Show Label instead of Value

Hi,

The value of $job_title is going to be a number from 1 to 12 given by the foreach loop. It doesn't contain job title label or anything else. If you want to get the full list of job labels and display them you will need to get the records from its table.

Let me know if you have questions!

Hi Leo

I understand what you are saying, not sure how to solve it though. If I were to move the line to within the foreach statement it would work. But outside of the foreach I would get a undefined index error. How do I define it without the foreach? Thx

Hi,

I assume what you want here is to show sections of job titles and show $operations_teamRecords under their category.

Do you have a section for job titles already? If you have it then grab job title records and display their labels first there. If not then you need to create an array to store the labels:

// With job title section
$jobTitleRecords = mysql_select('job_titles');


// Without job title section
$jobTitleLabels = array(
  1 => 'Job1',
  2 => 'Job2'
);

And the label will be:

// With job title section
foreach($jobTitleRecords as $jobTitleRecord){
  echo $jobTitleRecord['label_field'];
}

// Without job title section
foreach($jobTitleLabels as $jobTitleLabel){
  echo $jobTitleLabel;
}

This is just an example of how to get the label value. The exact code to put there will be different. Hope this helps.

Let me know if you have questions!

Jeff, 

Also, how are you loading the records?  The getRecords() function usually automatically loads labels for you.  You can see all the variables you have available for display with this code: 

<?php showme($record); ?>

Hi Dave, Hi Lee

Thanks for your assistance so far.

Lee: I'm with you on how to set up the arrays so that 1=> Job1, etc. But I still can't get the value to output.

Dave The getRecords() function isn't working here because I need to show the value outside of the foreach.

Here's what I am trying to achive:

I have a section called operations_teamRecords.
Each record has a  name, job title, bio and photo.
On my output page I would like to display all of the records, grouped under a job title heading. For example, lets say we have the the following names in the Operations_teamRecords and there are two different job titles:

Dave (title), Programmer (job_title)
Jeff (title), Designer (job_title)
Lee (title), Programmer (job_title)

My job_titles are set up as a pull down list, with value and label. i.e.:

1|Designer
2|Progammer

In the front end I would like it to display this as:

Designer
Jeff

Programmer
Dave
Lee

As things stand I have each record grouped under the correct heading. But instead of displaying the label it is displaying the value. i.e.:

1
Jeff

2
Dave
Lee

Here's my code again.

$jobTitleLabels = array(
1 => 'Designer',
2 => 'Programmer');

<?php foreach (range(1,12) as $job_title): ?>
<div class="row equal">
<div class="col-xs-12">

<h3><?php echo htmlencode($job_title) ?></h3> THIS IS THE PART THAT SHOWS THE VALUE. I WOULD LIKE TO DISPLAY THE LABEL

<?php foreach ($operations_teamRecords as $record): ?>
<?php if ($record['job_title'] != $job_title) { continue; } ?>
<div class="col-xs-12 col-sm-6 col-md-4 col-lg-3">
<div class="thumbnail">
<?php foreach ($record['photos'] as $index => $upload): ?>
       <?php if ($index >= 1) { continue; } ?>
<img src="<?php echo htmlencode($upload['urlPath']) ?>" width="100%" alt="<?php echo htmlencode($upload['info3']) ?>" />
         <?php endforeach ?>

<div class="panel panel-default">
<div class="panel-body">
<?php // output team member details
$details = ""; 
  if ($record['title'])  { $details .= "<strong>".htmlencode($record['title']).  "</strong><br/>\n"; }
if ($record['job_title'])  { $details .= "".htmlencode($record['job_title:label']).  "<br/>\n"; }
  if ($record['content_wysiwyg'])  { $details .= "".$record['content_wysiwyg'].  "\n"; }
if ($details) {
print "<div class='caption'>$details</div>\n";
}
   ?>
</div>


</div>
</div>
</div>
<?php endforeach ?>

</div>
</div>
<?php endforeach ?>

Thanks

Hi Jeff,

If you have already set up the list field in the section, you can use this function getListOptions($tablename, $fieldname) to get list options array, and the labels will be the values of the array.

Here are the steps:

// Get list options array
$jobTitleArray = getListOptions('operations_teamRecords', 'job_titles');

// The foreach loop
foreach(range(1,12) as $job_title){
  $label = $jobTitleArray[$job_title]
}


Let me know if you have any questions!