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How do I add an additional table to a PHP script

Posted on 2009-04-14
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Last Modified: 2013-12-13
I am modifying a piece of PHP code to include data from a third table.  I have tried to link the third table (mdl_user_info_data); however it returns no rows, when there should be the same no of rows as when it was without the link.

I have written the sql in MySql and it returns the correct rows.

I'm not sure how to proceed from here
Cheers
Bernard
In PHP:
 
$select .= 'qa.uniqueid AS attemptuniqueid, qa.id AS attempt, ' .
                    'u.id AS userid, u.idnumber, u.department, u.firstname, u.lastname, u.picture, u.imagealt,uid.data,  ' .
                    'qa.sumgrades, qa.timefinish, qa.timestart, qa.timefinish - qa.timestart AS duration ';
 
            // This part is the same for all cases - join users and quiz_attempts tables
            $from = 'FROM '.$CFG->prefix.'user u '; 
		$from .= 'LEFT JOIN '.$CFG->prefix.'quiz_attempts qa ON qa.userid = u.id AND qa.quiz = '.$quiz->id;
            $from .= 'LEFT JOIN '.$CFG->prefix.'user_info_data uid ON uid.userid = u.id AND uid.fieldid =2';
 
 
In MySql:
 
select  qa.uniqueid AS attemptuniqueid , qa.id AS attempt, u.id AS userid, u.idnumber, u.department
, u.firstname, u.lastname, u.picture, u.imagealt, uid.data as 'Site Location', qa.sumgrades, qa.timefinish
, qa.timestart, qa.timefinish - qa.timestart AS duration 
 FROM mdl_user u  
LEFT JOIN mdl_quiz_attempts qa ON qa.userid = u.id 
LEFT JOIN mdl_user_info_data uid ON uid.userid = u.id  and uid.fieldid=2
where qa.quiz =60

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Question by:BernardGBailey
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10 Comments
 
LVL 51

Expert Comment

by:Steve Bink
ID: 24144700
Can you post the final query from PHP?  It is hard to tell if the two queries are identical from the code you provided.
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Author Comment

by:BernardGBailey
ID: 24144754
Hi routinet,

The query from PHP is part of 15 pages of code.  I've reposted below all elements around the php query

Cheers
Bernard
 // Construct the SQL
            $select = 'SELECT '.sql_concat('u.id', '\'#\'', $db->IfNull('qa.attempt', '0')).' AS uniqueid, ';
            if ($qmsubselect) {
                $select .=
                    "(CASE " .
                    "   WHEN $qmsubselect THEN 1" .
                    "   ELSE 0 " .
                    "END) AS gradedattempt, ";
            }
 
            $select .= 'qa.uniqueid AS attemptuniqueid, qa.id AS attempt, ' .
                    'u.id AS userid, u.idnumber, u.department, u.firstname, u.lastname, u.picture, u.imagealt,uid.data, ' . 
                    'qa.sumgrades, qa.timefinish, qa.timestart, qa.timefinish - qa.timestart AS duration ';
 
            // This part is the same for all cases - join users and quiz_attempts tables
            $from = 'FROM '.$CFG->prefix.'user u '; 
		$from .= 'LEFT JOIN '.$CFG->prefix.'quiz_attempts qa ON qa.userid = u.id AND qa.quiz = '.$quiz->id;
            $from .= 'LEFT JOIN '.$CFG->prefix.'user_info_data uid ON uid.userid = u.id AND uid.fieldid =2';
 
		if ($qmsubselect && $qmfilter){
                $from .= ' AND '.$qmsubselect;
            }
            switch ($attemptsmode){
                case QUIZ_REPORT_ATTEMPTS_ALL:
                    // Show all attempts, including students who are no longer in the course
                    $where = ' WHERE qa.id IS NOT NULL AND qa.preview = 0';
                    break;
                case QUIZ_REPORT_ATTEMPTS_STUDENTS_WITH:
                    // Show only students with attempts
                    $where = ' WHERE u.id IN (' .$allowedlist. ') AND qa.preview = 0 AND qa.id IS NOT NULL';
                    break;
                case QUIZ_REPORT_ATTEMPTS_STUDENTS_WITH_NO:
                    // Show only students without attempts
                    $where = ' WHERE u.id IN (' .$allowedlist. ') AND qa.id IS NULL';
                    break;
                case QUIZ_REPORT_ATTEMPTS_ALL_STUDENTS:
                    // Show all students with or without attempts
                    $where = ' WHERE u.id IN (' .$allowedlist. ') AND (qa.preview = 0 OR qa.preview IS NULL)';
                    break;
            }
 
            $countsql = 'SELECT COUNT(DISTINCT('.sql_concat('u.id', '\'#\'', 'COALESCE(qa.attempt, 0)').')) '.$from.$where;

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Expert Comment

by:Steve Bink
ID: 24144925
There lies the problem.  We need to compare the query that does return data to the query that doesn't.

