How to remove preceding spaces before non space text begins + regex + sed +

Question

How do I remove the preceding spaces before 809 in the output of my echo??

here is my sample:

$ 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 using 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 what I want is my sed command to work on anything greater 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 do I write my regex in the sed to remove 3 or more spaces preceding 809?

Answer 1

If you want to remove 3 or more spaces at the start use

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

Details

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

If you need to remove 3+ spaces at the start before a specific value, say, 809, just add that value to the regex and replacement pattern:

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

or use a capturing group and placeholder:

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

Answer 2

xargs frees you from thinking about regular expressions.

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

---

It's a more or less unconventional trick of xargs. This tool is meant to be used to build up commands with long argument lists. From man page:

xargs reads items from the standard input, delimited by blanks (which can be protected with double or single quotes or a backslash) or newlines, and executes the command

However, it a part of the preprocess that white space will be trimmed, before sending to subsequent command. In this case, there isn't subsequent command, and therefore xargs just send the argument as it is, modulo with the spaces trimmed, to standard output, which does exactly what you are asking for.