Ubuntu – Remove everything before first occurence of specified word

sedtext processing

How can I remove everything before the first occurence of a specified word using sed?

What I have tried so far:

echo $(cat /etc/nginx/sites-enabled/default | sed 's/^*.server/server/') > /etc/nginx/sites-enabled/default

I need to cleanup all the trash before I find the first "server" word.

Mainly I'd need a sed regex…

I have a Nginx default file with proxying info that is getting some leading spaces while the script is copying it to a Docker container and therefore it cant be strated. I'd need to delete it with some command.

File:

       server
{
listen 80 default_server;
location / 
{
proxy_set_header X-Forwarded-Host $host;
proxy_set_header X-Forwarded-Server $host;
proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for;
proxy_set_header Connection "";
proxy_http_version 1.1;
proxy_pass http://127.0.0.1:5000;
}
}

I am not sure what kind of characters are there, so I'd like to remove everything before 'server'.

Best Answer

You simply remove every whitespace character (tabs, spaces, whatever) between the line start and the word "server" (only if there is nothing else but whitespaces) using the following simple command:

sed -i 's/^\s*server/server/' FILENAME

The -i option modifies the file in-place, that means all changes are applied and saved immediately and you don't get any output. If you don't want to rewrite the file and see the modified new version instead, omit the -i option and the command will print it to the STDOUT (standard output stream).

I use the regex ^\s*server to match any number of any kind of whitespaces between the line start and the word "server" inclusive, and let it replace those matches with the word "server" itself.

Related Solutions

Ubuntu – Remove the first part of a string using sed

I'd use a simple grep to look for user90:

$ echo "randomcollege-nt\user90" | grep -o user90
user90

If user90 is not constant, prefer this command:

$ echo "randomcollege-nt\user90" | grep -oP '(?<=randomcollege-nt\\)\w+'
user90

Finally using sed to edit the file in place:

$ sed -ri 's/randomcollege-nt\\(user[0-9]+)/\1/' my_file

Or to match all possible user accounts:

$ sed -ri 's/randomcollege-nt\\(\w+)/\1/' my_file

Ubuntu – Print word containing string and first word

Bash/grep version:

#!/bin/bash
# string-and-first-word.sh
# Finds a string and the first word of the line that contains that string.

text_file="$1"
shift

for string; do
    # Find string in file. Process output one line at a time.
    grep "$string" "$text_file" | 
        while read -r line
    do
        # Get the first word of the line.
        first_word="${line%% *}"
        # Remove special characters from the first word.
        first_word="${first_word//[^[:alnum:]]/}"

        # If the first word is the same as the string, don't print it twice.
        if [[ "$string" != "$first_word" ]]; then
            echo -ne "$first_word\t"
        fi

        echo "$string"
    done
done

Call it like so:

./string-and-first-word.sh /path/to/file text thing try Better

Output:

This    text
Another thing
It  try
Better

Best Answer

Related Solutions

Ubuntu – Remove the first part of a string using sed

Ubuntu – Print word containing string and first word

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