我有一个文本文件,其中包含大约150到200个文件名的列表
abc.txt
pqr.txt
xyz.txt
...
...
我需要一串逗号分隔的文件。 每个字符串不应超过20个文件。所以回声看起来像这样......
$string1="abc.txt,pqr.txt,xyz.txt..."
$string2="abc1.txt,pqr1.txt,xyz1.txt..."
...
字符串的数量将根据文件中的行数而有所不同。我写过这样的东西......
#!/bin/sh
delim=','
for gsfile in `cat filelist.txt`
do
filelist=$filelist$delim$gsfile
echo $filelist
done
翻译命令正在按预期工作,但如何将每个字符串限制为20个文件名?
cat filelist.txt | tr '\n' ','
答案 0 :(得分:5)
只需使用xargs
:
$ seq 1 50 | xargs -n20 | tr ' ' ,
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20
21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40
41,42,43,44,45,46,47,48,49,50
答案 1 :(得分:0)
这可能对您有用:
seq 41 | paste -sd ',,,,,,,,,,,,,,,,,,,\n'
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20
21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40
41
或GNU sed:
seq 41 | sed ':a;$bb;N;s/\n/&/19;Ta;:b;y/\n/,/'
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20
21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40
41
答案 2 :(得分:0)
使用sed
的一种方式:
将 Igor Chubin 50个数字添加到infile
:
seq 1 50 >infile
script.sed
的内容:
:b
## While not last line...
$! {
## Check if line has 19 newlines. Try substituting the line with itself and
## check if it succeed, then append next line and do it again in a loop.
s/\(\n[^n]*\)\{19\}/&/
ta
N
bb
}
## There are 20 lines in the buffer or found end of file, so substitute all '\n'
## with commas and print.
:a
s/\n/,/g
p
像以下一样运行:
sed -nf script.sed infile
使用以下输出:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20
21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40
41,42,43,44,45,46,47,48,49,50
答案 3 :(得分:0)
在sed的s
命令中使用一个标志,用换行符替换每个第20个逗号:
< filelist.txt tr '\n' , | sed ':a; s/,/\n/20; P; D; ta'; echo