Linux, Bash, Replace characters in a variable?

Question

linux

How to replace characters in a variable, for example replace spaces with %20

Answer 1

${mystring//pattern/replacement}

myurl="test toster"
echo ${myurl// /%20}

Answer 2

Judging by the number of questions on bash, I strongly advise you to read www.tldp.org/LDP/abs/html , there are answers to all your questions.

Answer 3

I will supplement the previous comments, to replace the first match, the construction is used

myurl="test toster testarossa"
echo ${myurl/ /%20}
test%20toster testarossa

For those who have trouble with English, there is a good dock on www.opennet.ru
https://www.opennet.ru/docs/RUS/bash_scripting_guide/

Answer 4

#!/bin/bash
sqlite3=`which sqlite3`
DB_FILE=words.db
echo "Enter word:"
read input_translate
input_translate=${input_translate//[^a-z0-9a-z,."'" ]/ }
$ sqlite3 $DB_FILE " insert into words (translate) values ​​(\"$input_translate\");"