Question

3天前我提出了这个问题，但是到目前为止，我无法解决我的问题。我将再次提出问题，希望有人帮助我。

我有以下JSON结构

{"Id":"1","Persons":[{"Name":"Luis","Time":"00:00:09","info":"","Timeext":"","Timeout":"","Timein":""}, {"Name":"Carl","Time":"00:00:03","info":"","Timeext":"","Timeout":"","Timein":""},{"Name":"Ben","Time":"00:00:08","info":"","Timeext":"","Timeout":"","Timein":""}]}

要访问元素Id不是问题。我可以得到这样的价值：

$arr['Id'] = $_POST['Id'];
echo $arr['Id'];

但是如果想要访问JSON内部的结构Persons，特别是每个Person的Time值，我喜欢这样：

 $arr['Persons'] = $_POST['Persons'];
    $jsdecode = json_decode($arr['Persons']);

    foreach ($arr['Persons'] as $p){
        echo "$p->Time <br/>";
    }

这是得到的结果：

<b>Warning</b>:  json_decode() expects parameter 1 to be string, array given in <b>C:\xampp\htdocs\Stopuhr\controller\prozess.controller.php</b> on line <b>38</b><br />
<br />
<b>Notice</b>:  Trying to get property of non-object in <b>C:\xampp\htdocs\Stopuhr\controller\prozess.controller.php</b> on line <b>41</b><br />
 <br/><br />
<b>Notice</b>:  Trying to get property of non-object in <b>C:\xampp\htdocs\Stopuhr\controller\prozess.controller.php</b> on line <b>41</b><br />
 <br/>

请问有人帮助我吗？

Answer 1

json_encode（） - PHP数组到JSON

json_encode()方法将采用PHP数组并将其编码为JSON，以供AJAX调用使用。

$myarray = array('Guitar' => 'Johnny', 'Vocals'=> 'Stephen', 'Bass' => 'Andy', 'Drums' => 'Mike');
$myJson = json_encode($myarray);
echo $myJson;

json_decode（） - JSON到PHP数组

json_decode()将获取JSON并将其转换为PHP数组。

$myJson = '{"Guitar" : "Johnny", "Vocals": "Stephen", "Bass" : "Andy", "Drums" : "Mike"}';
$myarray = json_decode($myJson, true);
print_r($myarray);

Answer 2

您遇到语法错误：

{"Id":"1","Persons":[
{"Name":"Luis","Time":"00:00:09","info":"","Timeext":"","Timeout":"","Timein":""}, 
{"Name":"Carl","Time":"00:00:03","info":"","Timeext":"","Timeout":"","Timein":""},
{"Name":"Luis","Time":"00:00:08","info":"","Timeext":"","Timeout":"","Timein":""}
]}

在Name Louis之前没有逗号。

您应始终在http://jsonlint.com/

上验证已编码的json

使用foreach PHP从JSON结构访问嵌套项会产生错误

2 个答案: