json_decode()返回空问题 [英] json_decode() returns null issues

问题描述

我的JSON有问题.它可以在PHP 5.3中正确返回(因此我不能使用json_last_error())，并且在我将字符串显式复制到json_decode(json_decode('{...}')中时成功返回.结果作为变量，我正在使用php 5.2，这就是我需要的.

I've an issue with my JSON. It works returns correctly in PHP 5.3 (so I can't use json_last_error()), and it returns successfully when I copy string explicitly into json_decode (json_decode('{...}'). It only returns null in when I pass the result as a variable and I'm using php 5.2, which is what I need it for.

输出来自PHPUnit中的JSON记录:

The output comes from JSON logging in PHPUnit:

[
    {
        "event": "suiteStart",
        "suite": "",
        "tests": 2
    },
    {
        "event": "suiteStart",
        "suite": "TagTestCase",
        "tests": 2
    },
    {
        "event": "test",
        "suite": "TagTestCase",
        "test": "TagTestCase::test_it",
        "status": "fail",
        "time": 0.00248718261719,
        "trace": [
            {
                "file": "\/UnitTest\/PHPUnit.php",
                "line": 98,
                "function": "run",
                "class": "PHPUnit_Framework_TestSuite",
                "type": "->",
                "args": [
                    {

                    }
                ]
            },
            {
                "file": "\/UnitTest\/PHPUnit.php",
                "line": 116,
                "function": "run",
                "class": "PHPUnit",
                "type": "->",
                "args": [

                ]
            },
            {
                "file": "\/UnitTest\/PHPUnit.php",
                "line": 212,
                "function": "__tostring",
                "class": "PHPUnit",
                "type": "->",
                "args": [

                ]
            }
        ],
        "message": "false assertionzzzzz.\nFailed asserting that <boolean:false> is true."
    },
    {
        "event": "test",
        "suite": "TagTestCase",
        "test": "TagTestCase::test_two",
        "status": "pass",
        "time": 0.00182914733887,
        "trace": [

        ],
        "message": ""
    }
]

这些是我一直在探索的路径-也许您是一个更好的探索者. 可能有帮助的三种路径:

These are the paths, I've been exploring - maybe you are a better explorer.. Three possible paths that could help:

• php 5.2和5.3中的json_decode()有何不同?他们做了什么改变?
• 其他人使用PHPUnit中的JSON及其解析方式.
• 将其包含在变量中与将其打印到屏幕并将其复制到json_decode()相比会发生什么变化

任何帮助将不胜感激(！).

Any help would be greatly(!) appreciated.

谢谢！ 马特

推荐答案

多么糟糕的调试会话..好消息是..我知道了..

What a HORRENDOUS debug session.. well there's good news.. I figured it out..

我开始使用AJAX进行查看，并使用Firebug对其进行记录...结果证明json_decode(或顺便说说是eval)无法处理"，这就是PHPUnit发送回去的内容(快来塞巴斯蒂安！)，所以解决它:

I started looking at it using AJAX and logging it with Firebug... and it turns out json_decode (or eval by the way) cannot handle ", which is what PHPUnit sends back (Come on Sebastian!), so to fix it:

$json = str_replace('"', '"', $json);

现在我以为他们是一样的..也许有人可以启发我..

Now I thought they were the same.. maybe someone can enlighten me..

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