Get another image if the main image is not available
-1
I have an API, which returns some data with image links.
The problem is that there is a type of image that is main but not always available.
When that happens, I want another image to appear, but it's also in the API.
API Extract
"images": {
"icon": "https://image.fnbr.co/outfit/5c2aad8560e93a37d9635605/icon.png", // secondary
"png": false,
"gallery": false,
"featured":
"https://image.fnbr.co/outfit/5c2aad8560e93a37d9635605/featured.png" // main
}
I have tried the label "onerror" but since it is not a error (because it says null) it does not replace the image.
PHP Code
<img src="<?php echo $image_data['data']['featured'][$i]['images']
['featured'] ?>";>
php api
asked Jan 2 at 23:48
Nacho AldamaNacho Aldama
1
add a comment |
-1
I have an API, which returns some data with image links.
The problem is that there is a type of image that is main but not always available.
When that happens, I want another image to appear, but it's also in the API.
API Extract
"images": {
"icon": "https://image.fnbr.co/outfit/5c2aad8560e93a37d9635605/icon.png", // secondary
"png": false,
"gallery": false,
"featured":
"https://image.fnbr.co/outfit/5c2aad8560e93a37d9635605/featured.png" // main
}
I have tried the label "onerror" but since it is not a error (because it says null) it does not replace the image.
PHP Code
<img src="<?php echo $image_data['data']['featured'][$i]['images']
['featured'] ?>";>
php api
asked Jan 2 at 23:48
Nacho AldamaNacho Aldama
1
You can solve that with a simple IF statement
– Second2None
Jan 2 at 23:54
add a comment |
-1
-1
-1
I have an API, which returns some data with image links.
The problem is that there is a type of image that is main but not always available.
When that happens, I want another image to appear, but it's also in the API.
API Extract
"images": {
"icon": "https://image.fnbr.co/outfit/5c2aad8560e93a37d9635605/icon.png", // secondary
"png": false,
"gallery": false,
"featured":
"https://image.fnbr.co/outfit/5c2aad8560e93a37d9635605/featured.png" // main
}
I have tried the label "onerror" but since it is not a error (because it says null) it does not replace the image.
PHP Code
<img src="<?php echo $image_data['data']['featured'][$i]['images']
['featured'] ?>";>
php api
asked Jan 2 at 23:48
Nacho AldamaNacho Aldama
1
I have an API, which returns some data with image links.
The problem is that there is a type of image that is main but not always available.
When that happens, I want another image to appear, but it's also in the API.
API Extract
"images": {
"icon": "https://image.fnbr.co/outfit/5c2aad8560e93a37d9635605/icon.png", // secondary
"png": false,
"gallery": false,
"featured":
"https://image.fnbr.co/outfit/5c2aad8560e93a37d9635605/featured.png" // main
}
I have tried the label "onerror" but since it is not a error (because it says null) it does not replace the image.
PHP Code
<img src="<?php echo $image_data['data']['featured'][$i]['images']
['featured'] ?>";>
php api
php api
asked Jan 2 at 23:48
Nacho AldamaNacho Aldama
1
asked Jan 2 at 23:48
Nacho AldamaNacho Aldama
1
asked Jan 2 at 23:48
Nacho AldamaNacho Aldama
1
asked Jan 2 at 23:48
Nacho AldamaNacho Aldama
1
asked Jan 2 at 23:48
Nacho AldamaNacho Aldama
1
1
You can solve that with a simple IF statement
– Second2None
Jan 2 at 23:54
add a comment |
You can solve that with a simple IF statement
– Second2None
Jan 2 at 23:54
You can solve that with a simple IF statement
– Second2None
Jan 2 at 23:54
You can solve that with a simple IF statement
– Second2None
Jan 2 at 23:54
add a comment |
2 Answers
2
active
oldest
votes
1
you need to check if the main image is empty:
$images = ...; // your current image object
$image = ($images['featured'] !== null) ? $images['featured'] : $images['icon']; // featured or icon image
answered Jan 2 at 23:55
Eriks KlotinsEriks Klotins
1,4031518
I've added it, but I'm getting an error 500. I don't know if something can be done, but I can change the class depending on the type of object. for example,<img class="card-img-top <?php echo $image_data['data']['featured'][$i]['type'] ?>"
– Nacho Aldama
Jan 3 at 9:20
add a comment |
0
$img = $image_data['data']['featured'][$i]['images']['featured'];
$alternativ = $image_data['data']['featured'][$i]['images']['icon'];
<img src="<?php echo ($img) ? $img:$alternativ ?>";>
or check null
<img src="<?php echo ($img === NULL) ? $alternativ:$img ?>";>
answered Jan 2 at 23:56
Mohammad bMohammad b
1,59121735
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
