[英]Removing part of a string in a query string in a URL
I have an input URL that looks something like this: 我有一个输入网址,看起来像这样:
http://localhost/20north/Numark/product/1/B$@!00$@!4JPPO94$@!
While redirecting this to a new URL, I need to find and remove all occurrences of "$@!" 将其重定向到新的URL时,我需要查找并删除所有出现的“ $ @!”。 from the last part of the url, so that it becomes:
从网址的最后一部分开始,这样它就变成了:
http://localhost/20north/Numark/product/1/B004JPPO94
Note: The last part can be anything and not just B$@!00$@!4JPPO94$@!
注意:最后一部分可以是任何内容,而不仅仅是
B$@!00$@!4JPPO94$@!
. 。 Also, the position of
$@!
另外,
$@!
的位置$@!
can be anywhere in that last part. 可以在最后一部分中的任何地方。
Using mod_rewrite, you just need this rule: 使用mod_rewrite,您只需要以下规则:
RewriteRule ^(.*)\$@!(.*)$ $1$2 [N]
Edit: 编辑:
Actually, there seems to be a problem when the $@!
实际上,当
$@!
时似乎存在问题$@!
is at the end of the URI. 在URI的末尾。 Adding an extra rule to remove the trailing match seems to fix it:
添加额外的规则以删除结尾的匹配似乎可以解决此问题:
RewriteRule ^(.*)\$@!$ $1
RewriteRule ^(.*)\$@!(.*)$ $1$2 [N]
Not quite sure why that was happening. 不太清楚为什么会这样。
If you're using php, you could do the following: 如果您使用的是php,则可以执行以下操作:
<?php
$this_url = $_SERVER['SERVER_NAME'] . $_SERVER['REQUEST_URI'];
if( strpos($this_url, '$@!') !== false )
die(header('Location: ' . str_replace('$@!', '', $this_url)));
?>
Edit: updated the code to become dynamic 编辑:更新代码以使其变为动态
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