bash 在bash中将多行输出捕获为数组
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Capture multiline output as array in bash
提问by djechlin
If inner.shis
如果inner.sh是
#...
echo first
echo second
echo third
And outer.shis
并且outer.sh是
var=`./inner.sh`
# only wants to use "first"...
How can varbe split by whitespace?
如何var用空格分割？
回答by sampson-chen
Try this:
尝试这个：
var=($(./inner.sh))
# And then test the array with:
echo ${var[0]}
echo ${var[1]}
echo ${var[2]}
Output:
输出：
first
second
third
Explanation:
解释：
- You can make an array in bash by doing
var=(first second third), for example. $(./inner.sh)runs theinner.shscript, which prints outfirst,second, andthirdon separate lines. Since we don't didn't put double quotes around$(...), they get lumped onto the same line, but separated by spaces, so you end up with what it looks like in the previous bullet point.
- 例如，您可以通过执行 bash 在 bash 中创建数组
var=(first second third)。 $(./inner.sh)运行该inner.sh脚本，它打印出来first，second和third在不同的行。由于我们没有在 周围放置双引号$(...)，因此它们会集中在同一行上，但由空格分隔，因此您最终会看到上一个要点中的样子。
回答by gniourf_gniourf
Don't forget the builtin mapfile. It's definitely the most efficient in your case: If you want to slurp the whole file in an array, the fields of which will be the lines output by ./inner.sh, do
不要忘记内置mapfile. 在您的情况下，这绝对是最有效的：如果您想将整个文件放入一个数组中，其字段将是由 输出的行./inner.sh，请执行
mapfile -t array < <(./inner.sh)
Then you'll have the first row in ${array[0]}, etc...
然后你会有第一行${array[0]}，等等......
For more info on this builtin and all the possible options:
有关此内置函数和所有可能选项的更多信息：
help mapfile
If you just need the first line in a variable called firstline, do
如果您只需要名为 的变量中的第一行firstline，请执行
read -r firstline < <(./inner.sh)
These are definitely the most efficient ways!
这些绝对是最有效的方法！
This small benchmark will prove it:
这个小基准将证明这一点：
$ time mapfile -t array < <(for i in {0..100000}; do echo "fdjsakfjds$i"; done)
real 0m1.514s
user 0m1.492s
sys 0m0.560s
$ time array=( $(for i in {0..100000}; do echo "fdjsakfjds$i"; done) )
real 0m2.045s
user 0m1.844s
sys 0m0.308s
If you only want the first word (with space delimiter) of the first line output by ./inner.sh, the most efficient way is
如果你只想要第一行输出的第一个单词（带空格分隔符）./inner.sh，最有效的方法是
read firstword _ < <(./inner.sh)
回答by Arnestig
You're not splitting on whitespaces but rather on newlines. Give this a whirl:
您不是在空格上拆分，而是在换行符上拆分。试一试：
IFS=$'
'
var=$(./echo.sh)
echo $var | awk '{ print }'
unset IFS
