SQL QUERY ERROR IN PHP.

[maazzarif]

SQL QUERY PROBLEM IN PHP.

I m using windows xp O.S. I have installed WAMP for learning and practicing PHP.

when I execute the following code i m getting an error.I M INCLUDING A FUNCTION FILE(functions.php)

IN MY content.php file.MY ALL OTHER FUNCTIONS AND SQL QUERIES ARE WORKING EXCEPT THIS ONE.

 

 

----------------------------------------------------------------------------------------------------------------

functions.php(c:/wamp/www/widget_corp/includes/functions.php)

 

function get_subject_by_id($subject_id){

 

global $connection;

$query = "SELECT * ";

$query .= "FROM subjects ";

$query .= "WHERE id=".$subject_id;

$query .= "LIMIT 1";

$result_set = mysql_query($query,$connection);

if(!$result_set){

 

die("database query failed:".mysql_error());

}

//if no rows are returned, then fetch array is going to return false

if($subject = mysql_fetch_array($result_set)) {

return $subject;

} else {

return NULL;

}

 

}

 

 

-----------------------------------------------------------------------------------------------------------------

 

 

content.php(c:/wamp/www/widget_corp/content.php)

 

<?php

require("constants.php");

 

//Creating a database connection

 

$connection = mysql_connect(DB_SERVER,DB_USER,DB_PASSWORD);

if(!$connection){

die("DATABASE CONNECTION FAILED:".mysql_error());

}

?>

 

 

<?php

 

//Selecting a database

 

$db_connect = mysql_select_db(DB_NAME,$connection);

if(!$db_connect){

die("Database selection failed:".mysql_error());

}

?>

 

 

<?php //including functions.php ?>

 

<?php require_once($_SERVER['DOCUMENT_ROOT']."/widget_corp/includes/functions.php") ?>

 

 

<?php //$_GET['subj'] and $_GET['page'] are returning my subjects and pages ids respectively. ?>

 

<?php

if(isset($_GET['subj'])){

$sel_subj = $_GET['subj'];

$sel_pg="";

}elseif(isset($_GET['page'])){

$sel_pg= $_GET['page'];

$sel_subj = "";

} else {

$sel_subj = "";

$sel_pg = "";

}

$sel_subject = get_subject_by_id($sel_subj);

$sel_page = get_page_by_id($sel_pg);

?>

 

 

------------------------------------------------------------------------------------------------------------------

 

ERROR(ON BROWSER):

 

database query failed: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 'LIMIT 1' at line 1

[denno020]

I think you might need to add quotes around subject_id

so like this:

 

$query .= "WHERE id=`$subject_id`";

[maazzarif]

@

denno020

 

I think it will make $subject_id as a string.Well i tried it and i got just a slight change in the error. Following is the error i got:

 

database query failed: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 '1' at line 1

 

database query failed: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 '1' at line 1

Edited February 17, 2013 by maazzarif

[Barand]

@denno, you should not quote number values.

 

@mazzarrif, echo $query to see what it looks like. I think you may have a spacing issue.

 

edit.

Create query code with multiple lines

 

$query = "SELECT *
FROM subjects
WHERE id = $subject_id
LIMIT 1";

 

Not only is its structure easier to read but when you get a syntax error it tells you which line of the query. This way everything is not on line 1.

Edited February 17, 2013 by Barand

[maazzarif]

@denno, you should not quote number values.

 

@mazzarrif, echo $query to see what it looks like. I think you may have a spacing issue.

 

edit.

Create query code with multiple lines

 

$query = "SELECT *
FROM subjects
WHERE id = $subject_id
LIMIT 1";

 

Not only is its structure easier to read but when you get a syntax error it tells you which line of the query. This way everything is not on line 1.

Sir i did as you said. and this is what i got

database query failed: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 'LIMIT 1' at line 4

 

database query failed: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 'LIMIT 1' at line 4

Edited February 17, 2013 by maazzarif

[Barand]

echo $query; ?

[j0nnie_m]

The error:

"Database query failed: 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 'LIMIT 1' at line 1"

Can Be fixed, try this:

change $query .= "LIMIT 1"; to $query .= "LIMIT 0,1";
and $query .= "WHERE id=" . $subject_id ." "; to $query .= "WHERE id=" . (int)$subject_id ." ";

regards!

[Barand]

j0nnie, LIMIT 1 and LIMIT 0,1 are identical.

 

It is more likely that $subject_id has no value but as the OP refuses to echo $query to verify then who knows. My view is that when more info (like echo $query) is requested but the OP can't be bothered then the only thing to do is walk away.

[priyankagound]

Best to echo the query and see what it looks like.

 

Probably $subject_id contains no value or an invalid value. If $subject_id is a string, you should escape it (using mysql_real_escape_string) and put it inside quotes in the query.

[Edit]

You know you can put enters in strings too, right?

// More readable
$query = "
SELECT *
FROM subjects
WHERE id = $subject_id
LIMIT 1";

[cyberRobot]

There needs to be a space between the WHERE and LIMIT clauses.

 

 

<?php
$query .= "WHERE id=$subject_id ";  //<-- added a space here
$query .= "LIMIT 1";
?>