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mysql_fetch_row issue. Please read.
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| sectionFOURTEEN 2006-03-29, 6:57 pm |
| Here's a snippet of my script:
$result = mysql_query("SELECT count(*) FROM cart WHERE
cookieId = '" . GetCartId() . "' AND prod_id = $prod_id"); //line 39
$row = mysql_fetch_row($result);
$numRows = $row[0];
I'm getting the "supplied argument is not a valid MySQL result resource
on line 39" error. I know it's getting returned because there aren't
any records that match this query. However, I would like it to return
zero records instead of this error. I have the rest of the script setup
to handle zero records returned but, this error is being returned
instead. This script works when there are records to return.
I was thinking about an if statement possibly but, I'm not sure which
direction to go with this.
Any help would be greatly appreciated!
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| Colin McKinnon 2006-03-29, 6:57 pm |
| sectionFOURTEEN wrote:
> Here's a snippet of my script:
>
> $result = mysql_query("SELECT count(*) FROM cart WHERE
> cookieId = '" . GetCartId() . "' AND prod_id = $prod_id"); //line 39
> $row = mysql_fetch_row($result);
> $numRows = $row[0];
>
> I'm getting the "supplied argument is not a valid MySQL result resource
> on line 39" error. I know it's getting returned because there aren't
> any records that match this query. However, I would like it to return
> zero records instead of this error. I have the rest of the script setup
> to handle zero records returned but, this error is being returned
> instead. This script works when there are records to return.
>
Off the top of my head...
$numRows = (integer)$row[0];
?
C.
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| diogo86@gmail.com 2006-03-29, 6:57 pm |
| this error message doesn't seem to be about no records found but a sql
error on the query.
try this:
$res = mysql_query("SELECT count(*) FROM cart WHERE
cookieId = '" . GetCartId() . "' AND prod_id = $prod_id") or
die(mysql_error());
$numRows = mysql_num_rows($res) ? mysql_result($res,0) : 0;
mysql_free_result($res);
when you correct the query and it's no longer returning an error you
show remove " or die(mysql_error())" to avoid relevant data to be show
to the user just in case it has an error.
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