bash shell脚本中的SPRINTF?
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SPRINTF in shell scripting?
提问by Manu R
I have an auto-generated file each day that gets called by a shell script. But, the problem I'm facing is that the auto-generated file has a form of:
我每天都有一个自动生成的文件,由 shell 脚本调用。但是,我面临的问题是自动生成的文件具有以下形式:
FILE_MM_DD.dat
... where MM and DD are 2-digit month and day-of-the-month strings.
... 其中 MM 和 DD 是两位数的月份和月份日期字符串。
I did some research and banged it at on my own, but I don't know how to create these custom strings using only shell scripting.
我进行了一些研究并自己进行了尝试,但我不知道如何仅使用 shell 脚本来创建这些自定义字符串。
To be clear, I am aware of the DATE function in Bash, but what I'm looking for is the equivalent of the SPRINTF function in C.
明确地说,我知道 Bash 中的 DATE 函数,但我正在寻找的是等效于 C 中的 SPRINTF 函数。
回答by Paused until further notice.
In Bash:
在 Bash 中:
var=$(printf 'FILE=_%s_%s.dat' "$val1" "$val2")
or, the equivalent, and closer to sprintf
:
或者,等价的,更接近于sprintf
:
printf -v var 'FILE=_%s_%s.dat' "$val1" "$val2"
If your variables contain decimal values with leading zeros, you can remove the leading zeros:
如果您的变量包含带前导零的十进制值,您可以删除前导零:
val1=008; val2=02
var=$(printf 'FILE=_%d_%d.dat' $((10#$val1)) $((10#$val2)))
or
或者
printf -v var 'FILE=_%d_%d.dat' $((10#$val1)) $((10#$val2))
The $((10#$val1))
coerces the value into base 10 so the %d
in the format specification doesn't think that "08" is an invalid octal value.
将$((10#$val1))
值强制为基数 10,因此%d
格式规范中的“08”不认为是无效的八进制值。
If you're using date
(at least for GNU date
), you can omit the leading zeros like this:
如果您正在使用date
(至少对于 GNU date
),您可以像这样省略前导零:
date '+FILE_%-m_%-d.dat'
For completeness, if you want to addleading zeros, padded to a certain width:
为了完整起见,如果要添加前导零,请填充到特定宽度:
val1=8; val2=2
printf -v var 'FILE=_%04d_%06d.dat' "$val1" "$val2"
or with dynamic widths:
或使用动态宽度:
val1=8; val2=2
width1=4; width2=6
printf -v var 'FILE=_%0*d_%0*d.dat' "$width1" "$val1" "$width2" "$val2"
Adding leading zeros is useful for creating values that sort easily and align neatly in columns.
添加前导零对于创建易于排序并在列中整齐对齐的值非常有用。
回答by Daniel B?hmer
Why not using the printf program from coreutils?
为什么不使用 coreutils 的 printf 程序?
$ printf "FILE_%02d_%02d.dat" 1 2
FILE_01_02.dat
回答by kenorb
Try:
尝试:
sprintf() { local stdin; read -d '' -u 0 stdin; printf "$@" "$stdin"; }
Example:
例子:
$ echo bar | sprintf "foo %s"
foo bar