2014 Commands out of sync; you can't run this command now

Hello All,

I was searching for a response about this topic but none get the right
answare, I hope i get help from you on this.

I trying to do an insert using a stored procedure to rely the charge
on the db but every type I run the call from my web page ussing PHP
5.X I get the same error.

This is what I'm doing:
______________________________________________________________________________________________
$link = mysql_connect( DB_HOST, DB_USER, DB_PASSWORD, true, 131074);
mysql_select_db(DB_NAME);

$sql = "call spTestCreate('" .
trim($Value1) . "', '" .
trim($Value2) . "', '" .
trim($Value3) . "', " .
$Value4 . ", " .
$Value5 . ", '" .
trim($Value6) . "', '" .
trim($Value7) . "', '" .
trim($Value8) . "') ; ";

$res = mysql_query($sql);

if (!$res) {
echo "Could not successfully run query (" . $sql . ") from DB: " .
mysql_error();
} else {
$row = mysql_fetch_row($res);
$LastId = $row[0];
}
______________________________________________________________________________________________

If I run the call from the SQLyog or any other tool it runs prefect
but that's not the case in the PHP.

Any Idea or Solution is welcome.

Thanks!

yaguirre.DeleteThis@gmail.com wrote:
> Hello All,
>
> I was searching for a response about this topic but none get the right
> answare, I hope i get help from you on this.
>
> I trying to do an insert using a stored procedure to rely the charge
> on the db but every type I run the call from my web page ussing PHP
> 5.X I get the same error.
>
> This is what I'm doing:
> ______________________________________________________________________________________________
> $link = mysql_connect( DB_HOST, DB_USER, DB_PASSWORD, true, 131074);
> mysql_select_db(DB_NAME);
>
> $sql = "call spTestCreate('" .
> trim($Value1) . "', '" .
> trim($Value2) . "', '" .
> trim($Value3) . "', " .
> $Value4 . ", " .
> $Value5 . ", '" .
> trim($Value6) . "', '" .
> trim($Value7) . "', '" .
> trim($Value8) . "') ; ";
>
> $res = mysql_query($sql);
>
> if (!$res) {
> echo "Could not successfully run query (" . $sql . ") from DB: " .
> mysql_error();
> } else {
> $row = mysql_fetch_row($res);
> $LastId = $row[0];
> }
> ______________________________________________________________________________________________
>
> If I run the call from the SQLyog or any other tool it runs prefect
> but that's not the case in the PHP.
>
> Any Idea or Solution is welcome.
>
> Thanks!
>
do an echo($sql) or print($sql) and post the results in here.