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[Bash] Remove everything after the first space in a string

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Gau36o -  
Hello,

Beginner in Bash, I have a string stored in a variable.

For example: maVar="ceci 001 est un exemple 5600"

To remove a strict occurrence such as the ceci (including the space following the word ceci) from the string above, no problem, I do:
line1=${maVar#*ceci }

After this, I would now like to remove the space and everything that follows this space, after the pattern 001. Knowing that 001 is necessarily a group of digits but with a variable length (between 2 and 6 characters).

I tried by doing:
line2=${line1#* }

As well as other attempts, but I can’t get the desired result...

Any help is welcome :-)

Thanks in advance!

Configuration: Windows / Chrome 58.0.3029.110

3 answers

  1.  dubcek Contributor
    hello
    or with a table 
    $ maVar="ceci 001 est un exemple 5600" 
    $ t=($maVar)
    $ echo ${t[0]}
    ceci
    $ echo ${t[1]}
    001
    2
    1. Gau36o Author
      Hi,
      Thanks for this answer!
      This method suits me much better :)
      0
  2. Gau36o Author
    Problem solved after all...

    If it can help someone... 
    line2=$(echo $line1 | cut -d " " -f 1)


    So the truncated result of the variable is stored in $line2
    0
    1. zipe31 Contributor
      Hi,

      We can do this as well: 

      $ maVar="ceci 001 est un exemple 5600"

      $ echo "${maVar}"
      ceci 001 est un exemple 5600

      $ echo "${maVar#* }"
      001 est un exemple 5600

      $ echo "${maVar##* }"
      5600

      $ echo "${maVar% *}"
      ceci 001 est un exemple

      $ echo "${maVar%% *}"
      ceci


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      0