Pass-phrase | Generate a random pass phrase | Router library

.bashrc configures your environment.

Apache is sometimes configured with a sudo to bind ports lower than 1024. And you probably have User and Group directives that define which user Apache will switch to once started.

All this to say that your Apache runs in a different environment than your user. It does not have access to the variables.

The "right" place to put such variables is in $APACHE_ROOT/bin/envvars.

I believe it's an issue with pass:xin line 1, x being only 1 character long.

pass:gsahdg etc should work.

Well you could google for this and as in the comment the tag john the ripper was mentioned, but is not there anymore, I still assume that you initially wanted to use jtr. A manual is given here.

It's basic steps are:

1. Download and compile the Jumbo version of John the Ripper from Github.

2. Use gpg2john to convert your rsa_key to a jtr understandable format refered as file1 now.

3. Use john --incremental file1 to start jtr in brute-force mode.

Note however that the third step will make jtr also use chars and not only digits. You could change this by via settings in the config or you could simply create a small script to generate a password list with all possible passwords.

This would be better done in a programming language that supports REGEX.

If you really have to do this in SQL, you can do it like this:

Use CHARINDEX to find the first occurrence of the first keyword you want to match.

Then use CHARINDEX to find the closest whitespace characters both before and after the keyword.

Then use SUBSTRING to get the characters between those two whitespace characters.

Loop to the next occurrence of the first keyword until you get to the end of the string.

Loop to the next keyword you want to match and repeat the above loop.

If you want to do this inline in a query, you'll want to encapsulate the above in a UDF.