今天用PHP读取一个接口的数据使用了iconv转换字符编码格式，出现Notice: iconv(): Unknown error (84) :
$this->debug_show("客户机命令：".iconv("GBK","UTF-8", $this->in));
上面这行摘自：http://justwinit.cn/post/1057/
最后，修改为如下：
$this->debug_show("客户机命令：".iconv("GBK","utf-8//TRANSLIT//IGNORE", $this->in));
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读其官方文档 http://www.php.net/manual/en/function.iconv.php对参数out_charset的解释：
The output charset.
If you append the string //TRANSLIT to out_charset transliteration is activated. This means that when a character can’t be represented in the target charset, it can be approximated through one or several similarly looking characters. If you append the string //IGNORE, characters that cannot be represented in the target charset are silently discarded. Otherwise, str is cut from the first illegal character and an E_NOTICE is generated.
大概的意思就是：
如果你加上 //TRANSLIT 到out_charset 的参数后面，意味着如果找不到目标编码，则程序会去找与其相近的编码。如果你加的是//IGNORE，则不会去找相近的编码，而且只要有一个字符是程序无法识别的则将会报错。
根据上面的解释我将代码
iconv('gb2312','utf-8', serialize($storeData));
改为
iconv('gb2312','utf-8//TRANSLIT//IGNORE', serialize($storeData));
这样就ok了！