如何从变量中删除前导零？

Question

I am trying to remove all the zeros in front of a variable in bash. But I am not able to save it in the variable. Some help

the problem is here: if (( $line > $lastMin )) ; then if (( $line > $lastMin )) ; then

a Bash error run because value too large for the base (the error element is "0455233")

That's why I'm trying to save that value by removing all the zeros I find in front of me

Any solutions please?

lastMin=0455233
zeroMin=$(( $lastMin | sed 's/^0*//' ))

Answer 1

lastMin='0455233'
zeroMin=$((10#$lastMin))

Answer 2

Removing all leading 0 's;

lastMin='0000455233'
zeroMin=$(echo "$lastMin" | sed 's/^0*//')

echo "$zeroMin"   # output : 455233

Your problem is caused by using $(( instead of $( , and not echo 'ing the old var.

Answer 3

The arithmetic expansion $(( ... )) is for, well, arithmetics. The | is "binary OR" operation inside $(( ... )) .

You just want to pipe the variable value to stdout of another command, ex. sed .

On posix compatible shells you can use a simple pipe | with process substitution $() :

zeroMin=$(printf "%s\n" "$lastMin" | sed 's/^0*//')
# or a little tiny bit less portable version with echo
zeroMin=$(echo "$lastMin" | sed 's/^0*//')

On bash you can use use here strings <<< :

zeroMin=$(<<<$lastMin sed 's/^0*//')

Answer 4

You don't need to spawn a separate process for this, just use parameter expansions :

$ a=01234
$ b=00001234
$ c=1234
$
$ echo "${a#${a%%[1-9]*}}"
1234
$ echo "${b#${b%%[1-9]*}}"
1234
$ echo "${c#${c%%[1-9]*}}"
1234

${var%%[1-9]*} will remove longest suffix beginning with a digit between 1 and 9 from var ; so it will leave leading zeroes, and now that you have leading zeroes you can remove them from var using ${var#${var%%[1-9]*}} .