Regular expression to replace text

Question

I am very new to regular expressions. I am using UltraEdit, and would like to use regular expressions to make the changes described below.

I have some text in the following pattern:

I need to add prefix 'Z' in front of href with extension .rock.

href="000760.rock" --> href="Z000760.rock"

The output should look like this:

What would be the regular expression to do this in UltraEdit?

Robert Mark Bram · Accepted Answer

Re-wrote my answer to

Add new use-case OP added where some values have the X prefix and must not be replaced.
I was initially putting the double quote character in brackets when there was no need.

The first case I answered is where none of the HREF values already have the X prefix.

Find:

href="([^"]*)\.rock"

And replace:

href="X\1.rock"

Start:

Finish:

Screen shot showing this first result is below.

Screen shot showing first result.

Breakdown of the regex:

Find: href="([^"]*)\.rock"
1. href=" - this finds href="
2. ([^"]*) - this creates the first backreference - tells the engine to look for and remember everything between the brackets: [^"]* so that we can reference it in the replace part.
  1. [^"] - this part of the pattern says any character that is not a double quote.
  2. And the asterisk at the end of [^"]* is a repetition pattern that says look for zero or more characters that matches the thing just before it (so find zero or more characters that are not a double quote).
3. \.rock" this defines the rest of the pattern which must be .rock"
4. Note that I have escaped the period character: \.. That is because period has a special meaning in a regex and we are telling the regex that we mean a literal dot or period.
Replace: href="X\1.rock"
1. href="X - says to output literally href="X..
2. \1 - says to replace \1 with the first backreference we created (zero or more characters that are not a double quote).
3. .rock" - says to output literally .rock".
  1. Note that I didn't need to escape the period here, because it doesn't have the same meaning in replace - it just means the literal dot.

The second case is in response to OP's comment that some of the HREF values already have the X prefix. In this case, change the regex as below.

Find:

href="([^X][^"]*)\.rock"

And replace:

href="X\1.rock"

Start:

Finish:

Screen shot showing this second result is below.

Screen shot showing second result.

Breakdown of the regex:

Find: href="([^X][^"]*)\.rock"
1. href=" - this finds href="
2. ([^X][^"]*) - this creates the first backreference - tells the engine to look for and remember everything between the brackets: ([^X][^"]*)* so that we can reference it in the replace part.
  1. [^X]* - this part of the pattern says any character that is not an X.
  2. [^"] - this part of the pattern says any character that is not a double quote.
  3. And the asterisk at the end of [^"]* is a repetition pattern that says look for zero or more characters that matches the thing just before it (so find zero or more characters that are not a double quote).
3. \.rock" this defines the rest of the pattern which must be .rock"
4. Note that I have escaped the period character: \.. That is because period has a special meaning in a regex and we are telling the regex that we mean a literal dot or period.
Replace: href="X\1.rock"
1. href="X - says to output literally href="X..
2. \1 - says to replace \1 with the first backreference we created (zero or more characters that are not a double quote).
3. .rock" - says to output literally .rock".
  1. Note that I didn't need to escape the period here, because it doesn't have the same meaning in replace - it just means the literal dot.

Answers (2)