AWK替换单引号 [英] Awk to replace single quote

问题描述

我想用include('替换一组文件中的所有include('./.我正在尝试如下使用awk:

I want to replace all include('./ in a set of files with include('. I am trying to use awk as follows:

awk '{gsub("include\('"'"'./", "include\('"'"'", $0); print > FILENAME}' *.php

它引发了我这个错误.

It throws me this error.

awk: (FILENAME=xyz.php FNR=1) fatal: Unmatched ( or \(: /include('.//

任何帮助将不胜感激.

推荐答案

@OP，您可以尝试将八进制代码用于单引号(\047)和正斜杠(\057)，例如

@OP, you can try using octal code for the single quote(\047) and forward slash(\057), eg

$ cat file
include('./
$ awk '{gsub(/include\(\047\.\057/ , "include(\047" ) }1' file
include('

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