How to check whether SELECT EXISTS returns a value or not?

Question

I am trying to quickly determine if a user_ID is the owner of a 'goal'. I believe my SQL query is good, but I'm trying to find a nice way of checking the result!

In this case, no matter what I put for $obj_id or $user_id, my function returns true. I assume it's because mysql_num_rows is counting even a false result as a row? So what PHP code should I use to check to see if the result exists or not?

Note that I want something short and elegant! I know I could do it the long way (check count(*), return mysql_assoc then check the count value...) but that is long winded and ugly.

Any ideas? Thanks!

$query = "SELECT EXISTS (SELECT * FROM goals WHERE goal_ID='$obj_id' AND user_ID='$user_id')"; if (@mysql_num_rows(mysql_query($query))!=1) { return false; } else { return true; }

Using count(*) should be preferred because it returns just one row instead of potentially millions. — Arjan
– Arjan, Commented Oct 9, 2013 at 15:55

gbn · Accepted Answer · 2011-11-20 09:32:05Z

Don't bother with EXISTS. The in-line exists will always give one row containing "true" or "false".

You're looking for either "zero rows" or "at least one row" so change the query to something like this and then check how many rows are returned

SELECT 1 FROM goals WHERE goal_ID='$obj_id' AND user_ID='$user_id' LIMIT 1

Brad Mace · Accepted Answer · 2010-12-19 20:54:09Z

I like gbn's answer the best, but I wanted to point out that this:

if (@mysql_num_rows(mysql_query($query))!=1) { return false; } else { return true; }

can be simplified to:

return @mysql_num_rows(mysql_query($query)) == 1;

I have some other stuff in my "if / else" sections in the real page, so I can't simplify quite that much, but I didn't realize you could just add the '==' like that. Thanks, that's useful!

Bas · Accepted Answer · 2013-07-08 14:54:09Z

This way is probably faster.

$query = "SELECT EXISTS (SELECT * FROM goals WHERE goal_ID='$obj_id' AND user_ID='$user_id')"; if(mysql_num_rows(mysqli_query($query)) < 1) { // Nothing found! }

This is indeed, faster. stackoverflow.com/questions/1676551/…

Jacek Laskowski · Accepted Answer · 2014-08-24 21:46:29Z

Counting how many rows match the criteria should be easier:

$sql = SELECT COUNT(*) FROM goals WHERE goal_ID='$obj_id' AND user_ID='$user_id' $query = mysql_query($sql); $result = mysql_fetch_row($query); return $result[0] >= 1;

Siebe Jongebloed · Accepted Answer · 2019-12-04 17:57:40Z

SELECT EXISTS always returns a row! The following code uses the fastest MySql query(according to this answer: https://stackoverflow.com/a/10688065/3710053) and gives in PHP the correct result:

$link = mysqli_connect($DB_SERV, $DB_USER, $DB_PASS, $DB_NAME); $query = "SELECT EXISTS (SELECT * FROM goals WHERE goal_ID='$obj_id' AND user_ID='$user_id' LIMIT 1) as `row_exists`"; if(mysqli_fetch_assoc(mysqli_query($link,$query))['row_exists'] ===0) { // Nothing found! }

ekochergin · Accepted Answer · 2021-03-05 10:47:12Z

Really working config for MySQL:

$sql="SELECT EXISTS (Select * FROM $dbname.$dbrel WHERE Index_r=$idx AND ... LIMIT 1)"; $r1=mysqli_fetch_row(mysqli_query($conn, $sql)); if (current($r1) == 0) {when no recoeds} else {when exist records }

tareco · Accepted Answer · 2011-01-14 15:49:00Z

what about this:

$query = "SELECT EXISTS (SELECT * FROM goals WHERE goal_ID='$obj_id' AND user_ID='$user_id')"; return mysql_query($query) ? false : true;

levent · Accepted Answer · 2014-09-03 10:48:52Z

mysql_result(mysql_query("SELECT EXISTS (SELECT * FROM goals WHERE goal_ID='$obj_id' AND user_ID='$user_id')"),0);

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