当Content-Type为HTML时,JSON_PRETTY_PRINT无法按预期工作

Tos*_*shi 10 php mysql sql json

我动态地收到一个JSON字符串,如下所示:

{"post":[{"id":"11","body":"","image":"images/rose.png","stamp":"2013-11-04 14:50:11" }]}

我正在尝试按如下方式打印此JSON字符串:

{
    "post": [
        {
            "id": "11",
            "body": "",
            "image": "images/rose.png",
            "stamp": "2013-11-04 14:50:11"
        }
    ]
}
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所以,我尝试了以下代码(仅用于演示目的):

<?php

$str = '{ "post": [ { "id": "11", "body": "", "image": "images\/rose.png", "stamp": "2013-11-04 14:50:11" } ] }';
$obj = json_decode($str);
echo json_encode($obj, JSON_PRETTY_PRINT);
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它只输出未格式化的JSON字符串:

{"post":[{"id":"11","body":"","image":"images/rose.png","stamp":"2013-11-04 14:50:11" }]}

但是,当我在json_encode()语句上方添加以下行时,它会按预期工作.

header('Content-Type: text/plain');
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可能导致此问题的原因是什么?为什么不工作时Content-Typetext/html

Tom*_*ess 20

JSON_PRETTY_PRINT使用空格来格式化JSON.当它显示为HTML时,将忽略空格.如果要保留格式,请将JSON包装在<pre>标记中.

例如:

<?php

$str = '{ "post": [ { "id": "11", "body": "", "image": "images\/rose.png", "stamp": "2013-11-04 14:50:11" } ] }';
$obj = json_decode($str);
$json = json_encode($obj, JSON_PRETTY_PRINT);

printf("<pre>%s</pre>", $json);
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如果您不想使用<pre>标记,那么您还可以在white-space: pre包含JSON的任何元素上使用CSS属性.


小智 6

您使用以下标头:

header('Content-Type: application/json; charset=utf-8');
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