?>' />

    <input type="text" name="HofT" disabled="" size="5" value='<?php $_POST[val2]; ?>' />

     

    the blue code is it right to called data from user values table?

  5. i didnt connect to my user database but i assume i did and i wrote this code >>> its right or ?

    <!DOCTYPE html><html><body bgcolor="#A3FFD1"><html dir='ltr'><meta charset='UTF-8'/><head><title>Home Works Page</title></head><body><br/>     <br/><br/>     <br/><font size='14' style='bold'><b>Hello:</b> <?php echo $username; ?> </font><form action=""  method="POST"><br/>     <br/><br/>     <br/> <table>  <tr>     <td>Calculating the area of a triangle if the length of the base of the triangle =</td>    <td>  <input type="text"  name="LofT"  disabled="" size="5" value='<?php $_POST[val1]; ?>' /> </td>    <td>  and the height or altitude of the triangle = <input type="text"  name="HofT" disabled="" size="5"  value='<?php $_POST[val2]; ?>' /> </td>  </tr></table>  <br/>  <br/>  <br/>  <br/>  <br/>  <br/>  <br/>  <br/><font size=14 style=bold>  The Solution will be</font><table><br/><br/>      <tr>        <td> The Result: <input type="text" name="result"/> <select name="sel1" id="sel1"><option></option><option id="op1">Meter</option> <option>Cm </option> <option>Mili-meter</option></select> Choose Measurement Unit</td>       </tr>            <td><input type="submit"  name="resultbtn" value="Send Solutions"/></td>     </tr></form></body></html></body></html>
  6. Yeah I did lookmmm it didn't Works with meMy Question was :I have some question in my page that has a 2 variable values (val1,val2) and these values changes by lookup table ..means that evry user in my database has a different values (val1,val2) appears in a FORM.... and thanks so much for ur reply

  7. 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 ?!

     

     

     

     

     

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