如何正确从 JSON PHP 中获取值?
phpserver side programmingprogramming更新于 2025/4/13 3:07:17
要从 JSON 中获取值，请使用 json_decode()。假设以下是我们的 JSON
$detailsJsonObject = '{"details":[{"name":"John","subjectDetails":{"subjectId":"101","subjectName":"PHP","marks":"58","teacherName":"Bob"}}]}';
我们需要获取特定值，即主题名称、标记等。
示例
PHP 代码如下
<!DOCTYPE html>
<html>
<body>
<?php
$detailsJsonObject = '{"details":[
{"name":"John","subjectDetails":
{"subjectId":"101","subjectName":"PHP","marks":"58",
"teacherName":"Bob"}
}]}';
$convertToArrayObject = json_decode($detailsJsonObject,true);
$actualSubjectName = $convertToArrayObject[details][0][subjectDetails][subjectName];
$actualTeacherName = $convertToArrayObject[details][0][subjectDetails][teacherName];
echo "The Subject Name is=",$actualSubjectName,"<br>";
echo "The Teacher Name is=",$actualTeacherName;
?>
</body>
</html>
输出
这将产生以下输出 −
The Subject Name is=PHP The Teacher Name is=Bob
