将 bash 中变量中的多行字符串转换为带有换行符的单行字符串

Question

I am looking for a way to convert a multiline string in a variable in bash to a single-line string that has each \n character escaped as the \n literal.

For example:

str="
Hello
World
"

I need this to become Hello\nWorld . I looked through the questions on SO and Unix StackExchange but I haven't been able to find a command yet that achieves what I need.

Answer 1

Bash has two built-in ways to quote values suitable for re-ingestion. These will handle not only newlines but also tabs, quotes, and backslashes:

❯ echo "${str@Q}"
$'\nHello\nWorld\n'

❯ printf '%q\n' "$str"
$'\nHello\nWorld\n'

Alternatively, if you simply want to replace newlines and nothing else you can use ${var//search/replace} syntax to do replacements:

❯ echo "${str//$'\n'/\\n}"
\nHello\nWorld\n

Answer 2

Try:

sed -zn 's/\n/\\n/p' <<< "$str"

Use -z to consume the variable as one line and substitute new lines for literal newlines.