从匹配正则表达式的结果中删除最后一个字符 [英] Remove last char from result of a match regex

问题描述

我正在尝试在jmeter测试中提取html的一部分.

I'm trying to extract a part of html in a jmeter test.

我只需要从<script src=""标签中提取一部分.

I need to extract just a part from a <script src="" tag.

• 完整脚本src: <script src="/Paginas/Inicializacao/AguardarAcao.aspx?_TSM_HiddenField_=ctl00_ToolkitScriptManager1_HiddenField&amp;_TSM_CombinedScripts_=%3b%3bAjaxControlToolkit%2c+Version%3d1.0.11119.38311%2c+Culture%3dneutral%2c+PublicKeyToken%3d28f01b0e84b6d53e%3apt-BR%3adf9c6e46-ef8c-4a3d-89af-f80adf22e9c2%3a865923e8%3a411fea1c%3ae7c87f07%3a91bd373d%3a1d58b08c%3a8e72a662%3aacd642d2%3a596d588c%3a77c58d20%3a14b56adc%3a269a19ae" type="text/javascript"></script>

• full script src: <script src="/Paginas/Inicializacao/AguardarAcao.aspx?_TSM_HiddenField_=ctl00_ToolkitScriptManager1_HiddenField&amp;_TSM_CombinedScripts_=%3b%3bAjaxControlToolkit%2c+Version%3d1.0.11119.38311%2c+Culture%3dneutral%2c+PublicKeyToken%3d28f01b0e84b6d53e%3apt-BR%3adf9c6e46-ef8c-4a3d-89af-f80adf22e9c2%3a865923e8%3a411fea1c%3ae7c87f07%3a91bd373d%3a1d58b08c%3a8e72a662%3aacd642d2%3a596d588c%3a77c58d20%3a14b56adc%3a269a19ae" type="text/javascript"></script>

，我只需要: %3b%3bAjaxControlToolkit%2c+Version%3d1.0.11119.38311%2c+Culture%3dneutral%2c+PublicKeyToken%3d28f01b0e84b6d53e%3apt-BR%3adf9c6e46-ef8c-4a3d-89af-f80adf22e9c2%3a865923e8%3a411fea1c%3ae7c87f07%3a91bd373d%3a1d58b08c%3a8e72a662%3aacd642d2%3a596d588c%3a77c58d20%3a14b56adc%3a269a19ae

现在我创建了此正则表达式:

Right now I created this regex:

%3b.*" 

符合的

%3b%3bAjaxControlToolkit%2c+Version%3d1.0.11119.38311%2c+Culture%3dneutral%2c+PublicKeyToken%3d28f01b0e84b6d53e%3apt-BR%3adf9c6e46-ef8c-4a3d-89af-f80adf22e9c2%3a865923e8%3a411fea1c%3ae7c87f07%3a91bd373d%3a1d58b08c%3a8e72a662%3aacd642d2%3a596d588c%3a77c58d20%3a14b56adc%3a269a19ae" 

但是我不希望最后两个字符(一个空格)和".
如何删除最后两个字符?

But I don't want the last two chars (one white space) and the ".
How to remove this last two chars?

推荐答案

使用正则表达式模式

%[^"]+

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