you need to check if the main image is empty:
$images = ...; // your current image object
$image = ($images['featured'] !== null) ? $images['featured'] : $images['icon']; // featured or icon image
answered Jan 2 at 23:55
Eriks KlotinsEriks Klotins
1,4031518
I've added it, but I'm getting an error 500. I don't know if something can be done, but I can change the class depending on the type of object. for example,<img class="card-img-top <?php echo $image_data['data']['featured'][$i]['type'] ?>"
– Nacho Aldama
Jan 3 at 9:20
add a comment |
1
you need to check if the main image is empty:
$images = ...; // your current image object
$image = ($images['featured'] !== null) ? $images['featured'] : $images['icon']; // featured or icon image
answered Jan 2 at 23:55
Eriks KlotinsEriks Klotins
1,4031518
I've added it, but I'm getting an error 500. I don't know if something can be done, but I can change the class depending on the type of object. for example,<img class="card-img-top <?php echo $image_data['data']['featured'][$i]['type'] ?>"
– Nacho Aldama
Jan 3 at 9:20
add a comment |
1
1
1
you need to check if the main image is empty:
$images = ...; // your current image object
$image = ($images['featured'] !== null) ? $images['featured'] : $images['icon']; // featured or icon image
answered Jan 2 at 23:55
Eriks KlotinsEriks Klotins
1,4031518
you need to check if the main image is empty:
$images = ...; // your current image object
$image = ($images['featured'] !== null) ? $images['featured'] : $images['icon']; // featured or icon image
answered Jan 2 at 23:55
Eriks KlotinsEriks Klotins
1,4031518
answered Jan 2 at 23:55
Eriks KlotinsEriks Klotins
1,4031518
answered Jan 2 at 23:55
Eriks KlotinsEriks Klotins
1,4031518
answered Jan 2 at 23:55
Eriks KlotinsEriks Klotins
1,4031518
1,4031518
I've added it, but I'm getting an error 500. I don't know if something can be done, but I can change the class depending on the type of object. for example,<img class="card-img-top <?php echo $image_data['data']['featured'][$i]['type'] ?>"
– Nacho Aldama
Jan 3 at 9:20
add a comment |
I've added it, but I'm getting an error 500. I don't know if something can be done, but I can change the class depending on the type of object. for example,<img class="card-img-top <?php echo $image_data['data']['featured'][$i]['type'] ?>"
– Nacho Aldama
Jan 3 at 9:20
I've added it, but I'm getting an error 500. I don't know if something can be done, but I can change the class depending on the type of object. for example,
<img class="card-img-top <?php echo $image_data['data']['featured'][$i]['type'] ?>"– Nacho Aldama
Jan 3 at 9:20
I've added it, but I'm getting an error 500. I don't know if something can be done, but I can change the class depending on the type of object. for example,
<img class="card-img-top <?php echo $image_data['data']['featured'][$i]['type'] ?>"– Nacho Aldama
Jan 3 at 9:20
add a comment |
0
$img = $image_data['data']['featured'][$i]['images']['featured'];
$alternativ = $image_data['data']['featured'][$i]['images']['icon'];
<img src="<?php echo ($img) ? $img:$alternativ ?>";>
or check null
<img src="<?php echo ($img === NULL) ? $alternativ:$img ?>";>
answered Jan 2 at 23:56
Mohammad bMohammad b
1,59121735
add a comment |
0
$img = $image_data['data']['featured'][$i]['images']['featured'];
$alternativ = $image_data['data']['featured'][$i]['images']['icon'];
<img src="<?php echo ($img) ? $img:$alternativ ?>";>
or check null
<img src="<?php echo ($img === NULL) ? $alternativ:$img ?>";>
answered Jan 2 at 23:56
Mohammad bMohammad b
1,59121735
add a comment |
0
0
0
$img = $image_data['data']['featured'][$i]['images']['featured'];
$alternativ = $image_data['data']['featured'][$i]['images']['icon'];
<img src="<?php echo ($img) ? $img:$alternativ ?>";>
or check null
<img src="<?php echo ($img === NULL) ? $alternativ:$img ?>";>
answered Jan 2 at 23:56
Mohammad bMohammad b
1,59121735
$img = $image_data['data']['featured'][$i]['images']['featured'];
$alternativ = $image_data['data']['featured'][$i]['images']['icon'];
<img src="<?php echo ($img) ? $img:$alternativ ?>";>
or check null
<img src="<?php echo ($img === NULL) ? $alternativ:$img ?>";>
answered Jan 2 at 23:56
Mohammad bMohammad b
1,59121735
answered Jan 2 at 23:56
Mohammad bMohammad b
1,59121735
answered Jan 2 at 23:56
Mohammad bMohammad b
1,59121735
answered Jan 2 at 23:56
Mohammad bMohammad b
1,59121735
1,59121735
add a comment |
add a comment |
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You can solve that with a simple IF statement
– Second2None
Jan 2 at 23:54