使用Shell从字符串中间删除第n个字符 [英] Remove nth character from middle of string using Shell

问题描述

我一直在搜索google，但找不到如何执行此操作的示例.我也不了解如何为SED构造正则表达式的概念，因此我希望有人可以向我解释一下.

I've been searching google for ever, and I cannot find an example of how to do this. I also do not grasp the concept of how to construct a regular expression for SED, so I was hoping someone could explain this to me.

我正在对包含以下文本行的文件运行bash脚本:2222，H，73.82,04,07,2012

I'm running a bash script against a file full of lines of text that look like this: 2222,H,73.82,04,07,2012

我需要使它们看起来都像这样:2222，H，73.82,04072012

and I need to make them all look like this: 2222,H,73.82,04072012

我需要删除最后两个逗号，即该行中的第16和19个字符.有人可以告诉我该怎么做吗?我本来要使用colrm，这很简单，但是我似乎无法在CYGWIN中安装它.请谢谢！

I need to remove the last two commas, which are the 16th and 19th characters in the line. Can someone tell me how to do that? I was going to use colrm, which is blessedly simple, but i can't seem to get that installed in CYGWIN. Please and thank you!

推荐答案

我将使用 awk :

awk -F',' -v OFS=',' '{ print $1, $2, $3, $4$5$6 }' inputfile

这将获取一个CSV文件，并打印第一个，第二个和第三个字段，每个字段后跟输出字段分隔符(，" )，然后将第四个，第五个和第六个字段串联起来.

This takes a CSV file and prints the first, second and third fields, each followed by the output field separator (",") and then the fourth, fifth and sixth fields concatenated.

我个人认为，这比 sed 中基于正则表达式的解决方案更易于阅读和维护，并且如果您的任何列变宽(变窄)，它将很好地应对.).

Personally I find this easier to read and maintain than regular expression-based solutions in sed and it will cope well if any of your columns get wider (or narrower!).

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