在 bash 中，如何打印列表的前 n 个元素？
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In bash, how can I print the first n elements of a list?
提问by Frank
In bash, how can I print the first nelements of a list?
在 中bash，如何打印n列表的第一个元素？
For example, the first 10 files in this list:
例如，此列表中的前 10 个文件：
FILES=$(ls)
UPDATE:I forgot to say that I want to print the elements on one line, just like when you print the whole list with echo $FILES.
更新：我忘了说我想在一行上打印元素，就像你用echo $FILES.
回答by Ayman Hourieh
FILES=(*)
echo "${FILES[@]:0:10}"
Should work correctly even if there are spaces in filenames.
即使文件名中有空格也应该正常工作。
FILES=$(ls)creates a string variable. FILES=(*)creates an array. See this page for more examples on using arrays in bash. (thanks lhunath)
FILES=$(ls)创建一个字符串变量。FILES=(*)创建一个数组。有关在 bash 中使用数组的更多示例，请参阅此页面。（感谢 lhunath）
回答by Niglas
FILE="$(ls | head -1)"
Handled spaces in filenames correctly too when I tried it.
当我尝试时，也正确处理了文件名中的空格。
回答by Pat
Why not just this to print the first 50 files:
为什么不只是这样打印前 50 个文件：
ls -1 | head -50
回答by pik3y
My way would be:
我的方法是：
ls | head -10 | tr "\n" " "
This will print the first 10 lines returned by ls, and then tr replaces all line breaks with spaces. Output will be on a single line.
这将打印 ls 返回的前 10 行，然后 tr 用空格替换所有换行符。输出将在一行上。
回答by FeatureCreep
echo $FILES | awk '{for (i = 1; i <= 10; i++) {print $i}}'
Edit: AAh, missed your comment that you needed them on one line...
编辑：啊，错过了你在一行上需要它们的评论......
echo $FILES | awk '{for (i = 1; i <= 10; i++) {printf "%s ", $i}}'
That one does that.
那个就是这样做的。
回答by Idelic
to do it interactively:
以交互方式执行此操作：
set $FILES && eval eval echo \\\${1..10}
set $FILES && eval eval echo \\\${1..10}
to run it as a script, create foo.sh with contents
要将其作为脚本运行，请使用内容创建 foo.sh
N=$1; shift; eval eval echo \\\${1..$N}
N=$1; shift; eval eval echo \\\${1..$N}
and run it as
并运行它
bash foo.sh 10 $FILES
bash foo.sh 10 $FILES
回答by sunny256
FILES=$(ls)
echo $FILES | fmt -1 | head -10
回答by Marco
An addition to the answer of "Ayman Hourieh" and "Shawn Chin", in case it is needed for something else than content of a directory.
“Ayman Hourieh”和“Shawn Chin”的答案的补充，以防目录内容以外的其他内容需要它。
In newer version of bash you can use mapfile to store the directory in an array. See help mapfile
在较新版本的 bash 中，您可以使用 mapfile 将目录存储在数组中。看help mapfile
mapfile -t files_in_dir < <( ls )
If you want it completely in bash use printf "%s\n" *instead of ls, or just replace lswith any other command you need.
如果您希望它完全在 bash 中使用printf "%s\n" *而不是ls，或者只是替换ls为您需要的任何其他命令。
Now you can access the array as usual and get the data you need.
现在您可以像往常一样访问数组并获取所需的数据。
First element:
第一个元素：
${files_in_dir[0]}
Last element (do not forget space after ":" ):
最后一个元素（不要忘记 ":" 后的空格）：
${files_in_dir[@]: -1}
Range e.g. from 10 to 20:
范围例如从 10 到 20：
${files_in_dir[@]:10:20}
Attention for large directories, this is way more memory consuming than the other solutions.
注意大目录，这比其他解决方案消耗更多内存。
