如何从JSON PHP正确获取值？
要从JSON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"}}]}';我们需要获取特定的值，例如主题名称，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>输出结果
这将产生以下输出
The Subject Name is=PHP The Teacher Name is=Bob
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