Grep: invalid content of \{\}

grepregular expression

Here's the command :

grep '\(2\)[[:digit:]]\{\1\}' numbers

What I want is to match exactly 2 digits after the digit '2', 3 digits if I change the digit '2' to '3' in my expression, 4 digits if …

I'm using back-referencing here but when I execute this command I have the following output :

grep: invalid content of \{\}

How could one change my expression ?

first solution :
As serge answered :

2[[:digit:]]\{1\}\|3[[:digit:]]\{2\}\|4[[:digit:]]\{3\}\|5[[:digit:]]\{4\}

That could actually work, but let's suppose I have $34$ as the first sub-expression, I wish I could back-referencing that number so I don't have to implement the 34 cases (or more).

Best Answer

Back-references are not allowed here. You have to write the expression that will describe all possible cases, e.g.:

2[[:digit:]]\{1\}\|3[[:digit:]]\{2\}\|4[[:digit:]]\{3\}\|5[[:digit:]]\{4\}

Related Solutions

bash grep – Why Quote Escaped Character in Regex

Why? because your shell interprets some special characters, such as \ in your example.

You are running into troubles because you do not protect the string that you try to pass as argument to grep via the Shell.

Several solutions:

singlequoting the string,
doublequoting the string (with doublequoting the shell will interpret several things, such as $variables , before sending the resulting string to the command),
or not use quoting (which I strongly advise against) but add backslashes in the right places to prevent the shell to interpret the next characters before sending it to the command.

I recommend to protect the string via single quotes, as it keeps almost everything literraly:

grep '9\.0' #send those 4 characters to grep in a single argument

The Shell pass the singlequoted string literally.

Note: The only thing you can't include inside a single quoted shell string is a single quote (as this ends the singlequoting). To include a singlequote inside a singlequoted shell string, you need to first end the singlequoting, immediately add an escaped singlequote \' (or one between doublequotes: "'" ) and then immediately reenter the singlequoting to continue the single quoted string : for exemple to have the shell execute the command grep a'b , you could write the parameter as 'a'\''b' so that the shell sends a'b to grep: so write: grep 'a'\''b' , or grep 'a'"'"'b'

If you insist on not using quoting, you need your shell to have a \\ to have it send a \ to grep.

grep 9\\.0  # ie: a 9, a pair \\, a ., and a 0 , and the shell interprets the pair \\ into a literal \

If you use doublequotes: you need to take into account that the shell will interprets several things first ($vars, \, etc). for exemple when it sees an unescaped or unquoted \, it waits the next character to decide how to interpret it. \w is seen as a single letter w, \\ is seen as a single letter \, etc.

grep "9\\.0"  # looks here the same as not quoting at all... 
    #but doublequoting allows you to have spaces, etc, inside the string

Best Answer

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bash grep – Why Quote Escaped Character in Regex

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