php json to object method: 1. Create a PHP sample file; 2. Define a "$json" variable; 3. Get the object object through the "json_decode($json);" method.
The operating environment of this article: Windows 7 system, PHP version 7.1, Dell G3 computer.
php How to convert json to object?
php json string to array or object
The method found on the Internet is to use get_object_vars to convert the class type into an array and then use foreach to traverse That's it
$array = get_object_vars($test); $json= '[{"id":"1","name":"\u5f20\u96ea\u6885","age":"27","subject":"\u8ba1\u7b97\u673a\u79d1\u5b66\u4e0e\u6280\u672f"},{"id":"2","name":"\u5f20\u6c9b\u9716","age":"21","subject":"\u8f6f\u4ef6\u5de5\u7a0b"}]';
First use json_decode to encode the JSON format string,
$students = json_decode($json);
Use $students directly in the PHP file:
for($i=0;$i<count($students);$i++){ echo "姓名:".$students[$i]['name']."年龄:".$students[$i]['age']."专业:".$students[$i]['subject']."<br/>"; }
The error will be reported as follows:
Fatal error: Cannot use objectof type stdClass as array in D:\wamp\www\test.phpon line 18
At this time, print $students:
var_dump($students);
will output:
array(2) { [0]=> object(stdClass)#2 (4) { ["id"]=> string(1)"1" ["name"]=> string(9)"张雪梅" ["age"]=> string(2)"27" object(stdClass)#3 (4) { 这个就说明转换的json字符串转为对象而非数组,请看下面的红色背景字 ["subject"]=>string(24) "计算机科学与技术" } [1]=> ["id"]=> string(1)"2" ["name"]=> string(9)"张沛霖" ["age"]=> string(2)"21" ["subject"]=> string(12) "软件工程" } }
It can be seen that the returned result is object instead of array. Should be accessed in object form:
foreach($students as $obj){ echo "姓名:".$obj->name."年龄:".$obj->age."专业:".$obj->subject."<br/>"; }
The output result is:
姓名:张雪梅 年龄:27 专业:计算机科学与技术 姓名:张沛霖 年龄:21 专业:软件工程
mixedjson_decode ( string$json [, bool$assoc ] )
Description : Accepts a JSON-formatted string and converts it into a PHP variable.
json_decode can receive two parameters:
json: the string in jsonstring format to be decoded.
assoc: When this parameter is TRUE, an array will be returned instead of an object.
$students = json_decode($json,true);
Print $students at this time:
var_dump($students);
Output:
array(2) { [0]=> array(4) { ["id"]=> string(1)"1" ["name"]=> string(9)"张雪梅" ["age"]=> string(2)"27" ["subject"]=>string(24) "计算机科学与技术" } [1]=> array(4) { ["id"]=> string(1)"2" ["name"]=> string(9)"张沛霖" ["age"]=> string(2)"21" ["subject"]=>string(12) "软件工程" } }
At this time, $students is an array and can be used directly:
for($i=0;$i<count($students);$i++){ echo "姓名:".$students[$i]['name']."年龄:".$students[$i]['age']."专业:".$students[$i]['subject']."<br/>"; }
The output result is:
姓名:张雪梅 年龄:27 专业:计算机科学与技术 姓名:张沛霖 年龄:21 专业:软件工程
Summary:
Two ways to process JSON format strings in PHP code:
Method one:
$json= '[{"id":"1","name":"\u5f20\u96ea\u6885","age":"27","subject":"\u8ba1\u7b97\u673a\u79d1\u5b66\u4e0e\u6280\u672f"},{"id":"2","name":"\u5f20\u6c9b\u9716","age":"21","subject":"\u8f6f\u4ef6\u5de5\u7a0b"}]'; $students= json_decode($json);//得到的是 object foreach($studentsas $obj){ echo "姓名:".$obj->name." 年 龄:".$obj->age." 专 业:".$obj->subject."<br />";}
Method two:
$json= '[{"id":"1","name":"\u5f20\u96ea\u6885","age":"27","subject":"\u8ba1\u7b97\u673a\u79d1\u5b66\u4e0e\u6280\u672f"},{"id":"2","name":"\u5f20\u6c9b\u9716","age":"21","subject":"\u8f6f\u4ef6\u5de5\u7a0b"}]'; $students= json_decode($json, true);//得到的是 array for($i=0;$i<count($students);$i++){ echo "姓名:".$students[$i]['name']." 年 龄:".$students[$i]['age']." 专 业:".$students[$i]['subject']."<br />";
Recommended learning: "PHP Video Tutorial"
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