Question

我有以下JSON结构

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

如何访问或读取嵌套内的项目？例如，如果我只想要Carl的时间价值或者有关Carl的所有信息。直到现在我可以毫无问题地收集集合'Id'中的单个项目。第一个嵌套的项目不是。我尝试使用json_decode：

if( $_POST ) {

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

            $jsdecode = json_decode($arr);

            foreach ($jsdecode as $values){
                echo $values->Time;
            }

请问有人帮助我吗？

Answer 1

if( $_POST ) {

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

        $jsdecode = json_decode($arr,true);

        foreach ($jsdecode as $values){
            echo $values->Time;
        }

添加＆＃39; true＆＃39; to json_decode将其转换为数组

Answer 2

您正在处理它是正确的，只需将true添加为json_decode的第二个参数，该参数将转换为数组，如

$jsdecode = json_decode($arr,true);

如何使用foreach PHP访问JSON中的嵌套项？

2 个答案: