Home > Backend Development > PHP Problem > How to convert PHP object to JSON string array

How to convert PHP object to JSON string array

PHPz
Release: 2023-04-05 11:30:02
Original
517 people have browsed it

JavaScript is increasingly important in modern web development, as more and more web applications require rich interactivity and responsiveness for web clients. JSON (JavaScript Object Notation) has become a popular and common data transmission format in web development because it is a text representation of JavaScript objects and can be easily transferred between the client and server. Therefore, converting objects to JSON string array in PHP is a very important task. In this article, we will explain how to convert PHP objects to JSON string arrays.

JSON Functions in PHP:

PHP provides a set of JSON functions that can easily convert PHP objects to JSON strings and JSON strings to PHP objects. The following are commonly used PHP JSON functions:

  • json_encode: Convert PHP objects to JSON strings
  • json_decode: Convert JSON strings to PHP objects

json_encode() function:

json_encode() function is a built-in function in PHP. It converts PHP objects and arrays into JSON encoded strings. It needs to be used when encoding PHP variables into JSON format. this function. The following is the syntax of the json_encode() function:

string json_encode ( mixed $value [, int $options = 0 [, int $depth = 512 ]] )
Copy after login

where the value parameter is the PHP variable to be encoded. The options parameter is an option used to control encoding behavior and can be ignored. The depth parameter is used to specify the maximum recursion depth and can be ignored.

Example of json_encode() function:

$obj = new stdClass();
$obj->name = 'Mike';
$obj->age = '30';
$obj->city = 'New York';

$json = json_encode($obj);

echo $json; //输出:{"name":"Mike","age":30,"city":"New York"}
Copy after login

In the above example, we first create a PHP object containing three properties (name, age, city), and then use json_encode( ) function encodes it into a JSON string and assigns it to the variable $json. Finally, use the echo statement to output the JSON string.

json_decode() function:

The json_decode() function is another built-in function of PHP that is used to decode JSON format strings into PHP objects. The following is the syntax of the json_decode() function:

mixed json_decode ( string $json [, bool $assoc = FALSE [, int $depth = 512 [, int $options = 0 ]]] )
Copy after login

where the json parameter is the JSON string to be decoded. The assoc parameter is an optional parameter. If set to true, the returned object will be an associative array; if set to false, the returned object will be a standard object. The depth parameter is used to specify the maximum recursion depth and can be ignored. The options parameter is an option used to control decoding behavior and can be ignored.

Example of json_decode() function:

$json = '{"name":"Mike","age":30,"city":"New York"}';

$obj = json_decode($json, false);

echo $obj->name; //输出:Mike
echo $obj->age; //输出:30
echo $obj->city; //输出:New York
Copy after login

In the above example, we first create a JSON string and then decode it into a PHP object using the json_decode() function and convert It is assigned to the variable $obj. Finally, use the echo statement to output the properties of these objects.

Summary:

In web development, JSON has become a popular data format because it can easily transfer and parse data between clients and servers. In PHP, we can use json_encode() function to convert PHP objects to JSON strings and use json_decode() function to convert JSON strings to PHP objects. These two functions are very useful when developing web applications.

The above is the detailed content of How to convert PHP object to JSON string array. For more information, please follow other related articles on the PHP Chinese website!

source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template