Question

echo $someJSON=json_encode($resultData);

输出为{"status":true,"postData":[{"post_id":"3","post_title":"JAVA","post_desc":"JAVA DESCRIPTION","status":"1"},{"post_id":"1","post_title":"PHP API","post_desc":"MAKING PHP API","status":"1"}]}

从输出中，如何使用PHP获取'post_title'的值？

Answer 1

$arr = json_decode('{"status":true,"postData":[{"post_id":"3","post_title":"JAVA","post_desc":"JAVA DESCRIPTION","status":"1"},{"post_id":"1","post_title":"PHP API","post_desc":"MAKING PHP API","status":"1"}]}');

foreach($arr->postData as $post) {
    $title = $post->post_title;
}

Answer 2

$resultData已经是一个对象，因此您无需转换它。只需像任何其他对象一样访问其属性。例如：

foreach($resultData->postData as $data) {
    echo $data->post_title, ', ';
}

演示：http://rextester.com/JHFH10035

Answer 3

您必须先使用json_decode

对其进行解码

$someJson = json_decode('{"status":true,"postData":[{"post_id":"3","post_title":"JAVA","post_desc":"JAVA DESCRIPTION","status":"1"},{"post_id":"1","post_title":"PHP API","post_desc":"MAKING PHP API","status":"1"}]}');

foreach ($someJson->postData as $data){
    echo $data->post_title;
}

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3 个答案: