capture the output of a shell script which is invoked from another shell script

Question

I have a script called a.sh, contents of which are:

//a.sh:

#!/bin/bash
temp=0
while [ "$temp" -ne 500 ]
do
    echo `date`
    temp=`echo "$temp+1" | bc`
    sleep 1
done
----------------------------------

Another script namedb.sh, contents of which are:

// b.sh:

#!/bin/bash

`a.sh`

exit
----------------------------------

When i execute a.sh separately, i'm able to see the output.. but, when i execute b.sh, i'm not able to see the output on the console.. (i tried a few times - to redirect the output of a.sh - but not being successful).

So, what i need is the redirection, which will enable me to see the output of a.sh's contents when i execute b.sh - on the console.

Thanks, Ravi.

Someone reformat, please, this is unreadable.

themel
– themel

2011-07-06 06:43:47 +00:00
Commented Jul 6, 2011 at 6:43 — themel
– themel, Commented Jul 6, 2011 at 6:43

Op De Cirkel · Accepted Answer · 2011-07-06 06:58:21Z

5

`a.sh`

in your b.sh means take the output of a.sh and use it as a command with arguments. you just have to execute a.sh in b.sh

$A_SH_PATH/a.sh instead of

`a.sh`

edited Jul 6, 2011 at 6:58

answered Jul 6, 2011 at 6:51

Op De Cirkel

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2 Comments

Ravishankar Over a year ago

Thanks a lot for the clarification.. this simply means, my understanding of backticks was wrong.. Thanks for enlightening

Ravishankar Over a year ago

Thanks a lot for the clarification.. so, this implies my understanding of backticks was not correct. Thanks for the help.

Szocske · Accepted Answer · 2011-07-06 06:51:07Z

0

This should "just work".

Is a.sh on your $PATH ?

If not, you need to invoke it with a path, e.g. ./a.sh

answered Jul 6, 2011 at 6:51

Szocske

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3 Comments

Op De Cirkel Over a year ago

a.sh outputs date in some format, in b.sh he is doing: $(a.sh) which will take the output of date and try to execute it.

Szocske Over a year ago

With proper formatting a lot more is visible :-) Drop the backticks, or do echo ./a.sh

Ravishankar Over a year ago

Thanks a lot for the clarification.. so, this implies my understanding of backticks was not correct. Thanks for the help.

Vijay · Accepted Answer · 2011-07-06 07:44:46Z

0

Try removing the backquotes from b.sh like below. this works for me.

#!/bin/bash

a.sh

exit

answered Jul 6, 2011 at 7:44

Vijay

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1 Comment

Ravishankar Over a year ago

Thanks a lot for the clarification.. so, this implies my understanding of backticks was not correct. Thanks for the help.

William Pursell · Accepted Answer · 2011-07-06 08:17:35Z

0

The following lines in a.sh:

    echo `date`
    temp=`echo "$temp+1" | bc`

can be rewritten as:

    date
    : $(( temp += 1 ))

It seems that the root problem you are having is a misunderstanding of backticks, and you should understand why "echo `date`" is (almost) exactly the same as just "date" (they differ in whitespace only). The change I made to the second line is just personal preference (it is also more efficient).

answered Jul 6, 2011 at 8:17

William Pursell

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1 Comment

Ravishankar Over a year ago

Thanks a lot for the clarification.. so, this implies my understanding of backticks was not correct. Thanks for the help.

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