Difficulty with regular expression syntax in Visual Studio

Question

I am using the this regular expression to find all occurrences of INT IDENTITY (1, 1) NOT NULL:

\bINT\s*IDENTITY\s*\(1,\s*1\)\s*NOT\s*NULL\b

A sample of where it is in the whole context is something like:

CREATE TABLE table1 (
    [col]                INT      IDENTITY (1, 1) NOT NULL,
    [col2]               INT      NOT NULL,
    [col3]               BIT      NOT NULL,
    [col4]               BIT      NOT NULL,
    [col5]               DATETIME NOT NULL,
    PRIMARY KEY CLUSTERED ([col] ASC) WITH (FILLFACTOR = 90, ALLOW_PAGE_LOCKS = ON, ALLOW_ROW_LOCKS = ON, PAD_INDEX = OFF, IGNORE_DUP_KEY = OFF, STATISTICS_NORECOMPUTE = OFF)
);

But the regular expression above does not find anything. When I try it online using Rubular, Gskinner, or Regex Hero, it works.

Oded · Accepted Answer · 2012-12-05 09:28:02Z

2

Visual Studio has its own specific regex flavour - for example there is no \b (for word boundary, I presume), so using it will not work.

I suggest reading up on Regular Expressions (Visual Studio) 2010 / 2012 for the syntax.

edited Dec 5, 2012 at 9:28

answered Jun 25, 2012 at 18:08

Oded

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3 Comments

Peter Mortensen Over a year ago

The question is tagged with Visual Studio 2010. The reference describes Visual Studio 2012 (at least as of today), not Visual Studio 2010: "In Visual Studio 2010 and earlier versions, Visual Studio used custom regular expression syntax in the Find and Replace windows. "

Peter Mortensen Over a year ago

Regular Expressions for Visual Studio 2010

Oded Over a year ago

@PeterMortensen - Thanks for the correction and link. Answer updated.

Peter Mortensen · Accepted Answer · 2012-12-09 20:33:07Z

2

Try this one:

:b+INT:b+IDENTITY:b+(1,:b*1):b+NOT:b+NULL

edited Dec 9, 2012 at 20:33

Peter Mortensen

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answered Jun 25, 2012 at 18:13

corn3lius

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2 Comments

Xaisoft Over a year ago

This was close, but I believe you need to escape the ( and ) characters or it treats it as a group.

corn3lius Over a year ago

@Xaisoft good to know. I didn't test it much I was hoping he would run with that.

Xaisoft · Accepted Answer · 2012-06-26 13:48:34Z

1

I ended up using the following expression:

INT:b+IDENTITY:b+\(1,:b*1\):b+NOT:bNULL

answered Jun 26, 2012 at 13:48

Xaisoft

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