如何拆分在bash字符串由制表符分隔 [英] How to split a string in bash delimited by tab

问题描述

我想在bash拆分标签delimitted场。

I'm trying to split a tab delimitted field in bash.

我知道这个答案：<一href=\"http://stackoverflow.com/questions/3162385/how-to-split-a-string-in-shell-and-get-the-last-field\">how拆壳串并获得最后一个字段

但是，这并不回答制表符。

But that does not answer for a tab character.

我想做获得制表符前的字符串的一部分，所以我这样做：

I want to do get the part of a string before the tab character, so I'm doing this:

x=`head -1 my-file.txt`
echo ${x%\t*}

但\\ t被上的字母T，而不是一个选项卡上的匹配。什么是做到这一点的最好方法是什么？

But the \t is matching on the letter 't' and not on a tab. What is the best way to do this?

感谢

推荐答案

如果你的文件看起来像这样（有标签作为分隔符）：

If your file look something like this (with tab as separator):

1st-field   2nd-field

您可以使用剪切来提取的第一个字段（默认选项卡上的操作）：

you can use cut to extract the first field (operates on tab by default):

$ cut -f1 input
1st-field

如果你使用 AWK ，也没有必要使用尾来获得最后一行，改变的输入：

If you're using awk, there is no need to use tail to get the last line, changing the input to:

1:1st-field     2nd-field
2:1st-field     2nd-field
3:1st-field     2nd-field
4:1st-field     2nd-field
5:1st-field     2nd-field
6:1st-field     2nd-field
7:1st-field     2nd-field
8:1st-field     2nd-field
9:1st-field     2nd-field
10:1st-field    2nd-field

解决方案用awk：

Solution using awk:

$ awk 'END {print $1}' input
10:1st-field

纯bash的解决方案：

Pure bash-solution:

#!/bin/bash

while read a b;do last=$a; done < input
echo $last

输出：

$ ./tab.sh 
10:1st-field

最后，利用溶液 SED

$ sed '$s/\(^[^\t]*\).*$/\1/' input
10:1st-field

在这里， $ 是范围操作;即，在最后一行仅操作

here, $ is the range operator; i.e. operate on the last line only.

有关你原来的问题，用文字标签，即

For your original question, use a literal tab, i.e.

x="1st-field    2nd-field"
echo ${x%   *}

输出：

1st-field

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