How to remove previous spaces before the start of the text without space + regex + sed +

Question

How to remove previous spaces before 809in my output echo??

here is my example:

$ echo '   809 23/Dec/2008:19:20'
   809 23/Dec/2008:19:20
^^^3 spaces here preceding the 809    

I can remove 3 or 4 spaces with sed:

        ...4 spaces here   
$ echo '    809 23/Dec/2008:19:20' | sed 's/^.\{3\,4\}//'
809 23/Dec/2008:19:20
0 spaces here preceding the 809    

But I want my sed team to work on something more than 3

        .....5 spaces here
$ echo '     809 23/Dec/2008:19:20' | sed 's/^.\{3\,4\}//'
 809 23/Dec/2008:19:20
^1 spaces here preceding the 809    

Ho, am I writing my regular expression in sed to remove 3 or more spaces preceding 809?

Answer 1

If you want to remove 3 or more spaces at startup, use

sed 's/^[[:blank:]]\{3,\}//'

More details

• ^ - start of input
• [[:blank:]]\{3,\} - 3 or more consecutive occurrences of any horizontal spaces.

3+ , , 809, :

sed 's/^[[:blank:]]\{3,\}809/809/'

:

sed 's/^[[:blank:]]\{3,\}\(809\)/\1/'

Answer 2

xargs .

$ echo '   809 23/Dec/2008:19:20' | xargs
809 23/Dec/2008:19:20

---

xargs. . :

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Answer 3

, {minimum, maximum}. , , , . , "\ {3, \}". !

Answer 4

:

$ echo '             809 23/Dec/2008:19:20' | sed -s 's/^ \{3,\}//'
809 23/Dec/2008:19:20