Trapping errors

Anon

Hi,

I was wondering if the code below would be a good way to trap errors. I grabbed the code off here and tried to fit it into my own code but it doesn't seem to be working:

----//begin query
$query="SELECT request.id, request.date, request.type, request.status, faculty.f_name, faculty.l_name, action.id, faculty.id FROM request, faculty, action WHERE request.requested_by=faculty.id AND action.request_id=request.id";

$result=mysql_query($query);

//trap errors
if (!$result=mysql_query($query))
{
echo mysql_error()." \n$query \n";
};

//begin while loop to grab all needed data out of the database

while ($query_data=mysql_fetch_array($result)) {

Anon

first... echo $query; just to see if it's ok 🙂
next, why not doing things like this:
if ($result=mysql_query($query)){
while ($query_data=mysql_fetch_array($result)) {
...
} // end while
} else { // or else the $result returned false
echo mysql_error();
} // end if

Try this, maybe some errors in the code, you fix it ! 🙂

and a bracket never ends with ;
//trap errors
if (!$result=mysql_query($query))
{
echo mysql_error()." \n$query \n";
};

ditch the ; in the end, if you decide to use this code...

Hope it helps, Arni

hawleyj

Remove the semicolon from the end of the if statement and then do one of the two following steps...

$result=mysql_query($query);

//trap errors
if (!$result=mysql_query($query))
{
echo mysql_error()." \n$query \n";
};

In the if declaration, remove the '=mysql_query($query)' statement. You have already done and are just repeating the step all over again. Depending on the size of the tables you are addressing, this could become a performance issue.
Get rid of the $results=mysql_query($query); line altogether and allow the if statement to do your work. Again, you are repeating the same call to the database twice.

HTH,

Jim Hawley