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Ayush Mishra
Ayush Mishra

Reputation: 1673

Replace spaces with underscores via BASH

Suppose i have a string, $str. I want $str to be edited such that all the spaces in it are replaced by underscores.

Example

a="hello world"

I want the final output of 

echo "$a"

to be hello_world

Upvotes: 49

Views: 104199

Answers (5)

Serge Stroobandt
Serge Stroobandt

Reputation: 31628

Multiple spaces to one underscore

This can easily be achieved with a GNU shell parameter expansion. In particular:

${parameter/pattern/string}

If pattern begins with /, all matches of pattern are replaced with string.

with +(pattern-list)

Matches one or more occurrences of the given patterns.

Hence:

$ a='hello world    example'

$ echo ${a// /_}
hello_world____example

$ echo ${a//+( )/_}
hello_world_example

However, for this to work in a bash script two amendments need to be made:

  1. The parameter expansion requires encapsulation in double quotes " " to prevent word splitting with the input field separator $IFS.
  2. The extglob shell option needs to be enabled using the shopt builtin, for extended pattern matching operators to be recognised.

The bash script finally looks like this:

#!/usr/bin/env bash
shopt -s extglob
a='hello world    example'
echo "${a//+( )/_}"

Upvotes: 10

William Hay
William Hay

Reputation: 2308

You could try the following:

str="${str// /_}"

Upvotes: 95

falsetru
falsetru

Reputation: 369454

$ a="hello world"
$ echo ${a// /_}
hello_world

According to bash(1):

${parameter/pattern/string}

Pattern substitution. The pattern is expanded to produce a pattern just as in pathname expansion. Parameter is expanded and the longest match of pattern against its value is replaced with string. If pattern begins with /, all matches of pattern are replaced
with string. Normally only the first match is replaced. If pattern begins with #, it must match at the beginning of the expanded value of parameter. If pattern begins with %, it must match at the end of the expanded value of parameter. If string is null, matches of pattern are deleted and the / following pattern may be omitted. If parameter is @ or *, the substitution operation is applied to each positional parameter in turn, and the expansion is the resultant list. If parameter is an array variable subscripted with @ or *, the substitution operation is applied to each member of the array in turn, and the expansion is the resultant list.

Upvotes: 36

fedorqui
fedorqui

Reputation: 290505

With sed reading directly from a variable:

$ sed 's/ /_/g' <<< "$a"
hello_world

And to store the result you have to use the var=$(command) syntax:

a=$(sed 's/ /_/g' <<< "$a")

For completeness, with awk it can be done like this:

$ a="hello my name is"
$ awk 'BEGIN{OFS="_"} {for (i=1; i<NF; i++) printf "%s%s",$i,OFS; printf "%s\n", $NF}' <<< "$a"
hello_my_name_is

Upvotes: 11

anubhava
anubhava

Reputation: 786291

Pure bash:

a="hello world"
echo "${a// /_}"

OR tr:

tr -s ' ' '_' <<< "$a"

Upvotes: 17

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