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

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 content开发者_开发知识库s when i execute b.sh - on the console.

Thanks, Ravi.

`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`

This should "just work".

Is a.sh on your $PATH ?

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

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

#!/bin/bash

a.sh

exit

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).