[PHP] QUery success, but blank results/variables

2002-09-09 Thread Patrick Hartnett 

Hello everyone..tryin to run this qry against a mysql db, but after it runs, 
it doesn't assign anything to the variables as it should. If i return all 
rows, and spit out each record in the result in an array, i have the same 
problem, but have 24 'blank' records instead of 1. Any ideas? Thanks for any 
input. I tried doing a print mysql_error(); after the query and the result, 
but it doesn't return anything. Column names, db name, and WHERE clause are 
all spelled correctly, and the $currenttaskid is populated (as 1)...

$detailqry = "SELECT id, parentitemid, itemtypeid, itemstatusid, 
itemlevelid, shortdescription,
createdby_memberid, assignedto_memberid, completedby_memberid, createddate, 
assigneddate,
estcompletiondate, completeddate, projectid, lastuserid, lastdate FROM item 
WHERE id=$currenttaskid";

$result = mysql_query($detailqry) or die("Failed finding task details");

   $taskid = $result["id"];
   $taskparentitemid = $result["parentitemid"];
   $taskitemtypeid = $result["itemtypeid"];
   $taskitemstatusid = $result["itemstatusid"];
   $taskitemlevelid = $result["itemlevelid"];
   $taskshortdescription = $result["shortdescription"];
   $createdbyid = $result["createdby_memberid"];
   $assignedtoid = $result["assignedto_memberid"];
   $completedbyid = $result["completedby_memberid"];
   $taskcreateddate = $result["createddate"];
   $taskassigneddate = $result["assigneddate"];
   $taskestcompletiondate = $result["estcompletiondate"];
   $taskcompleteddate = $result["completeddate"];
   $taskprojectid = $result["projectid"];
   $tasklastuserid = $result["lastuserid"];
   $tasklastdate = $result["lastdate"];

mysql_free_result($result);


-IrishBiotch
[EMAIL PROTECTED]


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Re: [PHP] QUery success, but blank results/variables

2002-09-09 Thread Tom Rogers 

Hi,

Tuesday, September 10, 2002, 1:41:23 PM, you wrote:
PH> Hello everyone..tryin to run this qry against a mysql db, but after it runs, 
PH> it doesn't assign anything to the variables as it should. If i return all 
PH> rows, and spit out each record in the result in an array, i have the same 
PH> problem, but have 24 'blank' records instead of 1. Any ideas? Thanks for any 
PH> input. I tried doing a print mysql_error(); after the query and the result, 
PH> but it doesn't return anything. Column names, db name, and WHERE clause are 
PH> all spelled correctly, and the $currenttaskid is populated (as 1)...

PH> $detailqry = "SELECT id, parentitemid, itemtypeid, itemstatusid, 
PH> itemlevelid, shortdescription,
PH> createdby_memberid, assignedto_memberid, completedby_memberid, createddate, 
PH> assigneddate,
PH> estcompletiondate, completeddate, projectid, lastuserid, lastdate FROM item 
PH> WHERE id=$currenttaskid";

PH> $result = mysql_query($detailqry) or die("Failed finding task details");

PH>$taskid = $result["id"];
PH>$taskparentitemid = $result["parentitemid"];
PH>$taskitemtypeid = $result["itemtypeid"];
PH>$taskitemstatusid = $result["itemstatusid"];
PH>$taskitemlevelid = $result["itemlevelid"];
PH>$taskshortdescription = $result["shortdescription"];
PH>$createdbyid = $result["createdby_memberid"];
PH>$assignedtoid = $result["assignedto_memberid"];
PH>$completedbyid = $result["completedby_memberid"];
PH>$taskcreateddate = $result["createddate"];
PH>$taskassigneddate = $result["assigneddate"];
PH>$taskestcompletiondate = $result["estcompletiondate"];
PH>$taskcompleteddate = $result["completeddate"];
PH>$taskprojectid = $result["projectid"];
PH>$tasklastuserid = $result["lastuserid"];
PH>$tasklastdate = $result["lastdate"];

PH> mysql_free_result($result);


PH> -IrishBiotch
PH> [EMAIL PROTECTED]


PH> _
PH> Join the world’s largest e-mail service with MSN Hotmail. 
PH> http://www.hotmail.com

You need to do this:

if($result = mysql_query($detailqry)){
   $row = mysq_fetch_array($result);
}
else{
 echo "Error: Failed finding task details ".mysql_error();
}

All your fields will be in the array $row. $result is just an
identifier not the actual result set.

-- 
regards,
Tom


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RE: [PHP] QUery success, but blank results/variables

2002-09-09 Thread David Freeman 


If you've typed your code in accurately then...

 > $detailqry = "SELECT id, parentitemid, itemtypeid, itemstatusid, 
[etc]

 > $result = mysql_query($detailqry) or die("Failed finding 
 > task details");

somewhere in about here you want to have something like:

$sqldata = mysql_fetch_array($result);

 >$taskid = $result["id"];
[etc]


... and if you are doing a query that will return more than one row then
you should also look at some sort of looping mechanism to step through
your result set.

CYA, Dave




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