但使用printf在bash脚本离奇的问题：＆QUOT; 09＆QUOT;和＆QUOT; 08＆QUOT;是无效的数字，＆QUOT; 07＆QUOT;和＆QUOT; 06＆QUOT;很好 [英] Bizarre issue with printf in bash script:"09" and "08" are invalid numbers, "07" and "06" are fine

问题描述

这是我的bash脚本 - 我只是想离开垫用零一组数字：

This is my bash script - I just want to left-pad a set of numbers with zeroes:

printf "%04d" "09"
printf "%04d" "08"
printf "%04d" "07"
printf "%04d" "06"

输出：

./rename.sh: line 3: printf: 09: invalid number 
0000
./rename.sh: line 4: printf: 08: invalid number 
0000 
0007
0006

......怎么办？

What...?

09只和08是导致该问题：在我的顺序每隔数似乎是确定。

Only 09 and 08 are causing the problem: every other number in my sequence seems to be OK.

推荐答案

如果你有你的09在一个变量，你可以做

If you have your "09" in a variable, you can do

a="09"
echo "$a"
echo "${a#0}"
printf "%04d" "${a#0}"

为什么此帮助？好了，一些字面出发与 0 但有没有 X 在第2位是国际preTED为八进制值。

Why does this help? Well, a number literal starting with 0 but having no x at the 2nd place is interpreted as octal value.

八进制值只有数字 0 .. 7 ， 8 和 9 是未知的。

Octal value only have the digits 0..7, 8 and 9 are unknown.

$ {A＃0}去掉一条通往 0 。结果值可给的printf 然后，它适当地打印了，以 0 prefixed，在4数字。

"${a#0}" strips one leading 0. The resulting value can be fed to printf then, which prints it appropriately, with 0 prefixed, in 4 digits.

如果您有期待，你得到的值，如009，事情就因为你将不得不使用一个循环，消除所有多余的<更为复杂code> 0 ，或在评论中提及的 extglob 前pression。

If you have to expect that you get values such as "009", things get more complicated as you'll have to use a loop which eliminates all excess 0s at the start, or an extglob expression as mentioned in the comments.

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