phpmyadmin query error in php part2

Question

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mysql_fetch_array() expects parameter 1 to be resource, boolean given in select

It says my syntax has errors, I cant seem to find the error.

$query_admin = "SELECT * FROM `user_accounts` WHERE `id` =1 AND `name` LIKE {$login_user_name}AND `password` LIKE {$login_password }";

while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
        echo $query_result;
    }

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\Database Manager\admin\sql_functions.php on line 32

You are missing a space here: `{$login_user_name}AND` Please always output the final generated queries with the actual values inserted - they make most errors very obvious. Also use `echo mysql_error()` to see the SQL error — Pekka, Mar 06 '11 at 09:19
are you executing the query using the mysql_query() function ? — Muhamad Bhaa Asfour, Mar 06 '11 at 09:28
Yes I am executing the query using the mysql_query() function. — NSanjay, Mar 06 '11 at 09:37

Nanne · Accepted Answer · 2011-03-06T10:30:20.737

1

@Pekka is right that there is an error in your query: you need a space there.

Also, you use $result in your mysql_fetch_array(), but that is never actually set. It can be you omitted it, and it is just false because of the query error ofcourse, but otherwise that's a second bug :)

Your code example is a bit incomplete. In the comments you do mention you are actually doing the query using mysql_query(). But it is not shown in the code.

In what var do you save the result? Are you saving it in $result (as you use that for the mysql_fetch_array) and if so, what is $query_result in your while?

Try this:

$query_admin = "SELECT * FROM `user_accounts` WHERE `id` =1 AND `name` LIKE {$login_user_name} AND `password` LIKE {$login_password }";

$result = mysql_query($query_admin);

if (!$result) {
    die('Invalid query: ' . mysql_error());
}

while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
    var_dump($row);
}

and work from there.

edited Mar 06 '11 at 10:30

answered Mar 06 '11 at 09:25

Nanne

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what I want to do is compare sql query result with a var result, do I need to format both before comparing the sql query result var with the my other var ? – NSanjay Mar 06 '11 at 10:05
I don't really understand the question, but maybe this edit helps? – Nanne Mar 06 '11 at 10:10
Well, try to debug a bit. For instance, I've added some debugging code for you, that will echo the SQL error in case of a problem. This will tell you what might be wrong with your sql. (see @Pekka 's first comment) – Nanne Mar 06 '11 at 10:30
are you testing this query in php my admnin? – Muhamad Bhaa Asfour Mar 06 '11 at 10:32
Try the new code, and look at the error the "die" reports. – Nanne Mar 06 '11 at 10:59
the game just changed, forget this question. – NSanjay Mar 06 '11 at 13:01
you can still award a vote or even accept is as a viable answer on _this_ question of course – Nanne Mar 06 '11 at 13:02

phpmyadmin query error in php part2

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