Re: Usiing FOREACH to loop through Array
| From: | Steve Edberg | Date: | Sat, 29 Jun 2002 23:53:59 +0000 |
| Subject: | Re: Usiing FOREACH to loop through Array | ||
| References: | 1 | Groups: | php.db php.general |
| Request: | Send a blank email to php-general+get-104538@lists.php.net to get a copy of this message | ||
At 3:27 PM -0700 6/29/02, Brad Melendy wrote:
Hi All, I've stumped myself here. In a nutshell, I have a function that returns my array based on a SQL query and here's the code: -------------begin code------------------- function getCourses($UID) { global $link; $result = mysql_query( "SELECT C.CourseName FROM tblcourses C, tblusers U, tblEnrollment E WHERE C.ID = E.CourseID AND E.UserID = U.ID AND U.ID = $UID", $link ); if ( ! $result ) die ( "getRow fatal error: ".mysql_error() ); return mysql_fetch_array( $result ); } ------------end code ---------------- I call this from a PHP page with the following code: ------------begin code-------------- $myCourses = getCourses($session[id]); foreach ($myCourses as $value) { print "<br>$value"; } ------------end code--------------- Now, when I test the SQL from my function directly on the database, it returns just want I want it to return but it isn't working that way on my PHP page. For results where there is a single entry, I am getting the same entry TWICE and for records with more than a single entry I am getting ONLY the FIRST entry TWICE. Now I know my SQL code is correct (I am testing it against a MySQL database using MySQL-Front) so I suspect I'm doing something stupid in my foreach loop.I think your problem lies in a misunderstanding of the mysql_fetch_array() function. It doesn't return the entire result set in an array - just one record at a time. You can fix this in one of two ways: (1) Loop though the entire result set in your function: function getCourses($UID)
{
global $link;
$ResultSet = array();
$result = mysql_query( "SELECT C.CourseName FROM tblcourses C, tblusers U,
tblEnrollment E WHERE C.ID = E.CourseID AND E.UserID = U.ID AND U.ID =
$UID", $link );
if ( ! $result )
die ( "getRow fatal error: ".mysql_error() );
while ($Row = mysql_fetch_array( $result ))
{
$ResultSet[] = $Row['C.CourseName'];
}
return $ResultSet;
}
...
$myCourses = getCourses($session[id]);
foreach ($myCourses as $value)
{
print "<br>$value";
}
or (2) set a flag in getCourses() so that the query is only executed once, otherwise returning a result line - something like:
function getCourses($UID)
global $link;
static $result = false;
if (!$result)
{
$result = mysql_query( "SELECT C.CourseName FROM tblcourses C, tblusers U,
tblEnrollment E WHERE C.ID = E.CourseID AND E.UserID = U.ID AND U.ID =
$UID", $link );
if ( ! $result )
die ( "getRow fatal error: ".mysql_error() );
}
{
return mysql_fetch_array( $result );
}
...
while ($Row = getCourses($session[id]) as $value)
{
print "<br>", $Row['C.CourseName'];
}
(standard caveats about off-top-of-head, untested code apply)
The reason you are getting the first record TWICE is becaouse of the default behaviour of the mysql_fetch_array() function. It returns both an associative array - ie, elements of the form <field-name> => <value> - and a numerically indexed array (0, 1, 2, etc.). You can alter this behaviour by the second parameter of the function: see
http://www.php.net/manual/en/function.mysql-fetch-array.php
-steve
I'm hoping someone will spot my dumb mistake. Thanks very much for any help at all on this. ....Brad -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php-- +------------------------------------------------------------------------+ | Steve Edberg sbedberg@ucdavis.edu | | University of California, Davis (530)754-9127 | | Programming/Database/SysAdmin http://pgfsun.ucdavis.edu/ | +------------------------------------------------------------------------+ | The end to politics as usual: | | The Monster Raving Loony Party (http://www.omrlp.com/) | +------------------------------------------------------------------------+