Do not display attributes if the field is empty

Question

On my page where there is an image, I added the alt and title attributes, and these fields are sometimes empty, sometimes they are not.

Now I'm taking it out like this:

<?php foreach ($images as $image) { ?> <img src="<?= $image->getImageUrl() ?>" alt="<?= $image->image_alt ?>" title="<?= $image->image_title ?>"> <?php } ?>

And I need to add a check here, when these fields are empty, then do not display the alt and title attributes on the page.

I try to do something like this, but they still show up:

<?php foreach ($images as $image) { ?> <img src="<?= $image->getImageUrl() ?>" <?php if (!empty($image->image_alt) { ?> alt="<?= $image->image_alt ?>" <?php } ?> <?php if (!empty($image->image_title) { ?> title="<?= $image->image_title ?>" <?php } ?>> <?php } ?>

Also tried like this:

<?php foreach ($images as $image) { ?> <img src="<?= $image->getImageUrl() ?>" <?php if (!empty($image->image_alt) && !empty($image->image_title)) { ?> alt="<?= $image->image_alt ?>" title="<?= $image->image_title ?>"> <?php } ?> <?php } ?>

skytorner · Accepted Answer · 2022-05-27 12:59:12Z

1

you can do it that way

<?php foreach ($images as $image) { if(!empty($image->getImageUrl() && !empty($image->image_alt) && !empty($image->image_title)) { ?> .....

i dont really know what to check but to check if a variable is empty, then use the empty function

If the value can be null

if(null !== $image->getImageUrl() && ....)

answered May 27, 2022 at 12:59

skytorner

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4 Comments

broken heart Over a year ago

When there are several pictures, it does not work, attributes are displayed with empty fields anyway

skytorner Over a year ago

that means that the content of each image's properties is not empty. Maybe blank space are not considered as empty. have you dump the content of each image's properties to see what they look like ? have you try to test with null?

broken heart Over a year ago

yes, I get NULL

yodamousta Over a year ago

have you tried isset() instead of empty() ?

Michel · Accepted Answer · 2022-05-27 14:37:52Z

Hopping in and out PHP is always quite messy. First sort the data out, only then output the html.

$myImages=[]; foreach ($images as $key=>$image) { $src = 'src="'.$image->getImageUrl().'"'; $alt = (empty($image->image_alt) ? '' : 'alt="'.$image->image_alt.'"' ); $title = (empty($image->image_title) ? '' : 'title="'.$image->image_alt.'"' ); $myImages[]="<img $src $alt $title>"; } echo implode('',$myImages);

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