Bash substitution; remove everything after the string

This is probably a very stupid question, but in Bash I want to do the following substitution:

I have in some files the line:

DATE: 12 jan 2009, weather 20°C

and I want to change it to

开发者_运维百科

DATE: XXXX

Using sed I guess I had to say "starting after "DATE:", please remove the rest and insert " XXXX".

Is this approach OK and how can I perform it?

Also, I want to learn. If you were in my situation what would you do? Ask here or spend one hour looking at the man page? Where would you start from?

---

Try:

sed -i 's/^DATE:.*$/DATE: XXXX/g' filename

Answer to your 2nd question depends on how often you need to write such things. If it's a one time thing and time is limited it's better to ask here. If not, it's better to learn how to do it by using an online tutorial.

---

while read line
do
    echo "${line/#DATE*/DATE:xxxxx}"
done < myfile

---

You can use the following code also

> var="DATE: 12 jan 2009, weather 20ºC" 
 > echo $var | sed -r 's!^(DATE:).*$!\1 XXXX!'

---

If your line isn't in a file, you can do it without sed :

var="DATE: 12 jan 2009, weather 20ºC"
echo "${var/DATE:*/DATE:xxxxx}"

---

You can use the shell

while IFS=":" read -r a b
do
  case "$a" in
   DATE* ) echo "$a:XXXX";;
   *) echo "$a$b" ;;
  esac
done < "file"

or awk

$ cat file
some line
DATE: 12 jan 2009, weather 20ºC
some line

$ awk -F":" '$1~/^DATE/{$2="XXXX"}1' OFS=":" file
some line
DATE:XXXX
some line