Replace " by \" inside a string

Question

How can I replace a " by \" inside a string using bash?

For example:

A txt-File contains text like:

Banana "hallo" Apple "hey"

This has to be converted into:

Banana \"hallo\" Apple \"hey\"

I tried

a=$(cat test.txt)
b=${a//\"/\"}}

But this didn't work.

How does that work?

e.g. a=$(cat test.txt) and b=${a//\"/\"}} this does not work of course — Anne K., Dec 15 '16 at 14:28
Your question is a bit vague about your requirements. In general, you might address a problem like this with `sed` or perhaps with `awk`. There are other possibilities. — John Bollinger, Dec 15 '16 at 14:28
The Problem is, that " is a Special character. So the solution that Julien Lopez proposes, is not possible. — Anne K., Dec 15 '16 at 14:30
@AnneK. You were actually very close. But both `\` and `"` need to be escaped if you want a literal `\` or `"` — sjsam, Dec 15 '16 at 14:52

sjsam · Answer 1 · 2016-12-15T15:00:53.433

Use [ parameter expansion ]

string='Banana "hallo" Apple "hey"'
echo "$string"
Banana "hallo" Apple "hey"
string=${string//\"/\\\"} # Note both '\' need '"' need to be escaped.
echo "$string"
Banana \"hallo\" Apple \"hey\"

A lil explanation

${var/pattern/replacement}

replaces one occurrence of pattern in var with replacement.

${var//pattern/replacement}

replaces all occurrences of pattern in var with replacement.

If the pattern or replacement contains characters like " or / with special meaning in shell, they need to be escaped to let shell treat them as literals.

Replace " by \" inside a string

1 Answers1