Cleaning up formatting after deletion using regex - regex

Question

I have a function similar to the one below appearing in multiple files. I want to use regex to get rid of all references to outputString, since clearly, they're wasteful.
... other functions, class declarations, etc
public String toString()
{
String outputString = "";
return ... some stuff
+ outputString;
}
... other functions, class declarations, etc
I'm happy to do this in multiple passes. So far I've got regexes to find the first and last line (String outputString = "";$ and ( \+ outputString;)$). However, I've got two problems: first, I want to get rid of the whitespace that results in deleting the two lines that refer to outputString. Second, I need the final ; on the second last line to move up to the line above it.
As a bonus, I'd also like to know what's wrong with adding the line start anchor (^) to either of the regexes I specified. It seems like doing so would tighten them up, but when I try something like ^( \+ outputString;)$ I get zero results.
After all's said and done the function above should look like this:
... other functions, class declarations, etc
public String toString()
{
return ... some stuff;
}
... other functions, class declarations, etc
Here's an example of what "some stuff" might be:
"name" + ":" + getName()+ "," +
"id" + ":" + getId()+ "]" + System.getProperties().getProperty("line.separator") +
" " + "student = "+(getStudent()!=null?Integer.toHexString(System.identityHashCode(getStudent())):"null")
Here's a concrete example:
Current:
public void delete()
{
Student existingStudent = student;
student = null;
if (existingStudent != null)
{
existingStudent.delete();
}
}
public String toString()
{
String outputString = "";
return super.toString() + "["+
"name" + ":" + getName()+ "," +
"id" + ":" + getId()+ "]" + System.getProperties().getProperty("line.separator") +
" " + "student = "+(getStudent()!=null?Integer.toHexString(System.identityHashCode(getStudent())):"null")
+ outputString;
}
public String getId()
{
return id;
}
Required:
public void delete()
{
Student existingStudent = student;
student = null;
if (existingStudent != null)
{
existingStudent.delete();
}
}
public String toString()
{
return super.toString() + "["+
"name" + ":" + getName()+ "," +
"id" + ":" + getId()+ "]" + System.getProperties().getProperty("line.separator") +
" " + "student = "+(getStudent()!=null?Integer.toHexString(System.identityHashCode(getStudent())):"null");
}
public String getId()
{
return id;
}

Answer 1

1st pass:
Find:
.*outputString.*\R
Replace with empty string.
Demo:
https://regex101.com/r/g3aYnp/2
2nd pass:
Find:
(toString\(\)[\s\S]+\))(\s*\R\s*?\})
Replace:
$1;$2
https://regex101.com/r/oxsNRW/3

Answer 2

Assuming that the wanted part of the return expression does not contain any semi colons (i.e. ;) then you can do it in one replace. Search for:
^ +String outputString = "";\R( +return [^;]+?)\R +\+ outputString;
and replace with:
\1;
The idea is to match all three lines in one go, to keep the wanted part and to add the ;.
An interesting point in this replacement. My first attempt had ... return [^;]+)\R +\+ ... and it failed whereas ... return [^;]+)\r\n +\+ ... worked. The \R version appeared to leave a line-break before the final ;. Turning on menu => View => Show symbol => Show end of line reveals that the greedy term within the capture group collected the \r and the \R matched only the \n. Changing to a non-greedy form allowed the \R to match the entire \r\n.