shortkid422
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Posts posted by shortkid422
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ya, it is. each number will increase by one, so it would be 1,2,3,4,5,6... and so on. It will be increased everytime a row is added, so no work.
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i got it fixed, somewhat. There needed to be a space by the user( part. But one problem that im having is the values are not being passed by the form. Here it is:
<?php session_start(); ?> <form action="register_member.php" method="post"> <table> <tr> <td>Please enter your desired username here:</td><td><input type="text" name="" value="<?php echo $username ?>"></td> </tr> <tr> <td>Enter your password that will be used:</td><td><input type="text" name="" value="<?php echo $user_password1 ?>"></td> </tr> <tr> <td>Please re-enter your password again:</td><td><input type="text" name="" value="<?php echo $user_password2 ?>"></td> </tr> <tr> <td>Now please enter your valid email address:</td><td><input type="text" name="" value="<?php echo $email ?>"></td> </tr> <tr> <td><input type="submit" value="Register!"></td><td><input type="reset" value="Clear"></td> </tr> </table> </form>
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Im having a problem with this query, can someone help me out?
$query1 = "INSERT INTO user(username, userpassword, email) VALUES(\'$username\',\'$userpassword\',\'$email\')"; $result1 = mysql_query($query1) or die(mysql_error());
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this is just from the database. Price is an integer value.
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Why wont the $price get echoed like i want it to when I select it?
<? include("misc.inc"); $connection = mysql_connect($host,$user,$password) or die ("couldn\'t connect to server"); $db = mysql_select_db($database,$connection) or die ("Couldn\'t select database"); $query = "SELECT * FROM flights ORDER BY \'price\'"; $result = mysql_query($query) or die ("Couldn\'t execute query."); echo "$price <br>"; ?>
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k, thanks i, i just didnt want to do 78 queries, which is what i did. I love copy and paste!
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How can i insert more than one value per variable WITHOUT using arrays?
Creating Table Problem
in MySQL Help
Posted
I think that the table needs to be in quotes.
[php:1:18aecfb026]<?php
if(mysql_list_tables($dbname) != news){
CREATE TABLE \'news\' (
author VARCHAR(30) NOT NUUL,
dateposted DATETIME NOT NULL,
title VARCHAR(100) NOT NULL,
content CHAR(99999) NOT NULL,
) }
?>[/php:1:18aecfb026]