bash 如何将 printf 语句的结果分配给变量（用于前导零）？
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How to assign results from printf statement to a variable (for leading zeroes)?
提问by Tom S?derlund
I'm writing a shell script (Bash on Mac OS X) to rename a bunch of image files. I want the results to be:
我正在编写一个 shell 脚本（Mac OS X 上的 Bash）来重命名一堆图像文件。我希望结果是：
frame_001
frame_002
frame_003
etc.
等等。
Here is my code:
这是我的代码：
let framenr=$[1 + (y * cols * resolutions) + (x * resolutions) + res]
echo $framenr:
let framename=$(printf 'frame_%03d' $framenr)
echo $framename
$framenrlooks correct, but $framenamealways becomes 0. Why?
$framenr看起来正确，但$framename总是变成0. 为什么？
回答by glenn Hymanman
The letcommand forces arithmetic evaluation, and the referenced "variable" does not exist, so you get the default value 0.
该let命令强制算术评估，并且引用的“变量”不存在，因此您获得默认值 0。
y=5
x=y; echo $x # prints: y
let x=y; echo $x # prints: 5
Do this instead:
改为这样做：
framenr=$(( 1 + (y * cols * resolutions) + (x * resolutions) + res ))
echo $framenr:
# if your bash version is recent enough
printf -v framename 'frame_%03d' $framenr
# otherwise
framename=$(printf 'frame_%03d' $framenr)
echo $framename
I recall reading somewhere that $[ ]is deprecated. Use $(( ))instead.
我记得在某个$[ ]已弃用的地方阅读过。使用$(( ))来代替。
