You are here: Re: Commands out of sync error « PHP Programming Language « IT news, forums, messages
Re: Commands out of sync error

Posted by Ratfish on 09/27/48 12:00

On Jan 16, 4:06 am, Jerry Stuckle <jstuck...@attglobal.net> wrote:
> Ratfish wrote:
> > I'm getting a "2014:: Commands out of sync; you can't run this command
> > now" error on a php page when I try to call a second stored procedure
> > against a MySQL db. Does anyone know why I might be getting this
> > error? The error doesn't occur on my development box where I use the
> > 'root' db user, but does occur in production where I'm using a non-
> > root user record to establish a connection. I'm essentially opening a
> > connection at the top of the php page and then calling multiple stored
> > procedures to fetch data. When I call the 2nd stored procedure I'm
> > getting the error. Any info would be greatly appreciated.
>
> > Aaron
>
> Sorry, my crystal ball is in the shop. Since you didn't post any code,
> I find it impossible to tell what's wrong.
>
> --
> ==================
> Remove the "x" from my email address
> Jerry Stuckle
> JDS Computer Training Corp.
> jstuck...@attglobal.net
> ==================


Here's the code that fails:

// load the entertainment record
$sql = "call sps_entertainment(?)";
$stmt = $link->prepare($sql);
if ($link->errno) {die($link->errno.":: ".$link->error);}

$stmt->bind_param("i", $EntId);

// execute the statement
$stmt->execute();
if ($link->errno) {die($db->errno.":: ".$link->error);}

//if ($result = $link->store_result()) {
if ($stmt->bind_result($FEntId,
$FCreateDate,
$FCreateUser,
$FModifyDate,
$FModifyUser,
$FEntTypeCode,
$FEntName,
$FURL,
$FPictureQualityCode,
$FPictureResolutionCode,
$FRevenueSourceCode,
$FEntDesc,
$FEntFullDesc,
$FSiteStatusCode,
$FZombiiRatingCode,
$FEntIconId))
{
if ($stmt->fetch())
{
$stmt->free_result();
$stmt->close();
?>
do some html here
<select id="ddlEntIcon" name="ddlEntIcon">
<option value="*" >(Not Specified)</option>
<option value="NEW" >(Upload new image...)</option>
<?php
$sql = "call sps_entertainment_icons()";
$stmt = $link->prepare($sql);
if ($link->errno) {die($link->errno.":: ".$link->error);}
// execute the statement HERE'S WHERE I THINK THE ERROR IS OCCURING!
$stmt->execute();
if ($link->errno) {die($db->errno.":: ".$link->error);}
$stmt->bind_result($EntIconId, $EntIconName);
while ($stmt->fetch())
{
if ($FEntIconId != "" && $FEntIconId == $EntIconId) {
echo "<option value=\"" . $EntIconId . "\" selected>" .
$EntIconName . "</option>\n";
}
else {
echo "<option value=\"" . $EntIconId . "\">" . $EntIconName . "</
option>\n";
}
}
$stmt->free_result();
$stmt->close();
?>
</select>

 

Navigation:

[Reply to this message]


Удаленная работа для программистов  •  Как заработать на Google AdSense  •  England, UK  •  статьи на английском  •  PHP MySQL CMS Apache Oscommerce  •  Online Business Knowledge Base  •  DVD MP3 AVI MP4 players codecs conversion help
Home  •  Search  •  Site Map  •  Set as Homepage  •  Add to Favourites

Copyright © 2005-2006 Powered by Custom PHP Programming

Сайт изготовлен в Студии Валентина Петручека
изготовление и поддержка веб-сайтов, разработка программного обеспечения, поисковая оптимизация