How to replace multiple patterns at once with sed?

Maybe something like this:

sed 's/ab/~~/g; s/bc/ab/g; s/~~/bc/g' 

from : https://stackoverflow.com/questions/26568952/how-to-replace-multiple-patterns-at-once-with-sed

reading-property-from-file.sh

This worked for me on all *nix platform:
PROP_VALUE=cat $PROPERTY_FILE | grep "$PROP_KEY" | cut -d'=' -f2

from : https://gist.github.com/marcelbirkner/9b133f800d7d3fc5d828

Remove the first character from a bash variable

$ VAR="Banana";

$ VAR=${VAR#?};

$ echo $VAR;

anana

from : https://bytefreaks.net/gnulinux/bash/remove-the-first-character-from-a-bash-variable

How to read a file into a variable in shell?

In cross-platform, lowest-common-denominator sh you use:

#!/bin/sh
value=`cat config.txt`
echo "$value"

In bash or zsh, to read a whole file into a variable without invoking cat:

#!/bin/bash
value=$(<config.txt)
echo "$value"

from : https://stackoverflow.com/questions/7427262/how-to-read-a-file-into-a-variable-in-shell

Bash command to sum a column of numbers

while read -r num; do ((sum += num)); done < inputfile; echo $sum

from : https://stackoverflow.com/questions/3096259/bash-command-to-sum-a-column-of-numbers

How to check if a file contains a specific string using Bash

if grep -q SomeString "$File"; then
  Some Actions # SomeString was found
fi

from : https://stackoverflow.com/questions/11287861/how-to-check-if-a-file-contains-a-specific-string-using-bash

Source .bashrc in zsh without printing any output

source ~/.bashrc > /dev/null 2>&1

from : https://unix.stackexchange.com/questions/213036/source-bashrc-in-zsh-without-printing-any-output