提取“image\u url”值并显示为image src

Question

是否可以从该链接中提取“image\u url”的值

http://theapache64.xyz:8080/gpix/v1/gpix?keyword=honda公民2009

并显示为




Answer 1

谢谢伙计们，我找到了解决办法。

我也会在这里发布我的解决方案，以防其他人遇到同样的问题

    <?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
}
?> 

Answer 2

您可以从json获取数据如下

<?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."'>";
?>