Determining if no rows are returned from a prepared statment

Question

I'm implementing prepared statements on my already working mysqli queries. I'm having trouble with a line if(mysqli_num_rows($result) == 0) as it's now a string instead of a mysqli_result.

if($nameAvailableStmt = mysqli_prepare($link, 'SELECT name FROM usetbl1 WHERE name=? LIMIT 1'))
{
    mysqli_stmt_bind_param($nameAvailableStmt, "s", $_POST['dangerous']);
    mysqli_stmt_execute($nameAvailableStmt);
    mysqli_stmt_bind_result($nameAvailableStmt, $result);
    mysqli_stmt_fetch($nameAvailableStmt);
}
if(mysqli_num_rows($result) == 0)

$result should be $nameAvailableStmt – Fabio Jun 11 '13 at 21:30 — Fabio, Jun 11 '13 at 21:30
it should bd `mysqli_num_rows($nameAvaialbleStmt)`. – Marc B Jun 11 '13 at 21:30 — Marc B, Jun 11 '13 at 21:30

Barmar · Accepted Answer · 2013-06-11T21:51:05.057

3

It should be:

mysqli_stmt_store_result($nameAvailableStmt);
if(mysqli_stmt_num_rows($nameAvailableStmt) == 0)

See the Documentation

edited Jun 11 '13 at 21:51

answered Jun 11 '13 at 21:30

Barmar

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Yes, it means that. `$result` will just contain the name that you were searching for. – Barmar Jun 11 '13 at 21:46
It still gives me the same error, I'll keep trying and clear the cache. – Celeritas Jun 11 '13 at 21:49
You need to call mysqli_stmt_store_result() first. – Barmar Jun 11 '13 at 21:51
Same ` Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, object given ` – Celeritas Jun 11 '13 at 21:52
Use `mysqli_stmt_num_rows()`. `mysqli_num_rows()` is for when you're not using prepared statements. – Barmar Jun 11 '13 at 21:53
thanks I think that does it, now the only problem is the statement is always true. – Celeritas Jun 11 '13 at 22:03
For some reason I had to add `mysqli_stmt_store_result($nameAvailableStmt)` for it to work – Celeritas Jun 13 '13 at 19:18
1

I know, I added that to the answer yesterday. – Barmar Jun 13 '13 at 19:58

Determining if no rows are returned from a prepared statment

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