如何拆分由tab分隔的bash中的字符串

Question

我正在尝试在bash中拆分制表符分隔字段.

我知道这个答案:如何在shell中拆分字符串并获取最后一个字段

但这并不能解决制表符.

我想在tab字符之前获取字符串的一部分,所以我这样做:

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

但是\ t匹配字母't'而不是标签上.做这个的最好方式是什么？

谢谢

Answer 1

如果您的文件看起来像这样(使用制表符作为分隔符):

1st-field   2nd-field

您可以使用cut提取第一个字段(默认情况下在选项卡上操作):

$ cut -f1 input
1st-field

如果您正在使用awk,则无需使用tail获取最后一行,将输入更改为:

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

使用awk的解决方案:

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

纯粹的bash解决方案:

#!/bin/bash

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

输出:

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

最后,使用的解决方案 sed

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

这里$是范围运营商; 即仅在最后一行操作.

对于您的原始问题,请使用文字选项卡,即

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

输出:

1st-field

Answer 2

在参数扩展中使用$'ANSI-C'字符串:

$ x=$'abc\tdef\tghi'
$ echo "$s"
abc     def     ghi
$ echo ">>${x%%$'\t'*}<<"
>>abc<<

Answer 3

使用awk.

echo $yourfield | awk '{print $1}'

或者,对于您的情况,来自文件最后一行的第一个字段

tail yourfile | awk '{x=$1}END{print x}'

Answer 4

read field1 field2 <<< ${tabDelimitedField}

要么

read field1 field2 <<< $(command_producing_tab_delimited_output)