Viewing an image from a related table

6 posts by 2 authors in: Forums > CMS Builder
Last Post: February 17, 2015   (RSS)

By northernpenguin - February 15, 2015 - edited: February 15, 2015

I have 2 multi-value tables.  One lists the personnel, while the other lists the positions (2nd field is an image).  How do I display the personnel information for each person plus the position image?

Right now, I'm stumped!

--
northernpenguin
Northern Penguin Technologies

"Any sufficiently advanced technology
is indistinguishable from magic."
........Arthur C. Clarke

By claire - February 16, 2015

Hey there - it depends on what the relationship between the two tables is. Can you give me any more information?

--------------------

Claire Ryan
interactivetools.com

Save time by getting our experts to help with your project.
http://www.interactivetools.com/consulting/

By northernpenguin - February 16, 2015

Claire:  Right now, one table just lists a set of Air Force ranks and the image (RANKS).  The main table has the personnel information, including name, position, email, rank.  The rank field in this table is a list linked to the RANKS table.  The only field it will let me display is the text rank field, not the uploaded image.

Does that help?

Ragi

--
northernpenguin
Northern Penguin Technologies

"Any sufficiently advanced technology
is indistinguishable from magic."
........Arthur C. Clarke

By northernpenguin - February 16, 2015 - edited: February 16, 2015

Claire:  I got everything except the "..create a new array of ranks....".  Not sure how to do that.

Ragi

P.S.  Is this what you mean?

<?php
  // load records from 'staff'
  list($staffRecords, $staffMetaData) = getRecords(array(
    'tableName'   => 'staff',
    'loadUploads' => true,
    'allowSearch' => false,
    'debugSql'    => false,      // optional, display SQL query, defaults to no
  ));

  // load records from 'ranks'
  list($ranksRecords, $ranksMetaData) = getRecords(array(
    'tableName'   => 'ranks',
    'loadUploads' => true,
    'allowSearch' => false,
  ));
  
  $rankStructure = getRecords(array(
    'tableName'   => 'ranks',
    'orderBy'   => 'num',
));
?>

--
northernpenguin
Northern Penguin Technologies

"Any sufficiently advanced technology
is indistinguishable from magic."
........Arthur C. Clarke

By claire - February 17, 2015

No no, there's a particular way of setting up an array so that each record is easily accessible. So you start with this:

// load records from 'ranks'
  list($ranksRecords, $ranksMetaData) = getRecords(array(
    'tableName'   => 'ranks',
    'loadUploads' => true,
    'allowSearch' => false,
  ));

The $ranksRecords come out as an array indexed in the order that they're picked from the database. So it looks like this:

array(
  0 => record 1,
  1 => record 2,
  2 => record 3...

etc. What you want to do is re-sort the array so that it's indexed to the num of the ranks records, and you do it like this:

$sortedRanks = array();
foreach ($ranksRecords as $rank) {
  $sortedRanks[$rank['num']] = $rank;
}

Now when you're looping through the $staffRecords, you can access the right rank record to display the picture by selecting the right index in the $sortedRanks, like this:

$rank = $sortedRanks[$record['rankNum']];

Change 'rankNum' to the right variable name, of course. This is a convenient piece of logic to use when you have a set of records that can be applied to multiple others.

--------------------

Claire Ryan
interactivetools.com

Save time by getting our experts to help with your project.
http://www.interactivetools.com/consulting/