Shell script to replace all occurrnces of string with a value - linux

Question

I'm new to shell script.
I have a YAML file that consists of placeholder . I need to replace all the occurrences of placeholder <test-name> with a value sweta
script.sh
Name="sweta"
Below is the example of YAML file
metadata:
name: <test-name>-svc
namespace: abc
spec:
selector:
app: <test-name>
ports:
- protocol: TCP
name: http
port: 80
targetPort: 8080
My expected output is:
metadata:
name: sweta-svc
namespace: abc
spec:
selector:
app: sweta
ports:
- protocol: TCP
name: http
port: 80
targetPort: 8080
Can someone help me?
Appreciate all your help. Thanks in advance!

Answer 1

There are two easy ways to accomplish your search and replace <test-name> string with sweta
Note: Assuming your yaml file name is config.yaml
Option-1: Using stream editor command 'sed' as below:
To overwrite the original file
sed -i 's/<test-name>/sweta/g' config.yaml
To write the output to a different file
sed 's/<test-name>/sweta/g' config.yaml > updated_config.yaml
The syntax of sed command is sed -i 's/original/new/g' file.txt
where
sed is the command name
-i stands for in-place edit. This option is used to overwrite the original file. If you don't use this, you have to redirect output to a different file as done in second command above.
s stands for substitute
original is the search string
new is the replace string
g is for global (replace all occurrences). Omitting this will just replace only the first occurence
file.txt is the text file (Linux doesn't use extensions like .txt but people often name files with such extensions as a convention to denote file type)
Option-2: Use 'awk' command:
awk is a very powerful text processing command with its own language syntax, but this search and replace is quite easy:
awk '{gsub("<test-name>", "sweta"); print $0}' config.yaml > updated_config.yaml