脚本和空格

Question

I have a simple bash find and replace script 'script.sh' :

#!/bin/bash 
find . -type f -name '.' -exec sed -i '' "s/$1/$2/" {} +

When I run my command ./script.sh foo bar, it works. Suppose now my two inputs strings $1 and $2 are sentences (with white spaces), how can I make the script to recognize each of them as whole strings ?

Answer 1

Add each sentence in quotes "

#!/bin/bash
echo $1
echo $2

output

$ ./tmp.sh "first sentence" "second sentence"
first sentence
second sentence

Edit:

Try this:

#!/bin/bash
find . -type f -exec sed -i "s/$1/$2/" {} +

Ouput:

$ cat test1.txt 
ola sdfd
$ ./tmp.sh "ola sdfd" "hello world"
$ cat test1.txt 
hello world
$ ./tmp.sh "hello world" "ols asdf"
$ cat test1.txt 
ols asdf

Answer 2

have you tried to screen spaces with \\? eg: this\\ is\\ sample\\ string\\ with\\ spaces

Answer 3

You just need to call your script as :

./myscript  "sentence 1 with spaces" "sentence 2"

But you may have to "escape" special characters (or use something else than sed)

You may want to add g to the sed s/.../.../ construct so that it replaces several occurence on the same line.

And you can't have a string with a / in it, as you use / as the sed's separator. (You can change that separator to anything, for example to the unlikely % : s%$1%$2%g )