mysql_query returns a row, but mysql_fetch_array does not do anything

Question

I have a strange problem this time. I've got a masking database setup that takes in a url and gives it a string mask so that I can redirect users to another site (direct download) by giving them a link to my site's redirection engine.

I'm having a problem in the administrator side (where I put in the urls) where my mysql_fetch_array command just doesn't run. I know the query is returning at least one row because when I run the following code:

$query = "SELECT * FROM `urls` WHERE `in`='$lks[$i]' ORDER BY `id` DESC LIMIT 1";

$r = mysql_query($query);

if(mysql_num_rows($r) < 1)
echo "Less than 1";
else
echo "At least 1";

it returns "At least 1", but when it gets down to the mysql_fetch_array while statement, nothing happens. It's like it just ignores it completely.

CODE:

$query = "SELECT * FROM `urls` WHERE `in`='$lks[$i]' ORDER BY `id` DESC LIMIT 1";

$r = mysql_query($query);

while($me = mysql_fetch_array($r))
{
    echo "Blah";
}

Nothing is echoed by this code. What's wrong?

By the way, I know that I am connected to the database and everything because I run other commands very similar to this before this one and they all execute without any problems.

Also this command is in a for loop (I don't know why this would be a problem because I have done this many times before and never had any problems).

please don't use mysql_* functions, it's deprecated (see red box) and vulnerable to sql-injection. Use PDO or MySQLi. — Nir Alfasi
– Nir Alfasi, Commented Aug 18, 2012 at 1:52
do a var_dump($me); also, echo out your $query variable to make sure it contains what you think it does. I'm not sure if array indexing inside "" will work properly. — databyss
– databyss, Commented Aug 18, 2012 at 1:57
where do I do the var_dump? After the while statement terminates, $me returns bool(false)? — Liftoff
– Liftoff, Commented Aug 18, 2012 at 2:02

Jared Drake · Accepted Answer · 2012-08-18 02:03:55Z

1

try echoing mysql_error(); I'm thinking it might be that you put your column and table names in quotes.

$query = "SELECT * FROM urls WHERE in='$lks[$i]' ORDER BY id DESC LIMIT 1";
$r = mysql_query($query);
if($r)
{
$rowCount = mysql_num_rows($r);
if($rowCount == 1)
{
echo "1 row";
}
}
else
{
echo mysql_error(); 
}

if this doesn't work there must be something wrong with your array. :/

edited Aug 18, 2012 at 2:03

answered Aug 18, 2012 at 1:51

Jared Drake

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8 Comments

Liftoff Over a year ago

echo mysql_error(); returned nothing. Changing 'in' (hash quotes, not apostrophes) to in gave me an invalid resource error in the mysql_fetch_array() line.

Liftoff Over a year ago

Same thing. if($r) does run, but mysql_error returns nothing.

Shackrock Over a year ago

I also recommend NOT quoting column names. AND on most servers they are case sensitive.

Jared Drake Over a year ago

lol Sorry for all the edits guys :( I can't figure this stupid indention out.

Liftoff Over a year ago

Same thing, the resource is valid, it has returned one row, but it will not run mysql_fetch_array($r)

|

rx80 · Accepted Answer · 2012-08-18 01:57:54Z

0

This is wrong:

$query = "SELECT * FROM urls WHERE in='$lks[$i]' ORDER BY id DESC LIMIT 1";

should be:

$query = "SELECT * FROM urls WHERE in='{$lks[$i]}' ORDER BY id DESC LIMIT 1";

But like stated earlier, use PDO.

answered Aug 18, 2012 at 1:57

rx80

1564 bronze badges

1 Comment

Liftoff Over a year ago

Adding that still did nothing. I'm still getting a valid result with 1 row in it, but the mysql_fetch_array command will not run.

Shehzad Bilal · Accepted Answer · 2012-08-18 02:02:25Z

0

You are checking for less then one row kindly echo mysql_num_rows($r) and check how many rows were returned.

or use die(mysql_error()) alongwith mysql_query to check for error.

answered Aug 18, 2012 at 2:02

Shehzad Bilal

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4 Comments

Liftoff Over a year ago

echo mysql_num_rows($r); returns 1

Shehzad Bilal Over a year ago

Then it is impossible that it will not enter the while loop. Check for any syntax error in your code.

Liftoff Over a year ago

Why do you think I'm so confused. I even took it out of the while loop and hard-coded a url into the query so that I could see if that was the problem and the same thing happens, every time.

Shehzad Bilal Over a year ago

use print_r(mysql_fetch_array($r)) to check the result... whats the output. (Not in while loop)

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