10 posts by 2 authors in: Forums > CMS Builder
Last Post: April 17, 2018   (RSS)

I have got a multi record displaying text and images, on the list page I have got the records displaying with text and an image which is fine but I want to show the first image from the first record separately, so I have the list on the left with small images and text then I want the first image form the first record on the right, at the moment it is showing the image from the second record, not the first. What am I doing wrong?

By leo - April 10, 2018

Hi,

When you get an array of records with their uploads, your first record in the array is $records[0] and the second will be $records[1] etc.. You can get specific record this way and put wherever you like on the page.

Hope that helps and let me know if you have any questions!

Thanks,

Leo - PHP Programmer (in training)
interactivetools.com

Thank you but where do I put that code? This is the code I have to pull in the image:

<!-- STEP 2a: Display Uploads for field 'summary_image' (Paste this anywhere inside STEP2 to display uploads) -->
        <!-- Upload Fields: extension, thumbFilePath, isImage, hasThumbnail, urlPath, width, height, thumbUrlPath, thumbWidth, thumbHeight, info1, info2, info3, info4, info5 -->
        <?php foreach ($record['summary_image'] as $index => $upload): ?>
          <?php if ($index >= 1) { continue; } // limit uploads shown ?>

          <img src="<?php echo htmlencode($upload['urlPath']) ?>" width="<?php echo $upload['width'] ?>" height="<?php echo $upload['height'] ?>" alt="<?php echo htmlencode($upload['info1']) ?>" />

        <?php endforeach ?>
      <!-- STEP2a: /Display Uploads -->

By leo - April 11, 2018 - edited: April 11, 2018

If you just want to get the first upload of the record, $record['summary_image'][0] is what you need instead of a foreach loop. If this code snippet is for displaying the rest images then the if statement should be if($index < 1).

Do you have a screenshot or example of what you want to accomplish so I can get a better understanding of the question?

Thanks,

Leo - PHP Programmer (in training)
interactivetools.com

By leo - April 12, 2018 - edited: April 12, 2018

According to your file, the first record of $latest_newsRecords is $latest_newsRecords[0]. After finishing the first foreach loop your $record variable will contain the last element of $latest_newsRecords. So just use $latest_newsRecords[0]['single_image'][0] should give you the first upload of the first record in $latest_newsRecords.

Leo - PHP Programmer (in training)
interactivetools.com

Sorry I still cannot get this to work, I need only one image to show on the page, so ONLY the image from the first record in a different DIV.

By leo - April 16, 2018

Hi,

Did you try using $latest_newsRecords[0]['single_image'][0]? If so what is the error that you are having?

Leo - PHP Programmer (in training)
interactivetools.com

Yes I did, I don't get an error it just doesn't work. Is there anyway you can try it and send me the code, the trouble is I want the image to be in a different area as you can see from the code I sent.

By leo - April 17, 2018

Sure! You can send a support request to support@interactivetools.com with the ftp information and the file and we will take a closer look at the problem!

Leo - PHP Programmer (in training)
interactivetools.com