Question

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"

产出:

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

什么?

只有09和08个国家造成这一问题:我顺序上的所有其他数字似乎都是科索沃。

Answer 1

如果您有<条码>09”,你可以做些什么。

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

为什么这样做? 确切地说,从0开始,但第2个地方没有x的编号被解释为中值。

编码<0。

http://www.un.org/Depts/DGACM/index_french.htm 由此得出的数值可加到f>,然后以4位数表示,以适当印制0。

如果您必须期望您获得诸如009”等值,则随着您在开始时不得不使用一种可消除<0<0<>0的所有超文本的通道,或如评论意见所述,extglob的表述,情况就变得更加复杂。

Answer 2

Bash s numeric arithmetic evaluation syntax (( ... )) can convert to base 10 (therefor ensuring correct interpretation) with the following syntax: (( 10#$var )). Or, in the case of a raw number: (( 10#08 )). Very simple & clean and can be used anywhere you re sure the base should be 10, but can t guarantee a leading zero won t be included.

So, in your example it would be as follows:

printf "%04d
" $(( 10#09 ))
printf "%04d
" $(( 10#08 ))
printf "%04d
" $(( 10#07 ))
printf "%04d
" $(( 10#06 ))

编制以下产出:

由于你接着以变数而不是变数本身的价值重新工作,因此,升温器(((var++ )) & decrementors (((var- ))赢得了t work,但仍可以相对清洁地作为和。

我首先碰到了这一解决办法:http://ubuntuforums.org/showthread.php?t=1402291#post8805742” rel=“noretinger”>here,但。

Answer 3

从“0”开始的编号作为内分数(即基数-8)。因此,“8”和“9”都是有效数字。

见http://www.gnu.org/software/bash/manual/bashref.html#Shell-Arithmetic>http://www.gnu.org/software/bash/manual/bashref.html#Shell-Arithmetic。

这种行为从C类语言继承。

Answer 4

处理点不同:

printf "%04.f" "009"

这样做可以得出正确的产出,而没有处理任何原始现金问题(每回答@Oli Charlesworth, 0***被作为中间值处理,但我认为Bash忽略了浮动点数字的中值/高值识别器)。

Answer 5

仅添加到Oli swer,以便在<>%<<<>/code>之后添加一个编号为零的编号。

printf "%04d" "9"

Answer 6

Why did “09” 以及 “08” are invalid numbers, “07” 以及 “06” are fine

由于之前按<代码>0进行分类,是一种常规的简短含义: 序号为。

printf "%d " 010 0100 8 64 printf "%o " 8 64 0100 10 100 100 printf "%04o " 8 64 0100 0010 0100 0100

How to work with this?

Using bash integer: preceding number by 10#

Under bash, you could precise by this way, which base is used for number

echo $(( 10#0100)) 100 echo $(( 10#0900)) 900 echo $(( 10#0900 )) 900

以及

echo $(( 8#100 )) 64 echo $(( 2#100 )) 4

当然!

There are only 10 types of people in the world: those who underst以及 binary, 以及 those who don t.

Using bash Parameter Expansion

为此,您必须使用一个变量。

a=09 echo ${a#0} 9

This will work fine until you just have only one 0 to drop.

a=0000090 echo ${a#*0} 000090 echo ${a##0} 000090

http://www.un.org/Depts/DGACM/index_french.htm

echo ${a##*(0)} 0000090 shopt -s extglob echo ${a##*(0)} 90

Using floating point with printf

printf "%.0f " 09 9

我愿使用<代码>-v 备选办法> printf,用于确定一些变量:

printf -v a %.0f 09 echo $a 9

or even if

a=00090 printf -v a %.0f $a echo $a 90

Answer 7

Adding * makes shell parameter expansion matching greedy (see, for example, Shell Tipps: use internal string handling)!

# strip leading 0s
- a="009"; echo ${a##0}
+ a="009"; echo ${a##*0}

Answer 8

if a=079 first, remove the trailing zeros by

a=`echo $a | sed  s/^0*// `

接着是零婚。

printf "%04d" $a

Answer 9

更简单地说,从职业转为男性:

nm=$((10#$nm))

Why did “09” 以及 “08” are invalid numbers, “07” 以及 “06” are fine

How to work with this?

Using bash integer: preceding number by `10#`

Using bash Parameter Expansion

Using `floating point` with `printf`

友情链接

Why did “09” 以及 “08” are invalid numbers, “07” 以及 “06” are fine

How to work with this?

Using bash integer: preceding number by 10#

Using bash Parameter Expansion

Using floating point with printf

友情链接

Using bash integer: preceding number by `10#`

Using `floating point` with `printf`