文本编辑器(Sublime Text，Geany，Notepad ++等)，用于从URL字符串中删除除一个参数值以外的所有参数 [英] Text Editor(Sublime Text, Geany, Notepad++ etc.) Regex to remove all parameters from URL string except one parameter-value

问题描述

我对Regex中的高级匹配模式不是很熟悉.

I am not very familiar with advanced matching patterns in Regex.

我有一些Google搜索URL，需要清理它们，而无需按住Backspace键5秒钟即可从URL中删除不必要的参数.

I have some Google Search URLs which I need to clean up without having to hold Backspace key for 5 seconds to remove unnecessary parameters from the URL.

比方说我有这个URL(下面的模式可以有许多不同的URL):

Let's say I have this URL(could many different URLs following patterns like below):

https://www.google.com/search?source=hp&ei=Ne4pXpSIHIW_9QOD-rmADw&q=laravel+crud+generator&oq=laravel+crud+generator&gs_l=psy-ab.3..0l8.1294.6845..7289...1.0..0.307.3888.0j20j2j1......0....1..gws-wiz.....6..0i131j0i362i308i154i357.PwlZ_932pXo&ved=0ahUKEwjU9pz4tJrnAhWFX30KHQN9DvAQ4dUDCAU&uact=5

我想将其转换为如下所示的干净网址:

And I want to turn that into nice clean Search URL as below:

https://www.google.com/search?q=laravel+crud+generator

我该如何在问题中提到的任何文本编辑器中使用正则表达式查找/替换?

How can I acheive that using Find/Replace with Regex of any of mentioned text editors in Question ?

推荐答案

我发布的消息是其他人使用该解决方案.

I'm posting that others use the solution.

在记事本++中，请按 CTRL + H ，然后在下面选择正则表达式.

in notepad++ please press CTRL+H then select Regular expression on below.

然后在上查找以下内容:此模式:.+&(q = [^&] +).+ 并在替换为插入"中: https://www.google.com/search?$1

Then place on Find what: this pattern: .+&(q=[^&]+).+ and in Replace with insert: https://www.google.com/search?$1

现在，轻松按 Replace 按钮进行一次替换或进行所有替换，按 ALT + A 或 Replace All 按钮.

Now, easily press the Replace button for single replace or for all replacements press ALT+A or Replace All button.

检查 Regex101

但是描述:

1- .+& 在 q 之后找到& 之前的所有字符.因此，这部分包括 https://www.google.com/search?source=hp&ei=Ne4pXpSIHIW_9QOD-rmADw&

1- .+& find all characters before & following a q. So this part includes https://www.google.com/search?source=hp&ei=Ne4pXpSIHIW_9QOD-rmADw&

2- (q = [^&] +)，我们的目标！我们希望在 q = 之后的所有内容都在下一个& 上.因此，我们搜索以q =开头的字符串，然后搜索不是& 的任何字符. [^&] 表示不是& 的字符，并且 + 表示任何不是& <的字符/code>大于零的时间.这部分将包括 q = laravel + crud + generator .请注意括号.

2- (q=[^&]+), our target! we want everything after q= up next &. So we search for a string which started with q= then any character which is not &. [^&] means a character that is not & and + is saying that any character that is not & more than zero time. this part will include q=laravel+crud+generator. Please notice the parentheses.

3- .+ 表示任何字符，包括& oq = laravel + crud + generator& gs_l = psy-ab.3..0l8.1294.6845..7289 ...1.0..0.307.3888.0j20j2j1 ...... 0 .... 1..gws-wiz ..... 6..0i131j0i362i308i154i357.PwlZ_932pXo& ved = 0ahahEWEjU9pz4tJrnAhWFX30KHQN9DvAQ4dUDCAU& uact = 5

3- .+ means any character and includes &oq=laravel+crud+generator&gs_l=psy-ab.3..0l8.1294.6845..7289...1.0..0.307.3888.0j20j2j1......0....1..gws-wiz.....6..0i131j0i362i308i154i357.PwlZ_932pXo&ved=0ahUKEwjU9pz4tJrnAhWFX30KHQN9DvAQ4dUDCAU&uact=5

好吧，还记得第2节中的()吗?那是一个小组.您可以通过此模式 $ groupNumber 在替换中使用组，其中groupNumber是括号的索引.这里我们只有一个()或实际上只有一组，所以我们的替换语句将为 $ 1 .

ok, remember () in section 2? that was a group. you can use groups in replacements by this pattern $groupNumber which groupNumber is the index of parentheses. Here we have just one () or actually just one group, so our replacement statement will be $1.

最后替换: https://www.google.com/search?$1 ，所以第一组中的所有内容都将替换为$ 1.

And finally replacement: https://www.google.com/search?$1 so everything is inside group one will replace with $1.

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