SQL request fails to retrieve data

Question

I'm trying to retrieve the email field from my database using the id associated with it:

$query2 = mysql_query("SELECT email FROM `users` WHERE `id` = '".$userID."' ") or die(mysql_error());

This query is always returning NULL and I can't work out why.

Have tested using a var_dump that $userID is indeed correct.

But when I use it with the hardcoded value instead of $userID it works fine:

$query2 = mysql_query("SELECT email FROM `users` WHERE `id` = '85' ") or die(mysql_error());

Why isn't the $userID variable being passed to my query? Is there a way to pass this correctly?

Edit:

Declaration of $userID as requested. var_dump of this variable works OK the line before the query2.

// Fetch ID for matching details
    $query = mysql_query("SELECT id FROM `users` WHERE `email` = '".$emailInput."' && `username` = '".$usernameInput."' ") or die(mysql_error());


        // Successful query - ID stored
        if(mysql_num_rows($query) > 0){
        $userID = mysql_fetch_array($query);}

        var_dump($userID);

Both var_dumps output the following on the page:

array(2) { [0]=> string(2) "85" ["id"]=> string(2) "85" } NULL

Hi Robert, yes I have done that as I said in the OP. It posts the correct result. — Francesca, Apr 01 '14 at 14:17
Francesca, Try storing the hole query on a var and dump it to see the result. And if Id is numeric you should not need the '' arround the 85 — Luis Tellez, Apr 01 '14 at 14:19
try `".$userID."` (that means passing the ID value as integer instead of string) — ponciste, Apr 01 '14 at 14:20
Post the output of echo $qry pls.... http://codepad.viper-7.com/aFyzU3 — Hackerman, Apr 01 '14 at 14:20
Id say the the fact that $userID is an array is your problem... Do `$userID['id']` instead in the query — superphonic, Apr 01 '14 at 14:24

score 3 · Accepted Answer · answered Apr 01 '14 at 14:26

3

Id say the the fact that $userID is an array is your problem... Do $userID['id'] instead in the query.

$query2 = mysql_query("SELECT email FROM `users` WHERE `id` = '" . $userID['id'] . "' ") or die(mysql_error());

answered Apr 01 '14 at 14:26

superphonic

7,954
6
30
63

Aous1000 · Answer 2 · 2014-04-01T14:28:34.087

0

try removing either double quotes or single quotes. because id is usually of data type int you are using twice which makes it a string value..

also PHP is quite smart in recognizing variables and data types so u can use "$userID" or "{$userID}" without concatination..

edited Apr 01 '14 at 14:28

answered Apr 01 '14 at 14:22

Aous1000

2,052
3
16
16

score 0 · Answer 3 · answered Apr 01 '14 at 14:26

Its probably a concatenation problem , if the $userid is string its should be fine , but if its an integer or double..etc it will be dealt with as a string

if its an integer try :

$query2 = mysql_query("SELECT email FROM `users` WHERE `id` = ".$userID) or die(mysql_error());

or

$query2 = mysql_query("SELECT email FROM `users` WHERE `id` = ".$userID['0']) or die(mysql_error());

score 0 · Answer 4 · answered Apr 01 '14 at 14:27

so, the problem is that your $userID is not 85 or a simple number but it's an array and you are still trying to concatenate it in the query.

The problem is somewhere else in your code, probably where you set $userID

Here goes the solution:

$sql = mysql_query("SELECT email FROM `users` WHERE `id` = " . $userID['id']) or die(mysql_error());

score 0 · Answer 5 · answered Apr 01 '14 at 14:32

0

Try this ....

$query2 = mysql_query("SELECT email FROM `users` WHERE `id` = $userID ") or die(mysql_error());

answered Apr 01 '14 at 14:32

Shashidhar Gr

420
2
6
15

SQL request fails to retrieve data

5 Answers5