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HomeBackend DevelopmentPHP Tutorial在php中判断一个请求是ajax请求还是普通请求的方法_PHP

Ajax

/path/to/pkphp.com/script.php?ajax
在php脚本中使用如下方法判断:
复制代码 代码如下:
if(isset($_GET['ajax'])) {
...这是一个ajax请求,然后...
}
else {
...这不是一个ajax请求,然后...
}

通过传递_GET参数的方法简单实现了网页请求的判断。但是如果需要这样的功能,这个方法可能就有弊端,功能需求如下:
1.通过ajax请求的网页与普通请求的网页内容是不相同的
2.通过ajax请求的网页是为了方便用户操作,两种方法请求打开的网页必须的内容是相同的,只是ajax请求到的网页内容比较简化和使用,去除了网页的大框架模板。
3.这么做的目的是:用户在网页操作时通过ajax实现,而搜索引擎访问网页时(相当于普通打开网页),得到的内容是一个完整的网页(包含了网页的大框架模板)。
要完成上面的这个功能,就不能使用前面介绍的通过GET参数传递来判断了,如果使用GET传递来判断的话,用户ajax请求和普通网页请求都会是一样的内容,因为你不可能为一个链接设置一个带ajax判断参数和不带的URL。那么如何才能实现这个功能呢?必须通过服务器端PHP判断解决这个问题。也就是今天要说的PHP如何判断ajax请求。这个问题要解决有一个先决条件,那就是你使用的ajax框架必须是jquery。在jquery框架中,对于通过它的$.ajax, $.get, or $.post方法请求网页内容时,它会向服务器传递一个HTTP_X_REQUESTED_WITH的参数,你可以利用如下方法判断某个请求是ajax请求还是普通请求:
if(isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest')
{
..这是一个ajax请求,然后...
}
else {
..这不是一个ajax请求,然后...
}
if(isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest')if(isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH
{
...这是一个ajax请求,然后...
}
else {
...这不是一个ajax请求,然后...
}
利用这个来进行判断操作,可以使网页端的URL保持一致,但是能够对两种不同的请求却能够得到不同内容的网页。即实现了用户操作优化,又不影响搜索引擎收录,我觉得是一个很棒的解决方案!
这里有一个另外需要注意的问题,就是如果你的jquery请求是通过iframe打开网页的,那么HTTP_X_REQUESTED_WITH参数不会被传递,也就是说你没有办法判断请求的类型。

主要内容是:

1.
这个问题要解决有一个先决条件,那就是你使用的ajax框架必须是jquery。在jquery框架中,对于通过它的$.ajax, $.get, or $.post方法请求网页内容时,它会向服务器传递一个HTTP_X_REQUESTED_WITH的参数,你可以利用如下方法判断某个请求是ajax请求还是普通请求:
if(isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest')
{

2.
这里有一个另外需要注意的问题,就是如果你的jquery请求是通过iframe打开网页的,那么HTTP_X_REQUESTED_WITH参数不会被传递,也就是说你没有办法判断请求的类型。
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