Regex question

[NancyM]

I want to extract a word out of this string

String = This loan is locked by soAndso.

I want to extract soAndso if the string matches: This loan is locked by...

When I've used regexp in the past I'd usually just enclose the part I want to capture in parens, like this:

pattern=This loan is locked by (\w+)

However, I end up with the entire pattern not the part in parenthesis. How do I do this using MS regex?

[dtaylor]

I would normally think that your code or at least the following code would return the results you are looking for.

Code: Select all

Let>String=This loan is locked by soAndso
RegEx>(?:this loan is locked by\s)(\w+),String,0,m,n,0,,
MessageModal>m_1

But, that is not the case. In the above RegEx, matching group #1 is supposed to be a non-capturing match, yet it is still returned in the match results.

The only way I have been able to get the RegEx results you are look for is use RegEx Replace as follows.

Code: Select all

Let>String=This loan is locked by soAndso
RegEx>(?:this loan is locked by\s)(\w+),String,0,m,n,1,$1,MatchedText
MessageModal>MatchedText

[hagchr]

Hi, The pattern shows what you are looking for so in both cases you are actually looking for the whole string

pattern=This loan is locked by (\w+)
pattern=(?:this loan is locked by\s)(\w+)

(?:this loan is locked by\s) still tries to match "this loan is locked by" followed by \s

To solve it, one approach could be what dtaylor suggests, to work with replacement (replace the full string (ie everything) with the text captured in (\w+)), alt just replace what you do not want. Another approach is to match into m_1 see code:

Code: Select all

//Replacement into string
//========================

//Use RegEx>
Let>string=This loan is locked by soAndso
Let>Pattern=This loan is locked by (\w+)
RegEx>Pattern,string,0,m,nm,1,$1,string
MDL>string

//Use StringReplace>
Let>string=This loan is locked by soAndso
StringReplace>string,This loan is locked by ,,string
MDL>string

//Match into m_1
//===============
//Use \K to discard what has been matched so far
Let>string=This loan is locked by soAndso
Let>Pattern=This loan is locked by \K\w+
RegEx>Pattern,string,0,m,nm,0
MDL>m_1

//Use look-behind to match a position
Let>string=This loan is locked by soAndso
Let>Pattern=(?<=This loan is locked by )\w+
RegEx>Pattern,string,0,m,nm,0
//This effectively matches a position (with certain text to the left of the position) followed by word characters
MDL>m_1

[NancyM]

Thanks!

I went for this solution:

Code: Select all

//Match into m_1
//===============
//Use \K to discard what has been matched so far
Let>string=This loan is locked by soAndso
Let>Pattern=This loan is locked by \K\w+
RegEx>Pattern,string,0,m,nm,0
MDL>m_1

Additional question: Is it possible to expand on this to capture more than one match into the array?

For instance, if I wanted to capture "locked" and soAndSo?

[PepsiHog]

This might help.

Code: Select all

//Match into m_1
//===============
//Use \K to discard what has been matched so far
Let>string=This loan is locked by soAndso
//Let>Pattern=This loan is locked by \K\w+
//RegEx>Pattern,string,0,m,nm,0
//MDL>m_1

RegEx>(.*)\bis\b(.*)?by\b(.*),string,0,match,nom,1,$2 $3,Result

mdl>%Result%

/*
() = a group to refer back to.
$1,$2, $3.... = group that has what you want to capture.
   -groups are counted from left to right
   -to count complex groups such as =====>(by(.*)is(a test))
        - the first outside bracket is $1 ^
        - $2 is the next inside bracket. ====^
        - $3 is the next one from $2  =============^
        - you always count from left to right.
()? = group is optional, does not have to be found for a match. You could replace this with...

   (locked)?      

I don't know your purpose, so just guessing at what you need.
        
\b = word boundary or space
*/

Good Luck,
PepsiHog

[NickD]

You can create non-capturing groups like this:

Code: Select all

RegEx>(?<=this loan is locked by\s)(\w+),String,0,m,n,0,,