smiles
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Everything posted by smiles
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thanks pulpfiction, but I sure that it hasn't been solved yet What I am trying to do is having a drop down form, its items are get from database, when you choose one item and click submit button, the information of that item will appear below it .Anyone has another way better than what I did ?
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uhm ... if the content in form is "hello"your code shows "SELECT Information FROM Person WHERE Name=' Hello 'it make me feel a bit better !
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here is the code, I get no error and no display <?php include("1stfile.php"); $result1 = $_POST["theName"];$sqlstr = "SELECT Information FROM Person WHERE Name='" . $result1 . "'";$result2 = mysql_query($sqlstr) or die(mysql_error());while ($row = mysql_fetch_array($result2)) { echo $row["Information"]; }?>
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still not work
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banner is good, but for making website ... more limited features
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make it in Flash 8 and try to publish as Flash 5 version, you will see
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uhm ... could you tell me how to display the content in $result1 while ($row = mysql_fetch_array($result1)) { echo $row['Information']; }// these above code not work ???
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In SQL we have this syntax, for i.e ... SELECT Information FROM Person WHERE Name = 'Peter' Is there anyway to change 'Peter' to an variable which I get from a formlike this $result = %_POST['Some_Name_Here'];$result1 = mysql_query("SELECT Information FROM Person WHERE Name = $result ") ;//// these above code not work thanks !
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A document ( introduction) about Web designer job
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No.Unless you upload your file from hard drive to a web page, vice versa, you must use browser to open it
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it refers to database, you have a password in it and compare with the password that guest entered, if similar then guest can enter ...
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I think he means a banner with buttons and when you click them, you will change the HTML contents (direct to another page) below that bannerI have no idea about this quality
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I start when I first choice this forum 1-December-2005, just more than 1 year
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Thanks! That command run well while($row = mysql_fetch_array($result2)) { echo $row['Name']." ".$row['Infor']; } Is there anyway to echo the first ( or 2nd,3rd) element of that array ???
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Well I understand but here is the new code and get that warningWarning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\xampp\htdocs\giup1.php on line 6 <?php include("giup.php"); $result1 = $_POST["theName"];echo $result1;$result2 = mysql_query("SELECT * FROM Person");while($row = mysql_fetch_array($result2)) // line 6 here { echo $row['Name']." ".$row['Infor']; }?> my Person's database isName | Infora.......| boyb.......| girl
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wow Your computer maybe very strong !!!
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That's so simple, thanks friends
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Hi I have just had a problem, show in my code <?php include("giup.php"); $result1 = $_POST["theName"];echo $result1; // I see the result, for example 'a'$result2 = mysql_query("SELECT Infor FROM PersonWHERE Name='$result1' "); // Person with Name 'a' has the Info is 'boy'echo $result2; // now don't see the result 'boy'?> How can I make it run ?
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I found it A commamysql_select_db("my_database", $connect);$sql = "CREATE TABLE Person(Name varchar(15),Infor varchar(15), // it's here)";mysql_query($sql,$connect);
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http://www.radiologyworkers.com I have an important topic waiting for their answer and what kind of this link mms://..................
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Just to try to create a database name my_database and create a table but ... database created, table isn't created ??? <html><body><?php$connect = mysql_connect("localhost","root","");if (!$connect) { die('Could not connect:'.mysql_error()); }if (mysql_query("CREATE DATABASE my_database",$connect)) { echo "Database created"; }else { echo "Error :".mysql_error(); };mysql_select_db("my_database", $connect);$sql = "CREATE TABLE Person(Name varchar(15),Infor varchar(15),)";mysql_query($sql,$connect);// Suppose insert some infomysql_query("INSERT INTO Person (Name,Infor) VALUE ('a','boy')");mysql_query("INSERT INTO Person (Name,Infor) VALUE ('b','girl')");// We get// ------------------// | Name | Infor |// | a | boy |// | b | girl |// ------------------// a,b live in drop down list $result = mysql_query("SELECT Name FROM Person");echo "<form>";echo "<select name='theName'>";while ($theArray = mysql_fetch_array($result)) { echo "<option value=' ".$theArray['Name']." '>".$theArray['Name']."</option>"; }echo "</select>";echo "</form>";?><hr/></body></html>
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I am doing with Xampp and could you help me with these questionwhat is MyISAM Type and beside text, can we put insert image into database ?
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.flv is the succesful file in anti-download field :)so .flv -> .mpeg is hard work
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delete image.gif and put another image into the folder contain the previous image and rename it as image.gifhope I understand you right ???