CODEWITHSUNDEEP
unixsh
Chor419
Chor419

Reputation: 1

How to substitute a value in a variable

Good day,

I have a simple UNIX script test.sh

I need to substitute the value of a variable. This variable contains a directory path.

My test.sh script

#!/bin/sh
filepath="/host/messages/in/documents"
archivePath=`${filepath/\/in/\/archive/}`
echo "archive path is " $archivePath

I get a "bad substitution" error when I run it.

The required output for archivePath should be:

/host/messages/archive/documents

What am I doing wrong and what could be a possible solution?

Upvotes: 0

Views: 156

Answers (2)

Chor419
Chor419

Reputation: 1

found a solution with sed:

#!/bin/sh
filepath="/host/messages/in/documents"
echo $filepath | sed -e %s/in/archive/g
archivePath=`echo $filepath | sed -e s/in/archive/g`
echo "archive path is $archivePath"

Upvotes: 0

fumiyas
fumiyas

Reputation: 367

Must use bash (or ksh or zsh), use correct syntax ${varname/pattern/replacement} and escape / by \ in pattern and replacement.

#!/bin/bash
filepath="/host/messages/in/documents"
archivePath="${filepath/\/in\//\/archive\/}"
echo "archive path is $archivePath"

Upvotes: 1

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