如果我在Bash中有一个这样的数组:
FOO=( a b c )
如何用逗号连接元素?例如,生成a b c。
如果我在Bash中有一个这样的数组:
FOO=( a b c )
如何用逗号连接元素?例如,生成a b c。
当前回答
现在我用的是:
TO_IGNORE=(
E201 # Whitespace after '('
E301 # Expected N blank lines, found M
E303 # Too many blank lines (pep8 gets confused by comments)
)
ARGS="--ignore `echo ${TO_IGNORE[@]} | tr ' ' ','`"
这是可行的,但是(在一般情况下)如果数组元素中有空格,将会严重破坏。
(对于那些感兴趣的人,这是一个围绕pep8.py的包装器脚本)
其他回答
感谢@gniourf_gniourf对我迄今为止的最佳世界组合的详细评论。很抱歉发布的代码没有完全设计和测试。这是一个更好的尝试。
# join with separator
join_ws() { local d=$1 s=$2; shift 2 && printf %s "$s${@/#/$d}"; }
这种概念上的美是
(still) 100% pure bash ( thanks for explicitly pointing out that printf is a builtin as well. I wasn't aware about this before ... ) works with multi-character delimiters more compact and more complete and this time carefully thought over and long-term stress-tested with random substrings from shell scripts amongst others, covering use of shell special characters or control characters or no characters in both separator and / or parameters, and edge cases, and corner cases and other quibbles like no arguments at all. That doesn't guarantee there is no more bug, but it will be a little harder challenge to find one. BTW, even the currently top voted answers and related suffer from such things like that -e bug ...
附加的例子:
$ join_ws '' a b c
abc
$ join_ws ':' {1,7}{A..C}
1A:1B:1C:7A:7B:7C
$ join_ws -e -e
-e
$ join_ws $'\033[F' $'\n\n\n' 1. 2. 3. $'\n\n\n\n'
3.
2.
1.
$ join_ws $
$
awk -v sep=. 'BEGIN{ORS=OFS="";for(i=1;i<ARGC;i++){print ARGV[i],ARGC-i-1?sep:""}}' "${arr[@]}"
or
$ a=(1 "a b" 3)
$ b=$(IFS=, ; echo "${a[*]}")
$ echo $b
1,a b,3
顶部答案的简短版本:
joinStrings() { local a=("${@:3}"); printf "%s" "$2${a[@]/#/$1}"; }
用法:
joinStrings "$myDelimiter" "${myArray[@]}"
liststr=""
for item in list
do
liststr=$item,$liststr
done
LEN=`expr length $liststr`
LEN=`expr $LEN - 1`
liststr=${liststr:0:$LEN}
这也可以处理结尾多余的逗号。我不是bash专家。只是我的2c,因为这更基本,更容易理解
也许,例如,
SAVE_IFS="$IFS"
IFS=","
FOOJOIN="${FOO[*]}"
IFS="$SAVE_IFS"
echo "$FOOJOIN"