在制表符分隔的文件linux中的封闭字符串中替换制表符 [英] replace tab in an enclosed string in a tab delimited file linux

问题描述

我有一个制表符分隔的txt文件，其中第三列包含附带的字符串，该字符串也可能具有制表符.由于有这个额外的标签，当我尝试读取此标签定界文件时，我得到了5列.所以我想用空格替换选项卡.

I have a tab delimited txt file in which third column contains enclosed string that might also has a tab. Because of this extra tab i am getting 5 columns when i try to read this tab delimited file. So i want to replace the tab with space.

以下是示例文件.

col1   col2   col3        col4  
1      abc    "pqr   xyz" asd  
2      asd    "lmn   pqr" aws  
3      abc    "asd"       lmn

我想要这样的输出

col1   col2   col3        col4  
1      abc    "pqr xyz"   asd  
2      asd    "lmn pqr"   aws  
3      abc    "asd"       lmn

这是我尝试过的

awk -F"\t" '{ gsub("\t","",$3); print $3 }' file.txt

在那之后我得到以下输出

after that i am getting following output

col3  
"pqr  
"lmn  
"asd"

请帮助

推荐答案

具有GNU awk(gawk)，您可以使用以下表达式:

Having GNU awk (gawk) you can use the following expression:

gawk '{gsub("\t"," ",$3)}1' OFS='\t' FPAT='"[^"]*"|[^\t]*' file

此处的键是FPAT变量.它定义了字段的外观，而不仅仅是指定字段定界符.

The key here is the FPAT variable. It defines how a field can look like instead of just specifying the field delimiter.

在我们的情况下，字段可以是用双引号"[^"]*"括起来的非双引号字符序列，或者是零个或多个非制表符[^\t]*的序列. (零，以正确处理空字段)

In our case a field can either be an sequence of non-double-quote chars enclosed in double quotes "[^"]*" or a sequence of zero or more non tab characters [^\t]*. (zero, to handle empty fields properly)

由于我们首先指定了非引号字符的序列，所以它具有优先级.

Since we are specifying the sequence of non quote characters first it has a precedence.

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