How to display something else if an SQL column is empty with php?

Question

I am having some trouble with getting this to work. I have an image gallery that is calling for the image link from the SQL database. Which works beautifully, however, if a user is only using 'pic1' and 'pic2' but 'pic3' and 'pic4' are empty in the database, I want to display a placeholder image instead. Any recommendations to do that?

    <li><a class="ns-img" href="<?php echo $row_Recordset1['pic1']; ?>"></a></li>
    <li><a class="ns-img" href="<?php echo $row_Recordset1['pic2']; ?>"></a></li>
    <li><a class="ns-img" href="<?php echo $row_Recordset1['pic3']; ?>"></a></li>
    <li><a class="ns-img" href="<?php echo $row_Recordset1['pic4']; ?>"></a></li>

@user3783243 Thank you for your reply! Dreamweaver is kicking the line back to me as broken, any suggestions? <li><a class="ns-img" href="<?php if(empty($row_Recordset1['pic4'])) { echo 'donate.png' }{ else echo $row_Recordset1['pic4'];?>"></a></li> — Cyndi
– Cyndi, Commented Apr 17, 2021 at 11:10
@Cyndi There were a few typos there. I've corrected and moved to an answer. — user3783243
– user3783243, Commented Apr 17, 2021 at 11:20
On second thought is $row_Recordset1['pic1'] simply a path to an image? — Steven
– Steven, Commented Apr 17, 2021 at 11:40
This question desperately needs more details about the actual problem. This is not simply a matter of displaying placeholder image. The questioner phrases the question completely differently in the comments to each answer. — Marcel
– Marcel, Commented Apr 17, 2021 at 12:48

user3783243 · Accepted Answer · 2021-04-17 14:01:34Z

2

I would do the check in SQL:

case when pic4 = '' then 'someimage.jpg' else pic4 end as pic4

or

case when pic4 is null then 'someimage.jpg' else pic4 end as pic4

if the data is null when not set.

This can be done in PHP as well:

if(empty($row_Recordset1['pic4'])) { 
    echo 'someimage.jpg';
} else {
    echo $row_Recordset1['pic4'];
    
}

or even

echo (empty($row_Recordset1['pic4']) ?  'someimage.jpg' : $row_Recordset1['pic4']);

for shorthand version.

Since your string has spaces you need to use trim as well:

echo (empty(trim($row_Recordset1['pic4'])) ?  'someimage.jpg' : $row_Recordset1['pic4']);

can be done in SQL as well:

case when trim(pic4) = '' then 'someimage.jpg' else pic4 end as pic4

edited Apr 17, 2021 at 14:01

answered Apr 17, 2021 at 11:19

user3783243

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10 Comments

user3783243 Over a year ago

var_dump($row_Recordset1['pic4']) gives what?

user3783243 Over a year ago

That must give back a variable type, length, and value.

user3783243 Over a year ago

What is the var_dump result. I need to know that to know why it isnt working.

user3783243 Over a year ago

NULL evalutes to true for empty() so you should have someimage.jpg returned in that scenario. Is it just the attribute just empty or what is there? Demo of how it should be working 3v4l.org/0CEZZ

user3783243 Over a year ago

So do you want no output if there isn't an image, or a placeholder? This should give placeholder.

|

Marcel · Accepted Answer · 2021-04-17 11:27:09Z

2

The PHP ternary or conditional operator could help you

<li>
    <a href="<?= $img ?: './images/default.jpg' ?>"></a>
</li>

This operator says, that if $img is empty or null the default image will be used.

edited Apr 17, 2021 at 11:27

answered Apr 17, 2021 at 11:01

Marcel

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5 Comments

Cyndi Over a year ago

Unfortunately this does not work. There is no original place holder image. I would need to either put a link to an image into the SQL database, which is fine, and name it something like $row_Recordset1['temp'], as an example, but I am not sure how to tell it to display $row_Recordset1['temp'] if $row_Recordset1['pic4'] is empty

Marcel Over a year ago

Quote from the original post: > if a user is only using 'pic1' and 'pic2' but 'pic3' and 'pic4' are empty in the database, I want to display a placeholder image instead.

Cyndi Over a year ago

The LINK to the images, are contained in columns that are named 'pic#'. If the 'pic#' column is empty, meaning there is NO LINK in the table, then I want to replace that with a specific place holder. Your recommendation ignores the use of SQL entirely, which is the critical component to the question.

Marcel Over a year ago

You should ask more detailed questions.

Cyndi Over a year ago

You don't seem to understand the question.

LIMAN MIA · Accepted Answer · 2021-04-17 11:40:15Z

0

This can be done in PHP as well:

<?php
   if($row_Recordset1['pic4'] == NULL){
       <li><a href="<?= $img ?: './images/default.jpg' ?>"></a></li>
   } else {
      <li><a href="<?= $img ?: './images/anyimages.jpg' ?>"></a></li>
   }
?>

You can also use while or forloop into same code.

answered Apr 17, 2021 at 11:40

LIMAN MIA

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