bash 如何在shell脚本中的模式之前提取字符串
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/24926606/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to extract string before pattern in shell script
提问by Kumaran
I have a variable which has contents like this
我有一个变量，它的内容是这样的
line 1
line 2
line 3
======
line 4
line 5
line 6
======
.........
line n
======
第 1
行 2
行 3
======
第 4
行 5
第 6 行
======
..........
第 n 行
======
Now i want to extract lines before all occurence of '======' pattern and i want to assign it to each variable.
Expected output
Var1=line 1
line 2
line 3
Var2=line 4
line 5
line 6
...........
Var_n=line n
Storing it in an array is also accepted
How to do it?
现在我想在所有 '======' 模式出现之前提取行，我想将它分配给每个变量。
预期输出
Var1=line 1
line 2
line 3
Var2=line 4
line 5
line 6
.....
Var_n=line n
也接受数组存储
怎么做？
回答by Jayesh Bhoi
You can achieve using parameter expansion as like
您可以像这样使用参数扩展来实现
str="mystring:yourstring" //here assume ":" is pattern
var=${str%:*}
echo $var
For your information you can use #instead %to get string after pattern
为了您的信息，您可以使用#，而不是%模式之后得到的字符串
var=${str#*:}
回答by David C. Rankin
#!/bin/bash
# original string
str="some lines1 ====== some lines2 ======"
# parse the string
v1="${str%% =*}" # beginning at right, remove all to ' ='
tmp="${str% =*}" # beginning at right, remove first occurrence of ' =*'
v2="${tmp##*= }" # beginning at left, remove all to '= '
# print the string and variable
echo "str: $str"
echo " v1: $v1"
echo " v2: $v2"
exit 0
output:
输出：
$ bash somelines.sh
str: some lines1 ====== some lines2 ======
v1: some lines1
v2: some lines2
