Unix. Call a variable inside another variable

Question

Currently I have a script like this. The intended purpose of this script is to use the function Getlastreport and retreive the name of lastest report in a folder. The folders name are typical a random generated number every night. I want to call the variable Getlastreport and put it inside Maxcashfunc.

Example :

Getlast report = 3473843.

Use MAXcashfunc grep -r "Max*" /David/reports/$Getlastreport[[the number 3473843 should be here ]]/"Moneyfromyesterday.csv" > Report`

Script:

#!bin/bash

Getlastreport()
{
cd /David/reports/ | ls -l -rt | tail -1 | cut -d' ' -f10-
}

MAXcashfunc()
{
grep -r "Max*" /David/reports/$Getlastreport/"Moneyfromyesterday.csv" > Report
}

##call maxcash func
MAXcashfunc

Did you really mean to pipe the output of cd into ls? s/|/\&\&/ — William Pursell
– William Pursell, Commented Jul 9, 2014 at 15:04
And why bother with -l followed by cut to remove everything that -l added? Try the simpler: cd /David/reports && ls -tr | tail -1 — William Pursell
– William Pursell, Commented Jul 9, 2014 at 15:09
Why are you doing a recursive grep on a file? Recursion is only useful for directories, but if Getlastreport is not doing what you think (eg if it introduces whitespace) then you are potentially doing a recursive grep that you do not want. — William Pursell
– William Pursell, Commented Jul 9, 2014 at 15:10
Given both the text of your question and the code sample, I think there is a confusion here between a variable (i.e.: $SOMETHING) that hold a value, and a function which is callable (i.e.: `cmd arg1 arg2 arg3`)... — Sylvain Leroux
– Sylvain Leroux, Commented Jul 9, 2014 at 22:44

anubhava · Accepted Answer · 2014-07-09 14:50:29Z

1

You can use:

MAXcashfunc() {
    grep -r "Max" /David/reports/`Getlastreport`/"Moneyfromyesterday.csv" > Report
}

`Getlastreport` - Call Getlastreport and get its output.

answered Jul 9, 2014 at 14:50

anubhava

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1 Comment

Ray Over a year ago

Nope. Not working. It just prints out /David/reports/Getlastreport/"Moneyfromyesterday.csv which leds me to a no file. 'Getlastreport' is suppose to print out the last file that was modify and get me a number like 343432

Elliott Frisch · Accepted Answer · 2014-07-09 14:56:50Z

0

If I follow your question, you could use

function Getlastreport() {
  cd /David/reports/ | ls -l -rt | tail -1 | cut -d' ' -f10-
}
function MAXcashfunc() {
  grep -r "Max" /David/reports/$(Getlastreport)/"Moneyfromyesterday.csv" > Report
}

answered Jul 9, 2014 at 14:56

Elliott Frisch

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2 Comments

Ray Over a year ago

This seems to send my script into a waiting for command phase. This was the first way I tried it actually and the way I learn it. Not sure why it isn't working.

Elliott Frisch Over a year ago

I don't think your cut command is correct. That function doesn't return just filenames here.

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