当数组在php中为空时，json_encode函数不返回大括号{} [英] json_encode function not return Braces {} when array is empty in php

问题描述

我有这个代码

$status = array(
                "message"=>"error",
                "club_id"=>$_club_id,
                "status"=>"1",
                "membership_info"=>array(),
                );

echo json_encode($status);

此函数返回json:
{"message":"error","club_id":275,"status":"1","membership_info":[]}

This function return json:
{"message":"error","club_id":275,"status":"1","membership_info":[]}

但是我需要这样的json:

But I need json like this:

{"message":"error","club_id":275,"status":"1","membership_info":{}}

推荐答案

使用json_encode的JSON_FORCE_OBJECT选项:

json_encode($status, JSON_FORCE_OBJECT);

文档

JSON_FORCE_OBJECT(整数) 使用非关联数组时，输出对象而不是数组.当输出的接收者期望对象并且数组为空时，此功能特别有用.自PHP 5.3.0起可用.

JSON_FORCE_OBJECT (integer) Outputs an object rather than an array when a non-associative array is used. Especially useful when the recipient of the output is expecting an object and the array is empty. Available since PHP 5.3.0.

或者，如果要在对象中保留其他"数组，则不要使用先前的答案，只需使用以下方法即可:

Or, if you want to preserve your "other" arrays inside your object, don't use the previous answer, just use this:

$status = array(
                "message"=>"error",
                "club_id"=>$_club_id,
                "status"=>"1",
                "membership_info"=> new stdClass()
                );

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