Extracting "image_url" value and display as image src

Question

Is it possible to extract the value of "image_url" from this link

http://theapache64.xyz:8080/gpix/v1/gpix?keyword=honda%20civic%202009&limit=1

and display as <img src=""> using php?

Yes, Download the result of the Link as a string, looks like its giving a JSON . so then access the JSON array "images" and get the "image_url" property. — cwishva, Dec 28 '16 at 04:13
take a look at this answer : http://stackoverflow.com/questions/15617512/get-json-object-from-url — Ahmad Mobaraki, Dec 28 '16 at 05:12

score 1 · Accepted Answer · answered Dec 28 '16 at 04:42

1

You can fetch data from json as follow

<?php
$content = file_get_contents('http://theapache64.xyz:8080/gpix/v1/gpix?keyword=honda%20civic%202009&limit=1');

$arr = json_decode($content,true);

$url = $arr['data']['images'][0]['image_url'];

echo "<img src='".$url."'>";
?>

answered Dec 28 '16 at 04:42

Bharat Dangar

527
4
20

Thanks for your reply – Bernie Carpio Dec 28 '16 at 05:04

Bernie Carpio · Answer 2 · 2016-12-28T05:34:29.620

Thanks guys, I found a solution.

I will also post my solution here just in case other will encounter the same problem

    <?php

$json_file = file_get_contents('http://theapache64.xyz:8080/gpix/v1/gpix?keyword='.$combineurl.'&limit=1');

$jfo = json_decode($json_file);


$posts = $jfo->data->images;



foreach ($posts as $post) {
    ?>
    <img src="<?php echo $post->image_url; ?>" alt="" width="500">
    <?php
}
?>

Extracting "image_url" value and display as image src

2 Answers2