url参数获取有关问题
url参数获取问题 有urlhttp://www.kaimen360.com/bargai.php?mod=showid=578 与 http://www.kaimen360.com/appraisals.php?mod=showid=79 怎么获取mod与id值？ 怎么判断这个网页是appraisals.php还是bargai.php页面？ ------解决思路----------------------
url参数获取问题
有urlhttp://www.kaimen360.com/bargai.php?mod=show&id=578
与
http://www.kaimen360.com/appraisals.php?mod=show&id=79
怎么获取mod与id值？
怎么判断这个网页是appraisals.php还是bargai.php页面？
------解决思路----------------------
怎么获取mod与id值？
$mode = $_GET['mode'];
$id = $_GET['mode'];
怎么判断这个网页是appraisals.php还是bargai.php页面？
无需判断，不是就进不来！
------解决思路----------------------
$filename = basename__FILE__);
这是没有意义的工作！
http://www.kaimen360.com/bargai.php?mod=show&id=578
只会进入 bargai.php
http://www.kaimen360.com/appraisals.php?mod=show&id=79
只会进入 appraisals.php
所以根本就不需要判断
你在 appraisals.php 中检查当面文件名是不是 bargai.php
这不是画蛇添足吗？
------解决思路----------------------
问题描述不完全。 听xuzuning 的 。
我估计楼主是把这两个url嵌入到一个另外一个脚本，然后再那里判断是来自哪个url。再决定title显示之类的。
赶紧设计的不是很合理。
------解决思路----------------------
是不是你的 bargai.php和appraisals.php的title是公共文件，所以要判断啊，要是只有部分不一样，就在公共里写title一样的地方，在具体页面用replace替换一下不同的，要是title完全不同，在自己页面直接写title就行，就像版主说的，不是apprai【本文来自鸿网互联 (http://www.68idc.cn)】sals.php肯定就不会进这个文件了，那就没有必要做判断了
------解决思路----------------------
怎么获取mod与id值？
$mode = $_GET['mode'];
$id = $_GET['mode'];
获取当前页面地址，可以这样写，不过既然能进入，当然就是当前页。
echo $_SERVER['REQUEST_URI'];
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