编写此Bash脚本的更好方法？

Question

This solution (listed below) works, but it doesn't look like the most practical of solutions. What I want to do is populate a folder with children with a name t_###_u and then fill those child directories with two files (one of which should be renamed to parent directory name and appended with a .c) Can anyone post a better, more elegant solution?

for i in {1..100}
do
    mkdir `printf "t_%03d_u" $i`;
    cp ./templates/* `printf "t_%03d_u" $i`;
    mv `printf "./t_%03d_u/template.c" $i` `printf "./t_%03d_u/t_%03d.c" $i $i`;
done

echo "Did it work?"

Thanks for any help in advance.

Answer 1

The most obvious way to improve the script is to assign a variable to the name you use:

name=`printf "t_%03d" $i`

so then you can do

mkdir ${name}_u
cp ./templates/* ${name}_u/
mv ./${name}_u/template.c ./${name}_u/$name.c

Answer 2

Bash 4.

for num in {001..100}
do        
    path="t_${num}_u"
    mkdir "$path"
    cp ./templates/* "$path";
    mv $path/template.c  ./$path/t_${num}.c
done

Answer 3

seq -w is hugely useful :)

#!/bin/bash

for i in `seq -w 1 100`
do
    mkdir t_${i}_u
    cp ./templates/* t_${i}_u
    mv ./t_${i}_u/template.c ./t_${i}_u/t_${i}_u.c
done

echo "Did it work?"

I'm sure it could be improved a bit further with computing the name of the directory once and re-using it, but this is nicer-enough that I'm fine stopping here. :)

Answer 4

My only advice would be to set the zero padded string to a variable to ease reading:

for i in {1..100}
do
  s=$(printf "%03d" $i)
  mkdir t_${s}_u
  cp ./templates/* t_${s}_u
  mv ./t_${s}_u/template.c ./t_${s}_u/t_${s}.c
done

Answer 5

for i in $(seq -w 1 100); do
 mkdir t_${i}_u
 cp templates/* t_${i}_u
 mv t_${i}_u/template.c t_${i}_u/t_${i}.c
done

haven't tested it.

Answer 6

You could save the result of the printf in a variable rather that re-doing it each time. Or just use seq -w 1 100 to generate the numbers:

for i in $(seq -w 1 100)
do
    d=t_${i}_u
    mkdir $d
    cp templates/* $d
    mv $d/template.c $d/t_$i.c
done