bash中的printf:"09"；和"08"表示是无效数字，"07"表示无效和"06"很好 [英] printf in bash: "09" and "08" are invalid numbers, "07" and "06" are fine

问题描述

这是我的bash脚本-我只想将一组带有零的数字左移:

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

什么...?

仅09和08引起了问题:我序列中的其他所有数字似乎都还可以.

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

推荐答案

如果将"09"放在变量中，则可以

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

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

为什么有帮助?好吧，以0开头但在第二位没有x的数字​​文字被解释为八进制值.

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

八进制值中只有数字0 .. 7，8和9是未知的.

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

"${a#0}"剥离一个前导0.所得的值可以输入到printf，然后以4位为前缀，适当地打印出带有0的值.

"${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.

如果必须期望得到诸如"009"之类的值，事情将变得更加复杂，因为您必须使用一个循环来消除开始时所有多余的0，或者使用extglob表达式作为在评论中提到.

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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