what is the correct way to handle empty database returns?

[lawrence]

If I do something like :

$result = mysql_query($qu ery);
$dbArray = dbResultIntoKey Array($result);
and this is the function:
function dbResultIntoKey Array($result) {
while ($row = mysql_fetch_obj ect($result, MYSQL_ASSOC)) {
$dbArray[] = $row;
}
return $dbArray;
}
then if the query came back empty I get an error message: not a valid
mysql resource.

What is the correct way of handling this? If I was dealing with an
array I'd start the function with something like:

if (is_array($arra y)) {

}

and then I wouldn't have to worry about error messages if it turned
the variable wasn't an array. What is the similar test for database
returns?

[Jochen Daum]

Hi lawrence!
On 14 Sep 2003 20:38:09 -0700, lk******@geocit ies.com (lawrence)
wrote:

If I do something like :

$result = mysql_query($qu ery);
$dbArray = dbResultIntoKey Array($result);
and this is the function:
function dbResultIntoKey Array($result) {
while ($row = mysql_fetch_obj ect($result, MYSQL_ASSOC)) {
$dbArray[] = $row;
}
return $dbArray;
}
then if the query came back empty I get an error message: not a valid
mysql resource.

What is the correct way of handling this? If I was dealing with an
array I'd start the function with something like:

if (is_array($arra y)) {

}

and then I wouldn't have to worry about error messages if it turned
the variable wasn't an array. What is the similar test for database
returns?

if (is_resource($r esult))
HTH, Jochen
--
Jochen Daum - CANS Ltd.
PHP DB Edit Toolkit -- PHP scripts for building
database editing interfaces.
http://sourceforge.net/projects/phpdbedittk/

[Andy Hassall]

On 14 Sep 2003 20:38:09 -0700, lk******@geocit ies.com (lawrence) wrote:

If I do something like :

$result = mysql_query($qu ery);
You've ignored possible errors here.
$dbArray = dbResultIntoKey Array($result);
and this is the function:
function dbResultIntoKey Array($result) {
while ($row = mysql_fetch_obj ect($result, MYSQL_ASSOC)) {
$dbArray[] = $row;
}
return $dbArray;
}

then if the query came back empty I get an error message: not a valid
mysql resource.
No, that'd be because an error occurred.

An empty result set is a perfectly valid result set, and will not give this
error.
What is the correct way of handling this? If I was dealing with an
array I'd start the function with something like:

if (is_array($arra y)) {

}

and then I wouldn't have to worry about error messages if it turned
the variable wasn't an array. What is the similar test for database
returns?

$result = mysql_query($qu ery);

if (!$result) { /* error, details available from mysql_error() */ }

mysql_query returns either a resource identifier after excuting the query
(which may or may not have any rows), or FALSE in the case of a fatal error.

--
Andy Hassall (an**@andyh.co. uk) icq(5747695) (http://www.andyh.co.uk)
Space: disk usage analysis tool (http://www.andyhsoftware.co.uk/space)

[lawrence]

Andy Hassall <an**@andyh.co. uk> wrote in message news:<c9******* *************** **********@4ax. com>...

On 14 Sep 2003 20:38:09 -0700, lk******@geocit ies.com (lawrence) wrote:

If I do something like :

$result = mysql_query($qu ery);
You've ignored possible errors here.

$dbArray = dbResultIntoKey Array($result);
and this is the function:
function dbResultIntoKey Array($result) {
while ($row = mysql_fetch_obj ect($result, MYSQL_ASSOC)) {
$dbArray[] = $row;
}

return $dbArray;

}

then if the query came back empty I get an error message: not a valid
mysql resource.

No, that'd be because an error occurred.

An empty result set is a perfectly valid result set, and will not give this
error.

Are you sure? The error only occurs when the result set is empty.
Otherwise everything works fine.

What is the correct way of handling this? If I was dealing with an
array I'd start the function with something like:

if (is_array($arra y)) {

}

and then I wouldn't have to worry about error messages if it turned
the variable wasn't an array. What is the similar test for database
returns?

$result = mysql_query($qu ery);

if (!$result) { /* error, details available from mysql_error() */ }

mysql_query returns either a resource identifier after excuting the query
(which may or may not have any rows), or FALSE in the case of a fatal error.

Thanks. I'll use this test you write out from now on.