String Formatting in VB.NET

[L. Scott M.]

Have a quick simple question:

dim x as string

x = "1234567890"

-------------------------------------------------------
VB 6

dim y as string

y = format(x,"(@@@) @@@ @@@@")
y is now ---------------------> (123) 456 7890

(think that is the correct VB 6 syntax)
-------------------------------------------------------
VB NET

How can I do this using formatting in VB.NET since the "@" and other
characters are no longer used. I would prefer to avoid verbose methods
- using LEFT, MID, RIGHT....

 

[Kevin Spencer]

http://msdn2.microsoft.com/en-us/library/system.string.format.aspx

--
HTH,

Kevin Spencer
Microsoft MVP
Professional Chicken Salad Alchemist

I recycle.
I send everything back to the planet it came from.

 

[L. Scott M.]

Thanks for the post.
I must be misreading something, without resorting to VB6 namespace - or
- writing my own formatting code. I would like to format the data in a
"string" object in the following manner - example phone number:

dim x as string
x = "7034449999"
(703) 444 9999

or

x = "703PRSWXYY"
(703) PRS WXYY

This does not appear to work ?
string.format("{0###) ### ####}", x)
neither does this ? (microsoft removed "@")
string format("{0@@@) @@@ @@@@}", x)

I know that I must be missing something simple in this, but from
Microsoft's documentation I do not understand the technique to format a
simple string object - without actually writing a function to format my
string which I could do in less time than I am writing this.

TIA
I appologize for my denseness,
Scott M.

 

[Kevin Spencer]

This might be helpful:

http://msdn2.microsoft.com/en-us/library/a292he7t.aspx

--
HTH,

Kevin Spencer
Microsoft MVP
Professional Chicken Salad Alchemist

I recycle.
I send everything back to the planet it came from.

 

[L. Scott M.]

Thanks for the post, with the link, taking a little while to get at it.
Apparently "msdn2" server has been busy for the last three and a half
hours. Would like to see a solution without manually having to break
up the string (left,right,mid) - almost be inclined to
override/overload the string object and add the old style formatting to
it, if that could be done to the string object. It would be a simple
replacement routine with a few masking rules.

A perplexed
Scott

This might be helpful:

http://msdn2.microsoft.com/en-us/library/a292he7t.aspx

--
HTH,

Kevin Spencer
Microsoft MVP
Professional Chicken Salad Alchemist

I recycle.
I send everything back to the planet it came from.

 

[L. Scott M.]

I am not too pleased with this solution. First I could not inherit
from string in the hopes of adding my own formatting to the string
object. But this is what I can glean to "format" this strings data:

dim x as string
x = "20060622163321"
y = String.Format("{0}\{1}\{2} {3}:{4}:{5}", Left(x, 4), Mid(x, 5, 2),
Mid(x, 7, 2), Mid(x, 9, 2), Mid(x, 11, 2), Mid(x, 13, 2))

with the "y" result being:
2006\06\22 16:33:21

Of course this will pop if the string is less than 14 characters.

Thanks again,
Scott

 

[Kevin Spencer]

Hi Scott,

Since you want to ensure that the format is correct, another option would be
to use a Regular Expression. A Regular Expression matches patterns in a
string. So, for example, if you wanted to format a phone number as "(@@@)
@@@ @@@@" you could use something ike the following:

(\d{3})[^\d]*(\d{3})[^\d]*(\d{4})

What this regular expression says is: Find 3 digits followed by 0 or more
non-digits, followed by 3 digits, followed by 0 or more non-digits, followed
by 4 digits. It puts the first 3 digits into Group 1, the second 3 into
Group 2, and the last 4 into Group 3. I tested it against the following:

1234567890
123 456-7890
(123) 456 - 7890
12 Baker's Street 19005
123 45th Ave.
12-23-456-789-10
123 (456) abcde 890

It found matches in the first 3 items, but none in any of the others,
because they did not have the right number of digits in the right sequence.
The last item *will* match if you add one more digit at the end.

You can then use the Regex Class to add your formatting, by formatting each
Group in each Match.

--
HTH,

Kevin Spencer
Microsoft MVP
Professional Chicken Salad Alchemist

I recycle.
I send everything back to the planet it came from.

 

[L. Scott M.]

Thanks Kevin,

Thought that I had seen something on string patterns when I started
..NETting from VB6 this last year. It would have been nice if they
(Microsoft) referenced "regular expressions" in their "format"
documentation (I could have overread it ). Guess I have some
more reading to do, kind of reminds me of what I remember of "awk". It
would be helpful if someone documented the formatting abilities of
"regular expressions" and "Regex". For example I see from your and
MS's examples the use of the "\d" escape but it is not referenced in
their "character escapes" documentation - ah - there it is in the
"character classes". Now to implement it, see how it plays .......

Kevin thanks for your assitance,
Scott

Hi Scott,

Since you want to ensure that the format is correct, another option would be
to use a Regular Expression. A Regular Expression matches patterns in a
string. So, for example, if you wanted to format a phone number as "(@@@)
@@@ @@@@" you could use something ike the following:

(\d{3})[^\d]*(\d{3})[^\d]*(\d{4})

What this regular expression says is: Find 3 digits followed by 0 or more
non-digits, followed by 3 digits, followed by 0 or more non-digits, followed
by 4 digits. It puts the first 3 digits into Group 1, the second 3 into
Group 2, and the last 4 into Group 3. I tested it against the following:

1234567890
123 456-7890
(123) 456 - 7890
12 Baker's Street 19005
123 45th Ave.
12-23-456-789-10
123 (456) abcde 890

It found matches in the first 3 items, but none in any of the others,
because they did not have the right number of digits in the right sequence.
The last item *will* match if you add one more digit at the end.

You can then use the Regex Class to add your formatting, by formatting each
Group in each Match.

--
HTH,

Kevin Spencer
Microsoft MVP
Professional Chicken Salad Alchemist

I recycle.
I send everything back to the planet it came from.

I am not too pleased with this solution. First I could not inherit
from string in the hopes of adding my own formatting to the string
object. But this is what I can glean to "format" this strings data:

dim x as string
x = "20060622163321"
y = String.Format("{0}\{1}\{2} {3}:{4}:{5}", Left(x, 4), Mid(x, 5, 2),
Mid(x, 7, 2), Mid(x, 9, 2), Mid(x, 11, 2), Mid(x, 13, 2))

with the "y" result being:
2006\06\22 16:33:21

Of course this will pop if the string is less than 14 characters.

Thanks again,
Scott