英文:
scripting bash for loop awk output in argument
问题
I'm just going to provide the translated code part as requested:
我正在编写一些Bash练习，特别是要检查所有以file-a(数字)命名的文件并将其移动到名为dirA的目录。但是for循环存在一些问题。
我可以在没有for循环的情况下使用以下简单的方式来完成：
mv $(ls -la | awk '$9 ~ /-a[0-9]+/ {print $9}') dirA
但我想使用for循环来完成，但我无法做到。以下是脚本：
FILES=$(ls -la | awk '$9 ~ /-a[0-9]+/ {print $9}')
for arg in "$FILES"
do
mv $arg dirA
done
Please note that I've made some corrections to the code to address the issues in the original script.
英文:
im writing some bash exercises, this in particular wants to check all the files named file-a(sonenumber) and move it to directory named dirA
But there's some problem with for loop
i can do that without for loop just simply writing
mv $(ls -la | awk '$9 - /-a[0-9]+/ {print $9}') dirA
but i wanna do it with a for and i cant do that
there's the script
FILES=(ls -la | awk '$9 - /-a[0-9]+/ {print $9}')
for arg in "$FILES"
do
mv $arg dirA
done
答案1
得分: 3
不需要使用循环，只需使用 shell glob：
mv -- *-a[0-9]* dirA
如果坚持要使用循环：
files=( *-a[0-9]* )
for f in "${files[@]}"; do
mv "$f" dirA/
done
或者
for f in *-a[0-9]*; do
mv "$f" dirA/
done
但请避免解析 ls 输出。
ls 是一个用于交互式查看目录元数据的工具。任何尝试使用代码解析 ls 输出的尝试都是错误的。使用通配符更简单且正确：for file in *.txt。阅读 http://mywiki.wooledge.org/ParsingLs
不要使用大写变量名，参考 Don't use UPPER case variables。
英文:
No need a for loop, just a shell glob:
mv -- *-a[0-9]* dirA
If you insist for a loop:
files=( *-a[0-9]* )
for f in "${files[@]}"; do
mv "$f" dirA/
done
or
for f in *-a[0-9]*; do
mv "$f" dirA/
done
But please, avoid parsing ls output.
ls is a tool for interactively looking at directory metadata. Any attempts at parsing ls output with code are broken. Globs are much more simple AND correct: for file in *.txt. Read http://mywiki.wooledge.org/ParsingLs
答案2
得分: 1
You're not setting FILES to the output of the command. For that, you need to use $(command). To make it an array, wrap another set of () around that.
然后，要循环遍历数组内容，请使用 "${FILES[@]}"。
I'm not sure why you're using ls -la when you could just use ls -a. Then each filename will be a single line, you don't have to extract $9.
然而，最好不要解析ls的输出。
英文:
You're not setting FILES to the output of the command. For that, you need to use $(command). To make it an array, wrap another set of () around that.
Then to loop over the array contents, use "${FILES[@]}.
I'm not sure why you're using ls -la when you could just use ls -a. Then each filename will be a single line, you don't have to extract $9.
FILES=($(ls -a | awk '/-a[0-9]+/'))
for arg in "${FILES[@]}"
do
mv "$arg" dirA
done
However, it's best not to parse the output of ls.
答案3
得分: -6
我修改了你的代码以在for循环中运行。
FILES=$(ls -la | awk '/-a[0-9]+/ {print $9}')
for arg in $FILES
do
mv $arg dirA
done
英文:
I modified your code to work in the for loop.
FILES=$(ls -la | awk '/-a[0-9]+/ {print $9}')
for arg in $FILES
do
mv $arg dirA
done
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