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| Thread title: SQL Not working |
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10-21-2006, 09:34 PM
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#1
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Status: Dediport Hosting
Join date: Jul 2006
Location: Berkshire
Expertise: programming, business
Software: Dreamweaver
Posts: 1,316
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 SQL Not working
Hi,
I have the code below and it doesnt seem to be inserting anything into the db, Could anyone help and is tehre a way to make sure that it has entered correctly without selecting?
Steven
PHP Code:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>New Admin</title>
<link type="text/css" rel="stylesheet" href="style.css" />
</head>
<body>
<?php
include_once('connect.php');
if (empty($_POST['submit']))
{
} else {
if (empty($_POST['name']) || empty($_POST['pass']))
{
echo "<div class=\"text\">Please fill in all fields.</div>";
} else {
$name = stripcslashes($_POST['name']);
$pass = stripcslashes($_POST['pass']);
$permission = '1';
$fpass = md5($pass);
$ip = $_SERVER['REMOTE_ADDR'];
$query = mysql_query("INSERT INTO users (`username` , `password` , `permissions` , `ip`) VALUES ('$name', '$fpass', '$permission', '$ip'");
if (isset($query)) {
echo "<div class=\"text\">User successfully added.</div>";
}
}
}
?><div class="text">
Enter the details for the new admin below. <br /><br />
<form name="newuser" method="POST" action="newadmin.php">
Username: <input type="text" name="name" /><br /><br />
Password: <input type="password" name="pass" /><br /><br />
<input type="submit" name="submit" value="Add user" />
</form></div>
</body>
</html>
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10-21-2006, 10:10 PM
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#2
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Status: design rockstar
Join date: Jan 2005
Location: guelph, ontario
Expertise:
Software:
Posts: 2,246
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well you never actually tell it to write to the databse, you're just storing the query in the $query variable. you need to add mysql_query($query).
PHP Code:
<?php
if(isset($_POST['submit'])) {
if (empty($_POST['name']) || empty($_POST['pass'])) {
echo "<div class=\"text\">Please fill in all fields.</div>";
exit();
} else {
$name = stripcslashes($_POST['name']);
$pass = stripcslashes($_POST['pass']);
$permission = '1';
$fpass = md5($pass);
$ip = $_SERVER['REMOTE_ADDR'];
$query = mysql_query("INSERT INTO users (`username` , `password` , `permissions` , `ip`) VALUES ('$name', '$fpass', '$permission', '$ip'");
if (mysql_query($query)) {
echo "<div class=\"text\">User successfully added.</div>";
} else {
echo "div class=\"text"\>User was not added, please check your code<./div>";
}
}
}
?>
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10-22-2006, 05:08 AM
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#3
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Status: Dediport Hosting
Join date: Jul 2006
Location: Berkshire
Expertise: programming, business
Software: Dreamweaver
Posts: 1,316
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Its not making the user now...
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10-22-2006, 01:01 PM
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#4
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Status: Member
Join date: May 2006
Location:
Expertise:
Software:
Posts: 119
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*sigh* Change it to this:
PHP Code:
$query = mysql_query("INSERT INTO users (`username` , `password` , `permissions` , `ip`) VALUES ('$name', '$fpass', '$permission', '$ip'") or die(mysql_error());
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10-22-2006, 01:49 PM
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#5
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Status: design rockstar
Join date: Jan 2005
Location: guelph, ontario
Expertise:
Software:
Posts: 2,246
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$query should just equal the string, it doesn't need mysql_query in there.
PHP Code:
$query = "INSERT INTO users (username , password , permissions , ip) VALUES ('$name', '$fpass', '$permission', '$ip')";
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10-22-2006, 02:57 PM
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#6
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Status: Request a custom title
Join date: Dec 2005
Location: Arizona
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Software:
Posts: 5,200
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Originally Posted by derek lapp
$query should just equal the string, it doesn't need mysql_query in there.
PHP Code:
$query = "INSERT INTO users (username , password , permissions , ip) VALUES ('$name', '$fpass', '$permission', '$ip')";
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Yup, because otherwise you guys would be using mysql_query twice.
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10-22-2006, 03:51 PM
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#7
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Status: Dediport Hosting
Join date: Jul 2006
Location: Berkshire
Expertise: programming, business
Software: Dreamweaver
Posts: 1,316
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Yehm Sorry i didnt post i fixed about 5 minutes after the second post i made.
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