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":""}]}
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如何在巢内访问或读取项目?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;
}
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请问有人帮帮我吗?
小智 5
if( $_POST ) {
$arr['Id'] = $_POST['Id'];
$arr['NP'] = $_POST['NP'];
$jsdecode = json_decode($arr,true);
foreach ($jsdecode as $values){
echo $values->Time;
}
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将"true"添加到json_decode会将其转换为数组
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