八次出现后，从一行中删除所有标签 [英] Remove all tabs from a line after the eight occurance

问题描述

我正在尝试创建一个正则表达式，在八次出现后删除所有标签但我无法弄明白。
示例行：
kdasjla <标签> sadkj ... ksjs< tab nr。 8> ksdjlak djsakd askdj< tab> jsdhskj< tab>

因此，选项卡nr 8之后可能存在的所有选项卡都需要用空格替换。其他字符需要保留。

有没有人有正则表达式模式的建议？
谢谢。

解决方案

我没有在这台计算机上安装Visual Studio，所以正则表达式可能有点不对劲。它将创建一个名为"tabs_to_be_replaced"的命名组。其中包含您要替换的标签。

（[^ \\t] * \\t）{8}（[^ \\t] *（？< tabs_to_be_replaced> \\t））*结果

Hi,


I'm trying to create a regular expression that removes all tabs after the eight occurance but I can't figure it out.
example line:
kdasjla <tab> sadkj...ksjs <tab nr. eight> ksdjlak  djsakd askdj <tab> jsdhskj <tab>

So all tabs that might exist after tab nr eight needs to be replaced by spaces. The other characters needs to be preserved.

Has anyone got a suggestion for the regular expression pattern?
Thanks.

解决方案

I don't have Visual Studio installed on this computer so the regex might be a little bit wrong. It will create a named group called "tabs_to_be_replaced" which contains the tabs you want to replace.

([^\\t]*\\t){8}([^\\t]*(?<tabs_to_be_replaced>\\t))*

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