How do I fix the "syntax" issue?

Question

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You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ''information' WHERE 'id' = 'SELECT `id` FROM `account` WHERE `username` ='stswil' at line 1

PHP

<?php
include "connection.php";
$username =$_POST['username'];
$password =$_POST['password'];
$query = " SELECT `username`, `password` FROM `account` WHERE `username` = '$username' AND `password` = '$password' ";
$id = "SELECT `id` FROM `account` WHERE `username` ='$username' ";
$IDquery = " SELECT 'id' FROM 'information' WHERE 'id' = '$id' ";
$IDresults = mysqli_query($conn, $IDquery) or die(mysqli_error($conn));
$result="";
$results = mysqli_query($conn, $query) or die(mysqli_error($conn));

if(mysqli_num_rows($results)!=1){
	$result="Non-Vaild user";
}
else if(mysqli_num_rows($IDresults)!=1){
	header( 'Location: https://xesusanproject.000webhostapp.com/information.html' ) ;
}
else{
	header( 'Location: https://xesusanproject.000webhostapp.com/information.php' ) ;
}
?>
<!DOCTYPE html>
	<head>
		<link rel="apple-touch-icon" sizes="180x180" href="images/favicons/apple-touch-icon.png">
		<link rel="icon" type="image/png" href="images/favicons/favicon-32x32.png" sizes="32x32">
		<link rel="icon" type="image/png" href="images/favicons/favicon-16x16.png" sizes="16x16">
		<link rel="manifest" href="manifest.json">
		<link rel="mask-icon" href="images/favicons/safari-pinned-tab.svg" color="#741372">
		<meta name="apple-mobile-web-app-title" content="Medical Database">
		<meta name="application-name" content="Medical Database">
		<meta name="theme-color" content="#741372">
		<link rel="stylesheet" href="style/main.css" type="text/css">
		<title>Sucess-Login</title>
	</head>
	<body>
		<menu>
			<ul>
				<li><a href="logout.php">Log Out</a></li>
			</ul>
		</menu>
		<span style="margin:0px 0px 0px 10px"><?php echo $result ?></span>
	</body>
</html>

Update

New error : You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'stswilliams15' '' at line 1

from updating

PHP

$IDquery = " SELECT `id` FROM `information` WHERE `id` = '$id' ";

Wanted to get the id from account (The person just logged in with that information). Then find the matching id in the information to see if there data already in information table. If there no matching id in the information table then I want to go to "https://xesusanproject.000webhostapp.com/information.html" to added the data. If there is matching id then I want the data to be shown

What I have tried:

checking vairables, asking for people in real life

Posted 12-Jan-17 10:51am

Katherynliza

Updated 12-Jan-17 15:43pm

v3

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Comments

j snooze 12-Jan-17 17:40pm

You don't need parenthesis around SELECT `id` FROM `account` WHERE `username` ='$username'? Most SQL I know of needs parenthesis when doing a subquery like that.

Katherynliza 12-Jan-17 18:53pm

What do you mean?

Mohibur Rashid 12-Jan-17 17:47pm

Your IDquery statement is wrong, you have mixed up between grave accent(`) and single quote('),

Katherynliza 12-Jan-17 18:53pm

Thanks, but now is is this error. You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'stswilliams15' '' at line 1

Mohibur Rashid 12-Jan-17 19:41pm

Echo your sql, run that from console and try to understand the result.

Katherynliza 12-Jan-17 20:14pm

<?php echo $query ?>
<?php echo $id ?>
<?php echo $IDquery ?>
<?php echo $IDresults ?>
<?php echo $result ?>
Add these does nothing

Mohibur Rashid 12-Jan-17 22:19pm

This is not supposed to solv your prblem. This will help you to debug. What is query you are getting?

Katherynliza 13-Jan-17 7:05am

Nothing, no query

1 solution

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Patrice T · Accepted Answer · 2017-01-12T14:44:00

Quote:
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ''information' WHERE 'id' = 'SELECT `id` FROM `account` WHERE `username` ='stswil' at line 1

The problem comes from

PHP

$id = "SELECT `id` FROM `account` WHERE `username` ='$username' ";
$IDquery = " SELECT 'id' FROM 'information' WHERE 'id' = '$id' ";

In second line, the $id is replaced by its contain, second line become.

PHP

$IDquery = " SELECT 'id' FROM 'information' WHERE 'id' = 'SELECT `id` FROM `account` WHERE `username` ='$username' ' ";

Need to know what was your intend there to understand what is wrong.