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SOLVED

RegEx issue with string

daophuongtrinh
8 - Asteroid

Hello everyone,

 

I'm having the string below:

 

XYZ.12.99]

XYZ.0]

XYZ.1]

XYZ.22]

 

I built two expression for RegEx tool which are:

\w{3}\.(.*[^\]])

 

\w{3}\.(\d.*[^\]])

 

The first expression matched all four strings, however when I run the second expression, the string "XYZ.0]" and "XYZ.1]" will result as Null.

 

I wonder why \d would come out as Null in this case.

 

If anyone have experience on this, please let me know.

 

Thank you very much,

Trinh

6 REPLIES 6
DiegoParker
10 - Fireball

Hi @daophuongtrinh 

 

I do believe is because (\d.*[^\]]) is marking two characters, the \d and the [ ] because your string only has one character then the rest is null as it can't match them.

 

Not 100% sure but that is my hunch.

 

Best,

Diego

VianneyM
Alteryx
Alteryx

hi @daophuongtrinh,

 

could you please let me know what you try to do? are you trying to match some strings?

 

what is the expected output?

 

Best,

Vianney

Best,
Vianney
daophuongtrinh
8 - Asteroid

Hi Vianney

 

My expected output is the number after "XYZ.", so I wanted to parse the digits after XYZ. and exclude the bracket (]) behind. In this example it would be 12.99 ; 0 ; 1 ; 22.

 

So I was wondering why this expression \w{3}\.(\d.*[^\]]) will result Null for some data such as "XYZ.0]" and "XYZ.1]"

 

Thank you

Trinh

daophuongtrinh
8 - Asteroid

Hi Diego,

 

Thank you for the explanation, but I'm sorry I don't really get what you mean. I wrote the expression \w{3}\.(\d.*[^\]]) in order to parse the number after "XYZ." but exclude the bracket (]) in the end of this string. It came out that [^\]] did exclude the bracket as I wanted, but I still don't get why \d.* will result in Null for "XYZ.0]" or ""XYZ.1]". 

 

Thank you for your support!

Trinh

fmvizcaino
17 - Castor
17 - Castor

Hi @daophuongtrinh ,

 

It took me a while to realized what was the problem with your regular expression.

(\d.*[^\]])

In this specific part of your expression, the problem is that you are demanding 2 characters, meaning that \d is a digit but [^\] is also anything other that a bracket.

 

So XYZ.0] won't be a match since 0 = \d but nothing is equal to [^\]

This expression will work for any other string with more that one digit

 

Best,

Fernando Vizcaino

 

daophuongtrinh
8 - Asteroid

Dear Fernando @fmvizcaino ,

 

Thank you very much for your explanation, which is very clear for me now. ^^

 

Best regards

Trinh

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