April 1, 2012
SQL syntax error while using MySQL and PHP
Question by liamwli
I am trying to make a thing, but I have hit a problem. I have tried all I know, but I am new to MySQL, so I have hit a dead end.
This code:
<?php
require('cfg.php');
mysql_connect($server, $user, $pass) or die(mysql_error());
mysql_select_db($database) or die(mysql_error());
if (isset($_GET['name'])){
$name = $_GET['name'];
}
else
if (isset($_POST['submit'])){
$name = $_POST['name'];
$name1 = $_POST['name1'];
$name2 = $_POST['name2'];
$name3 = $_POST['name3'];
mysql_query("INSERT INTO data (name, name1, name2, name3) VALUES($name, $name1, $name2, $name3 ) ") or die(mysql_error());
echo ("Data entered successfully!");
}
?>
<html>
<head>
<title>Random giffgaff simmer</title>
</head>
<body>
<form action="" method="post">
<p>Your Username: <input type="text" name="name"></p>
<p>Username 1: <input type="text" name="name1"></p>
<p>Username 2: <input type="text" name="name2"></p>
<p>Username 3: <input type="text" name="name3"></p>
<p>Username 4: <input type="text" name="name4"></p>
<p>Username 5: <input type="text" name="name5"></p>
<p>Username 6: <input type="text" name="name6"></p>
<p><input type="submit" name="submit" value="Submit"></p>
</form>
</body>
</html>
Brings this error:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ‘ )’ at line 1
Now, that would tell me that this SQL code has a syntax error:
INSERT INTO data (name, name1, name2, name3) VALUES($name, $name1, $name2, $name3 )
But I don’t think I can see one?
Answer by Starx
You haven’t quoted your query. You should quote every field like this
INSERT INTO data (name, name1, name2, name3) VALUES('$name', '$name1', '$name2', '$name3' )
As an tribute to TheCommonSense, I am providing a mysqli version using correct prepared statement for data safety
$db = new mysqli(...);
$stmt = $db -> prepare("INSERT INTO data (name, name1, name2, name3) VALUES(?, ?, ?, ?)");
$stmt -> bind_param("ssss", $name, $name1, $name2, $name3);
$stmt -> execute();
$db -> close()
