将变量设置为PHP中json_encode()输出的内容 [英] Set a variable to what json_encode() outputs in PHP

问题描述

json_encode()返回一个字符串...

json_encode() returns a string...

下面的代码不行吗?

class TestClass {
    private $test = json_encode("test");
}

PHP输出

Parse error: syntax error, unexpected '(', expecting ',' or ';' in /home/testuser/public_html/test.php on line 10

推荐答案

在声明类属性时不能分配表达式或任何变量.此处允许使用文字常量，例如__FILE__.

You can not assign expression or any variable while declaring class property. The literal constants such as __FILE__ are allowed here.

它们必须是文字值，例如字符串或常量.

they have to be a literal value, such as a string, or a constant.

所有工作.

private $test= 98;
private $test= "test value";
private $test= CONSTANT;
private $test= __FILE__;

但是这些不会

private $test= 98*2;
private $test= "test value"."some other value";

您可以使用构造函数

 function __construct() {
        $this->test = json_encode("test");
    }

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