Bash: issue with sed on encountering '&'

Question

I have a config file which I use to store key values which is then over written on actual properties file. I have below code which I use for this purpose and this works fine for key values which does not have '&' in it.

Get the desired value from the config file:

postgresql_url=$(grep "^postgresql.url=" ./<file_name> |cut -d= -f2,3,4,5)

the above code gives me:

jdbc:postgresql://ec2-22-222-2-22.eu-central-1.compute.amazonaws.com:111/d3bxxxx2cj1?password=xyz&user=user1&sslmode=require

I am trying to over write the below value with the one from above

app1.postgresql.url=jdbc:postgresql://ec2-11-111-1-139.eu-west-1.compute.amazonaws.com:222/d53hyyyy928s0?password=abc&user=user1&sslmode=require

sed: however gives me this output after replacing:

sed -i -e '/postgresql.url[[:blank:]]*=/ s|=.*|='$postgresql_url'|' $property_file

Result:

app1.postgresql.url=jdbc:postgresql://ec2-22-222-2-22.eu-central-1.compute.amazonaws.com:111/d3bxxxx2cj1?password=xyz=jdbc:postgresql://ec2-11-111-1-139.eu-west-1.compute.amazonaws.com:222/d53hyyyy928s0?password=abc&user=user1&sslmode=requireuser=user1=jdbc:postgresql://ec2-11-111-1-139.eu-west-1.compute.amazonaws.com:222/d53hyyyy928s0?password=abc&user=user1&sslmode=requiresslmode=require

As you can see the replace it not what is required. How to ignore '&' in sed replacement and get below result:

app1.postgresql.url=jdbc:postgresql://ec2-22-222-2-22.eu-central-1.compute.amazonaws.com:111/d3bxxxx2cj1?password=xyz&user=user1&sslmode=require

Vinay Vissh · Accepted Answer · 2018-10-04T11:34:07.527

0

How about escaping & (a.k.a adding \ before each &) in the replacement string for sed?

A bit explanation about escaping special characters inside a variable using sed can be found here.

Here is a sample script for your case:

#!/bin/bash

mkdir temp && cd temp

# Create the property file (in which url needs to be replaced)
echo 'app1.postgresql.url=jdbc:postgresql://ec2-11-111-1-139.eu-west-1.compute.amazonaws.com:222/d53hyyyy928s0?password=abc&user=user1&sslmode=require' >> property


# The new url
postgresql_url='jdbc:postgresql://ec2-22-222-2-22.eu-central-1.compute.amazonaws.com:111/d3bxxxx2cj1?password=xyz&user=user1&sslmode=require'

# Escape &
postgresql_url=$(echo "${postgresql_url}" | sed -r 's/[&]/\\&/g')


# For sed
pattern='app1\.postgresql\.url=.*$'
replacement="app1.postgresql.url=${postgresql_url}"

# Replace
sed -i -r "s;${pattern};${replacement};g" property

edited Oct 04 '18 at 11:34

answered Oct 04 '18 at 11:26

Vinay Vissh

457
2
9
12

excellent..this is exactly what i want, escaping does the trick. Thanks. – VN- Oct 04 '18 at 11:47
@VV- Glad that it helped you. Can you mark my answer as **accepted** if the problem is solved? – Vinay Vissh Oct 04 '18 at 11:55
1

Since we're in Bash, we can just perform substitution in the parameter expansion `${postgresql_url//&/\\&}` - much cheaper than a subshell for `sed`. – Toby Speight Oct 04 '18 at 12:12
See my (related) [answer to *sed replace with variable with multiple lines*](/a/39269241). – Toby Speight Oct 04 '18 at 12:19
Agree. This eliminates additional subshell. Thanks. – VN- Oct 04 '18 at 12:22

Bash: issue with sed on encountering '&'

1 Answers1