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Posted by Jerry Stuckle on 06/13/25 11:40
coder wrote:
> I am new to programming in PHP however, this should be a pretty
> straight forward answer. I have three queries that I am pulling for a
> content form page.
>
> 1) The Author List
> 2) The Content Page (pulls the PK, Title and the Description of the
> content.
> 3) The Category List
>
> When I attempt to pull the data the only result set I get back is from
> the Item #1 query above. Now, I made sure that there is data in the
> tables, but I still get no data back for the other two queries(????)
>
> Is there something that I missing when connecting to PHP/MYSQL? Do I
> need to take some additional steps??
>
> Any help would be greatly appreciated. I've been struggling with this
> problem for several hours now scratching my head why this is.
>
> :-)
>
> // here is the PHP Code I am using
>
> <?php
> $db = mysql_connect("$DB_SERVER", "$DB_SERVER_USERNAME",
> "$DB_SERVER_PASSWORD");
> mysql_select_db("$DB_DATABASE",$db);
> //result set 1 - get members
> $result_M=mysql_query("SELECT * FROM tblMembers",$db);
> $myrow_M=mysql_fetch_assoc($result_M);
> //result set 2 - get content ID
> $result_C=mysql_query("SELECT * FROM tblContent WHERE
> contentID=1",$db);
> $myrow_C=mysql_query($result_C);
> //result set 3 - get categories
> $result_Cat=mysql_query("SELECT * FROM tblCat",$db);
> $myrow_Cat=mysql_query($result_Cat);
>
>
> //result set 1 - output
> $memID =$myrow_M['memberID'];
> $fname =$myrow_M['fname'];
> $lname =$myrow_M['lname'];
> //result set 2 - output
> $contentID =$myrow_C['contentID'];
> $Title =$myrow_C['Title'];
> $e1m1 =$myrow_C['Descr'];
> //result set 3 - output
> $CatID = $myrow_Cat['CatID'];
> $CatName = $myrow_Cat['CatName'];
>
>
> echo ($memID);
> echo ($fname);
> echo ($lname);
> echo ($contentID);
> echo ($Title);
> echo ($e1m1);
> echo ($CatID);
> echo ($CatName);
>
> return false;
>
> ?>
>
Check the results from your queries. If FALSE is returned, you have an
error. Call mysql_errno() and mysql_error() to find out what it is, i.e.
$result_C=mysql_query("SELECT * FROM tblContent WHERE contentID=1",$db);
if ($result_C) (
$myrow_C=mysql_query($result_C);
}
else {
echo {"MySQL Error detected: " . mysql_error() . "<br>\n");
}
ALWAYS check the results of ANY call to MySQL!
--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================
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