Home php教程 php手册 如何正确解决PHP缺少JSON包问题

如何正确解决PHP缺少JSON包问题

Jun 13, 2016 am 11:10 AM
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这一问题的替代办法就是重写PHP json 的json_decode 和 json_encode方法。

这两个PHP缺少JSON包的解决方法主要就是把数组,字符串,或者对象 转换成 json格式的数据(特殊的一种结构体格式) 。首先需要的是json_encode方法, 此方法 主要将 数组,字符串,对象等转换成json格式。解决此问题的方法:

a. 简单版:

直接针对自己的问题需要(自己只需要将简单的数组进行操作),可以直接将此数组进行拼串。

b。完善的 encode方法:

解决PHP缺少JSON包的问题的简单方法就是 利用别人以完成的开源代码:

首先想到 zend framework。 拿来 zf_json 包中的 Decoder.PHP, 和 Encoder.PHP。 Decoder.PHP其实只要将其中某个protect方法的 protect去掉就行。 具体操作忘了,读者自行调试来解决。改写好Decoder 类后,调用如下:

<ol class="dp-xml">
<li class="alt"><span><span>$</span><span class="attribute">filepath</span><span> = $_SERVER['DOCUMENT_ROOT'];  </span></span></li>
<li>
<span>include $filepath.'/inc/json/Encoder.</span>PHP<span>';   </span>
</li>
<li class="alt">
<span>$</span><span class="attribute">encoder</span><span> = </span><span class="attribute-value">new</span><span> Zend_Json_Encoder(false);  </span>
</li>
<li>
<span>$</span><span class="attribute">json</span><span> = $encoder -</span><span class="tag">></span><span>encode($result);   </span>
</li>
<li class="alt"><span>header("X-JSON: $json"); </span></li>
</ol>
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json_decode方法同上,如果是PHP使用。

由于很多使用json的地方还是在ajax调用中, 所以一些ajax '框架' 提供了decode方法来解决PHP缺少JSON包这一问题:

比如在prototype中, 如果PHP页面包含了prototype.js, 那么就可以直接来调用 function(transport, json) { 。。操作} 在其中取json数据 就是 json.a , json.b 等了。

除了PHP缺少JSON包这一问题的出现外,可能遇到问题,编码。

zend framework返回的汉字编码估计不大对应(不是说它不是utf-8)我的页面编码(utf-8),因此会有问题出现。因此对于某些验证返回的报错信息(汉字)会有影响。在这里的替代方法就是使用传统的接口返回数字代码形式(通信网关经常采用的方法),比如返回 json.code 有 0,1,2,3等多种可能。则在js端处理时, 0表示成功,1表示非法的客户端数据,等等。 问题得到完美解决。


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