Linux 在 bash 脚本中连接输入
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concatenate inputs in bash script
提问by zpesk
I would like to concatenate all the arguments passed to my bash script except the flag.
我想连接传递给我的 bash 脚本的所有参数，除了标志。
So for example, If the script takes inputs as follows:
例如，如果脚本采用如下输入：
./myBashScript.sh -flag1 exampleString1 exampleString2
I want the result to be "exampleString1_exampleString2"
我希望结果是“exampleString1_exampleString2”
I can do this for a predefined number of inputs (i.e. 2), but how can i do it for an arbitrary number of inputs?
我可以为预定义数量的输入（即 2）执行此操作，但是如何为任意数量的输入执行此操作？
回答by mvds
This is an ugly but simple solution:
这是一个丑陋但简单的解决方案：
echo $* | sed -e "s/ /_/g;s/[^_]*_//"
回答by Tyilo
function concatenate_args
{
string=""
for a in "$@" # Loop over arguments
do
if [[ "${a:0:1}" != "-" ]] # Ignore flags (first character is -)
then
if [[ "$string" != "" ]]
then
string+="_" # Delimeter
fi
string+="$a"
fi
done
echo "$string"
}
# Usage:
args="$(concatenate_args "$@")"
回答by Jonathan Leffler
flag=""
shift
oldIFS="$IFS"
IFS="_"
the_rest="$*"
IFS="$oldIFS"
In this context, "$*"is exactly what you're looking for, it seems. It is seldom the correct choice, but here's a case where it really is the correct choice.
在这种情况下"$*"，这似乎正是您要寻找的。这很少是正确的选择，但这里有一个案例，它确实是正确的选择。
Alternatively, simply loop and concatenate:
或者，只需循环并连接：
flag=""
shift
the_rest=""
pad=""
for arg in "$@"
do
the_rest="${the_rest}${pad}${arg}"
pad="_"
done
The $padvariable ensures that you don't end up with a stray underscore at the start of $the_rest.
该$pad变量确保您不会在$the_rest.
回答by Jo So
Here's a piece of code that I'm actually proud of (it is very shell-style I think)
这是一段我引以为豪的代码（我认为它是非常 shell 风格的）
#!/bin/sh
firsttime=yes
for i in "$@"
do
test "$firsttime" && set -- && unset firsttime
test "${i%%-*}" && set -- "$@" "$i"
done
IFS=_ ; echo "$*"
I've interpreted your question so as to remove allarguments beginning with -
我已经解释了你的问题，以删除所有以-
If you only want to remove the beginning sequence of arguments beginnnig with -:
如果您只想删除参数 beginnnig 的开头序列-：
#!/bin/sh
while ! test "${1%%-*}"
do
shift
done
IFS=_ ; echo "$*"
If you simply want to remove the first argument:
如果您只想删除第一个参数：
#!/bin/sh
shift
IFS=_ ; printf %s\n "$*"
回答by user unknown
#!/bin/bash
paramCat () {
for s in "$@"
do
case $s in
-*)
;;
*)
echo -n _${s}
;;
esac
done
}
catted="$(paramCat "$@")"
echo ${catted/_/}
回答by nickl-
You can also use formatted strings to concatenate args.
您还可以使用格式化字符串来连接 args。
# assuming flag is first arg and optional
flag=
[[ = ${1#-} ]] && unset $flag || shift
concat=$(printf '%s_' ${@})
echo ${concat%_} # to remove the trailing _
nJoy!
快乐！
