php如何获取json⾥的值，如何从JSONPHP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html>
$detailsJsonObject = '{"details":[
{"name":"John","subjectDetails":
{"subjectId":"101","subjectName":"PHP","marks":"58",
"teacherName":"Bob"}
}]}';
$convertToArrayObject = json_decode($detailsJsonObject,true);
$actualSubjectName = $convertToArrayObject[details][0][subjectDetails][subjectName];
phpjson格式化输出
$actualTeacherName = $convertToArrayObject[details][0][subjectDetails][teacherName];
echo "The Subject Name is=",$actualSubjectName,"
";
echo "The Teacher Name is=",$actualTeacherName;
>
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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