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lang-bash
|in the string :P . Could you think of a way to escape it (without breaking something else)?|in the above) from a list of characters, finding the first that isn't in the string. Then use that instead of always using|. E.g. characters to use/test:|;,&%etc.splitin one line (as the OP wants?).var=$'foo\0bar'; echo "$var"). This is what my suggested answer does.\0doesn't work, I think IFS needs something apart from NULL character. But I tried with\1instead, and it seems to work, and good to learn more about$''syntax, thanks. I will do a last update, but perhaps the OP doesn't have a real problem to solve, it feels more like a quiz.