In your code, you'll have a line to execute the query.  It might look something like this:

$result = mysql_query($query);

Change that to this:

echo $query;  // or error_log($query)
$result = mysql_query($query);

Post that query here, as well as the query that you know returns good results.
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Author Comment

by:BernardGBailey
ID: 24146466
Hi routinet,

I'm sorry if I cannot see what you asked me to provide.

I'm trying to add one field from an additional table.   When I use this snippet

 // Construct the SQL
            $select = 'SELECT '.sql_concat('u.id', '\'#\'', $db->IfNull('qa.attempt', '0')).' AS uniqueid, ';
            if ($qmsubselect) {
                $select .=
                    "(CASE " .
                    "   WHEN $qmsubselect THEN 1" .
                    "   ELSE 0 " .
                    "END) AS gradedattempt, ";
            }

            $select .= 'qa.uniqueid AS attemptuniqueid, qa.id AS attempt, ' .
                    'u.id AS userid, u.idnumber, u.department, u.firstname, u.lastname, u.picture, u.imagealt, ' . //uid.data, ' .
                    'qa.sumgrades, qa.timefinish, qa.timestart, qa.timefinish - qa.timestart AS duration ';

            // This part is the same for all cases - join users and quiz_attempts tables
            $from = 'FROM '.$CFG->prefix.'user u ';
            $from .= 'LEFT JOIN '.$CFG->prefix.'quiz_attempts qa ON qa.userid = u.id AND qa.quiz = '.$quiz->id;
//            $from .= 'LEFT JOIN '.$CFG->prefix.'user_info_data uid ON uid.userid = u.id AND uid.fieldid =2';


The correct result comes up.

When I use this  (I removed the // from lines 12 & 17) I get no records returned

Does this make more sense?

Cheers
Bernard    


    // Construct the SQL
         $select = 'SELECT '.sql_concat('u.id', '\'#\'', $db->IfNull('qa.attempt', '0')).' AS uniqueid, ';
            if ($qmsubselect) {
                $select .=
                    "(CASE " .
                    "   WHEN $qmsubselect THEN 1" .
                    "   ELSE 0 " .
                    "END) AS gradedattempt, ";
            }
 
            $select .= 'qa.uniqueid AS attemptuniqueid, qa.id AS attempt, ' .
                    'u.id AS userid, u.idnumber, u.department, u.firstname, u.lastname, u.picture, u.imagealt, uid.data, ' . 
                    'qa.sumgrades, qa.timefinish, qa.timestart, qa.timefinish - qa.timestart AS duration ';
 
            // This part is the same for all cases - join users and quiz_attempts tables
            $from = 'FROM '.$CFG->prefix.'user u '; 
		$from .= 'LEFT JOIN '.$CFG->prefix.'quiz_attempts qa ON qa.userid = u.id AND qa.quiz = '.$quiz->id;
            $from .= 'LEFT JOIN '.$CFG->prefix.'user_info_data uid ON uid.userid = u.id AND uid.fieldid =2';

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Author Comment

by:BernardGBailey
ID: 24146477
The second line is line 18, not 17 as mentioned.
0
 

Author Comment

by:BernardGBailey
ID: 24153307
Some more detail here,

I've looked right through the php script and not found any relationship from the newly selected field to the table it is to show in.

This is probably why I get "Nothing to display".

Obviously the link from the select statement to the display is somewhere else or obsured so I cannot recognise it somehow in the script.  

Comments?

Cheers
Bernard
0
 
LVL 51

Expert Comment

by:Steve Bink
ID: 24154164
Do you know if the query is returning an error?  I think it might be.  Change line 18 to add a space before the LEFT JOIN:

 $from .= ' LEFT JOIN '.$CFG->prefix.'user_info_data uid ON uid.userid = u.id AND uid.fieldid =2';

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Author Comment

by:BernardGBailey
ID: 24154213
routinet,

That is fantastic.  It works fine, I've been able to pull the data through to the export table as required.

But why the space?

Cheers
Bernard

0
 
LVL 51

Accepted Solution

by:
Steve Bink earned 2000 total points
ID: 24154378
Without the space, that section of the query looks like this:

ts qa ON qa.userid = u.id AND qa.quiz = 60LEFT JOIN user_info_data uid ON uid.userid = u.id AND uid.fieldid =2

That would result in a syntax error, since qa.quiz looks to be a numeric field testing against unquoted string data (60LEFT).  Even assuming the syntax went through, the parser would consider this an INNER JOIN (since LEFT would no longer be considered a join modifier), which would eliminate any rows not matching the final table.
0
 

Author Closing Comment

by:BernardGBailey
ID: 31570273
Routinet,

 I read your profile and yes you are worthy of the title "GENIUS"

Thank you for putting your experience out there for us.

Cheers
Bernard  
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