[SOLVED] Bash error 1: command not found

pankajk2526 · 01-25-2018, 01:10 AM

There is an Error in

Code:

ch=$ch$($i)

in the following code:

Code:

#!/bin/bash

origclip="open documents open music"
i=0
set -- $origclip
ch=$1


while [[ ch != "" ]]; do
    ((i++))
    ch=$ch$($i)                      #this does not execute even ch=$($i) doesn't execute
    echo $ch
done

I want to achieve following output.

Code:

open
open documents
open documents open
open documents open music

new line and space are not required, presence of space don't matter in output.

In short I want to achieve output of following code, with a variable 'i' which increments as in previous code

Code:

origclip="open documents open music"
set -- $origclip
echo $1
echo $1$2
echo $1$2$3
echo $1$2$3$4

Also there is a problem in achieving end of line

Code:

while [[ ch != "" ]]; do

i am not sure of this.

jsbjsb001 · 01-25-2018, 02:05 AM

While I'm no expert in scripting, the "command not found" error message means that the shell (bash) cannot find the executable for that command.

I'd say it's in the following line where your problem lies:

Code:

while [[ ch != "" ]]; do

Probably inside those brackets somewhere.

I've asked for this question to be moved to the Programming forum, so you can get better answers.

pankajk2526 · 01-25-2018, 02:13 AM

The problem lies here:

Code:

origclip="open documents open music"
set -- $origclip
ch=$1
echo $ch                                #this executes well
i=2
ch=$ch$($i)                             #this gives error 2: command not found

The last line in above code will not execute in terminal
even last line in below code will not execute, which is simpler

Code:

origclip="open documents open music"
set -- $origclip
ch=$1
echo $ch                                #this executes well
i=2
ch=$($i)                                #this gives error 2: command not found

jsbjsb001 · 01-25-2018, 02:14 AM

Just noticed the following:

Code:

#!/bin/bashorigclip="open documents open music"

should be:

Code:

#!/bin/bash
origclip="open documents open music"

up the top there.

pankajk2526 · 01-25-2018, 02:18 AM

Quote:

Originally Posted by jsbjsb001

Just noticed the following:

Code:

#!/bin/bashorigclip="open documents open music"

should be:

Code:

#!/bin/bash
origclip="open documents open music"

up the top there.

No, that was just a typing error in this post, i have edited it now,.... please consider this while answering the question...

Code:

origclip="open documents open music"
set -- $origclip
ch=$1
echo $ch                #this executes well
i=2
ch=$($i)                #this gives error 2: command not found

descendant_command · 01-25-2018, 02:30 AM

Doing the substitution for the last line gives

Code:

ch=$(2)

which clearly shows where your error message is coming from.

Likely you want

Code:

ch=${$i}

pankajk2526 · 01-25-2018, 02:34 AM

Quote:

Originally Posted by descendant_command

Doing the substitution for the last line gives

Code:

ch=$(2)

which clearly shows where your error message is coming from.

Likely you want

Code:

ch=${$i}

That results in error bash: ${$i}: bad substitution

descendant_command · 01-25-2018, 03:03 AM

Yeah, I thought that might happen.

You have further research to do on nesting bash variables, but at least you're not trying to execute a command called '2' any more ...

astrogeek · 01-25-2018, 03:07 AM

It is not clear what your actual goal is, but that is a hard way to go about getting what you have described.

You may have some reason for setting the values of positional parameters, but if not something like this will do what you have described:

Code:

origclip="open documents open music"
for wd in $origclip; do
        ch="${ch}${wd}"
        echo "$ch"
done

The reason for the error in your example code is this...

Code:

ch=$ch$($i)                      #this does not execute even ch=$($i) doesn't execute

$($i) attempts to execute $i, so the command does not exist. Were you thinking that this was equivalent to an array subscript?

The problem with detecting the end of the string is that you are testing whether "$ch" is empty, when you continue to append more to it - so it is never empty.

You need to read up on how shell scripting works.

I highly recommend the bash man page as an excellent place to get the details - well organized and well written (uncommon among man pages!).

You might also like to look over the Bash Guide For Beginners, and the Advanced Bash Scripting Guide.

pankajk2526 · 01-25-2018, 04:52 AM

Thankyou!

I can't believe you solved this... answer accepted...

MadeInGermany · 01-27-2018, 01:11 AM

Yes, $($i) runs the contents of i as a command. Not wanted here.
For a space-separated list a for loop is more appropriate than a while loop.

Code:

origclip="open documents open music"
set -- $origclip

ch=""
for c; do
    ch=$ch$c
    echo "$ch"

done

The for takes the positional parameters as default.
Perhaps you do not even need to set the positional parameters?
This was suggested before. Here is how to add a space separator in the output.

Code:

origclip="open documents open music"

ch="" sep=""
for c in $origclip; do
    ch=$ch$sep$c
    echo "$ch"
    sep=" "
done