如何在bash中提取子字符串?

I have the following string in bash with length > 4

我在bash中有以下字符串，长度为>4

str = "abcdefghijklmno"

and I want to extract into str2 the 5 first charachter of str. So

我想提取str2, str的第5个特征

str2="abcde"

How to do it with bash?

如何使用bash?

1 个解决方案

#1

Do use the expression

使用表达式

{string:position:length}

So in this case:

所以在这种情况下:

$ str="abcdefghijklm"
$ echo "${str:0:5}"
abcde

See other usages:

看到其他的用法:

$ echo "${str:0}"      # default: start from the 0th position
abcdefghijklm
$ echo "${str:1:5}"    # start from the 1th and get 5 characters
bcdef
$ echo "${str:10:1}"   # start from 10th just one character
k
$ echo "${str:5}"      # start from 5th until the end
fghijklm

Taken from:
- wooledge.org - How can I use parameter expansion? How can I get substrings? How can I get a file without its extension, or get just a file's extension?
- Shell Command Language - 2.6.2 Parameter Expansion

引用自:- wooledge.org -我如何使用参数扩展?如何得到子字符串?如何获得一个没有扩展名的文件，或者仅仅获得一个文件的扩展名?- Shell命令语言- 2.6.2参数扩展

#1