PHP: Where clause will not execute when using a variable

Question

For the user I am testing with, their org_id column value is "student_life"

I am trying to have this function display whatever rows have the student_life column = 1. (so yes there is a column student_life which is a boolean, and then I also have a separate column named org_id and in this case has the value student_life)

I am pretty sure there is a syntax error but I cannot figure it out.

function org_id_users_table() 
{
$org_id = mysql_real_escape_string($_POST["org_id"]);

$sql = $this->query("SELECT * FROM ".DBTBLE." WHERE '$org_id' = '1'");
$result = $sql['sql'];
$num_rows = $sql['num_rows'];
$this->create_table($result, $num_rows);
}

(when I replace $org_id in the "$sql=..." line with student_life the code works.

So `$org_id` contains the column name that your searching through your table with for a value of `1` — Daryl Gill, Apr 19 '13 at 00:52
Yest it does, if I put '$org_id = 'student_life';' it works but the current code doesn't work. — user2296393, Apr 19 '13 at 01:24

Mark Parnell · Answer 1 · 2013-04-19T01:39:19.467

1

You're quoting the column name, which makes MySQL think it's a string.

$sql = $this->query("SELECT * FROM ".DBTBLE." WHERE $org_id = '1'");

Edit:

Based on your comments, I think what you actually want is this:

$sql = $this->query("SELECT * FROM ".DBTBLE." WHERE org_id = '$org_id'");

edited Apr 19 '13 at 01:39

answered Apr 19 '13 at 00:51

Mark Parnell

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That will make the query run as expected (assuming `DBTBLE` is defined and `$org_id` has the expected value). What that function actually returns will depend on what `$this` is. Are you seeing any errors? When you say it doesn't work, what exactly does that mean? Which part doesn't work? – Mark Parnell Apr 19 '13 at 01:08
I've narrowed the problem down to the first part where I am giving '$org_id' a value. For the user I am signed into, the org_id column in mysql is student_life. When I manually input '$org_id='student_life'' it works. – user2296393 Apr 19 '13 at 01:28
The value for $org_id needs to be pulled from the database column, org_id . That's what I'm having trouble with. – user2296393 Apr 19 '13 at 01:31
Huh? You're using the value of `$org_id` as the column name. You're saying now that's not what you want to do? – Mark Parnell Apr 19 '13 at 01:37
Sorry for the confusion, I am trying my best. I would like the value of my column, `org_id` to become the value of my newly created variable, `$org_id` – user2296393 Apr 19 '13 at 01:41

score 0 · Answer 2 · edited May 23 '17 at 11:43

0

Change quotes.

$sql = $this->query("SELECT * FROM ".DBTBLE." WHERE `$org_id` = '1'");

P.S. Why shouldn't I use mysql_* functions in PHP?

edited May 23 '17 at 11:43

Community

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answered Apr 19 '13 at 00:51

sectus

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No, the database just comes up empty, but I have many users that have student_life = 1 – user2296393 Apr 19 '13 at 01:29
The value for $org_id needs to be pulled from the database column, org_id . That's what I'm having trouble with. – user2296393 Apr 19 '13 at 01:32

score 0 · Answer 3 · answered Jul 15 '14 at 14:34

Where is this coming from? $_POST["org_id"]

Do you have a form on the page posting that? Or are you just trying to get that from the database? If so, wouldn't you need another query to obtain that first?

$row_MyFirstQuery['org_id']

Otherwise if it is $_POST["org_id"], wouldn't it be single quotes not double? $_POST['org_id']

PHP: Where clause will not execute when using a variable

3 Answers3