PHP利用header跳转失效的解决方法，phpheader跳转失效
PHP利用header跳转失效的解决方法，phpheader跳转失效
本文实例讲述了PHP利用header跳转失效的解决方法，分享给大家供大家参考。具体方法分析如下：
一、问题：
今天header(\"Location: $url\"),以往跳转总是可以的,今天却不动,只是输出结果,以往自己要确认检查，$url的值获取的是否正确,所以在前面加了echo $url；来调试用,结果就导致了header函数的无效.
二、解决方法：
在PHP中用header("location:test.php")进行跳转要注意以下几点:
1、location和“:”号间不能有空格,否则会出错.//phpfensi.com
2、在用header前不能有任何的输出,包括include的页面中标签“?>”后不能有空格.
3、header后的PHP代码还会被执行.
PHP的 header 跳转之前不能有任何内容输出,因为PHP开始执行的时候就已经向浏览器送出HTTP头信息,之后就不再允许更改了.
但是如果必须要在输出之后再处理header信息的话可以使用 ob_start() ob_end_flush() 来缓存一下要发送的内容,等到header继续再发送内容.
或者一个更简单的办法,修改php.ini,找到 output_buffering=Off 修改为 output_buffering=4096.
希望本文所述对大家的PHP程序设计有所帮助。
如果你确定要使用post方式提交的话，那是一件很复杂的事，因为需要php来模拟post提交，你可以使用 fsockopen 来处理，具体方案我可以教你。
另外一种方法就是使用get方案，相对来言，就是一种超简单，直接转向的时候，把post来的数据转成get方式header过去就行了。
还有一种解决方案就是在前段使用ajax来处理。在a页面ajax提交数据到b.php，根据返回值代表不通过验证时，然后把页面用js转到b.php，如果返回值代表验证通过，就再次把当前post的数据再提交一次到c.php。
编码问题 把charset=UTF-8改为charset=GB2312
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