[英]Partial File Name as Shell Script Input
I am trying to create a shell script that takes a partial file name as its input and then does operations on the files with the matching names. 我正在尝试创建一个将部分文件名作为输入的shell脚本,然后对具有匹配名称的文件进行操作。 For example, I have three files
sample1.txt,sample2.txt and sample3.txt
and this is my script 例如,我有三个文件
sample1.txt,sample2.txt and sample3.txt
,这是我的脚本
#!bin/bash
VALUE=$1
VALUE2=$2
FILE_NAME=$3
echo $FILE_NAME
And I run it with this command 我用这个命令运行它
sh myscript.sh arg1 arg2 sample*
But I get this as the output 但是我得到这个作为输出
sample3.txt (sample2.txt)
But what I want is 但是我想要的是
sample1.txt
sample2.txt
sample3.txt
How could I do this? 我该怎么办?
sample*
get's expanded into (if the files exists, no more files with that name, etc): get扩展为(如果文件存在,则不再具有该名称的文件,等等):
sample1.txt sample2.txt sample3.txt
So when you write: 所以当你写:
sh myscript.sh arg1 arg2 sample*
What you really write, what your script sees is: 您实际编写的脚本看到的是:
sh myscript.sh arg1 arg2 sample1.txt sample2.txt sample3.txt
Your script get's 5 arguments, not 3 您的脚本得到5个参数,而不是3个
Then you can: 那么你也能:
#!bin/bash
VALUE=$1
VALUE2=$2
# shift the arguments to jump over the first two
shift 2
# print all the rest of arguments
echo "$@"
# ex. read the arguments into an array
FILE_NAME=("$@")
echo "${FILE_NAME[@]}"
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