Question

VB.NET: how to truncate a string to a certain length yet keeping words

Asked by: andy7789

Hi X-perts,

What is the simplest way to truncate a string to a certain length but keep the words intact?

If we use a direct string cut as

If (Len(fName) > 24) Then trName = Left(fName, 24) & "..."

it may cut words in the middle. I need to make the length max 24 characters, but, if it cuts a word, it should also remove this half-word completely.

Please, advise.

Thanks

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Asked On
2009-09-21 at 16:20:46ID24750162
Topics

Microsoft Visual Basic.Net

,

VB Script

,

VB Objects

Participating Experts
2
Points
500
Comments
13

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Answers

 

by: jeflabaPosted on 2009-09-21 at 17:21:24ID: 25388851

I would create a function to return a truncated string from a passed string.  The first step is to truncate the passed string to the desired size + 1.  Then loop deleting the last character until the last character is a space then delete the space and return the truncated string.  The reason you need to add 1 to the desired length is to handle one character words like "I" and "a."

 

by: jeflabaPosted on 2009-09-21 at 17:25:14ID: 25388864

Also, you need to take into account the other characters which might constitute an end of word like ".", "!", "?" and end of line markers like carriage return and linefeed.  If these characters could be part of your string you might consider replacing them with a space prior to processing (after the first truncation) so your loop only has to look for a space.

 

by: andy7789Posted on 2009-09-21 at 17:29:53ID: 25388882

thank you another option I am thinking is to create an array from words and delete words one-by-one. still looking for a simplest yet elegant solution

 

by: jeflabaPosted on 2009-09-21 at 17:39:04ID: 25388918

andy7789 That sounds quite reasonable.  First replace any potential end of word character with space, then use split to create the array, then loop deleting the last word until the string is less than or equal to the desired length.  Some prior knowledge about the average length of the original string compared to the desired length might lead you do some initial truncate if the average original size is much larger than the desired length.

 

by: Cayman21Posted on 2009-09-21 at 18:50:36ID: 25389111

Try:

str = str.substring(0, str.substring(0, 24).LastIndexOf(" "))

                                              
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by: Cayman21Posted on 2009-09-21 at 18:52:50ID: 25389120

That's about as simple as you get, but you can run into some issues with it.  For example if someone types in a 25 letter word to start it's going to error out.

In order to make sure it's solid do this:

str = str.Substring(0, IIf(str.Substring(0, 24).LastIndexOf(" ") < 0, 0, str.Substring(0, 24).LastIndexOf(" ")))

                                              
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by: Cayman21Posted on 2009-09-21 at 19:01:12ID: 25389136

A slight tweak.

We need to make sure we check that the string is over 24, otherwise we always lop off the last word.  We also need to make sure that it the truncated string isn't right 24 when were done we don't try and read past the end.

So here you go, try this one:

str = IIf(str.Length > 24, str.Substring(0, IIf(str.Substring(0, Math.Min(str.Length, 24)).LastIndexOf(" ") < 0, 0, str.Substring(0, Math.Min(str.Length, 24)).LastIndexOf(" "))), str)

                                              
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by: jeflabaPosted on 2009-09-21 at 19:21:36ID: 25389186

Quite elegant Cayman21!  I can see I need to get my head out of the VB6 sand and start to learn VB.NET.

 

by: Cayman21Posted on 2009-09-21 at 19:39:21ID: 25389235

Some would say elegant, some would say messy.  :)

 

by: andy7789Posted on 2009-09-21 at 19:44:40ID: 25389248

It looks quite elegant to me (though I don't like IIf as it could be tricky sometimes). Anyway I need two solutions - one for VB.NET and one for VBA excel.... so I could use both :)

 

by: Cayman21Posted on 2009-09-21 at 19:49:08ID: 25389262

Sorry, I can't help you with the VBA, I haven't done that for years.

 

by: jeflabaPosted on 2009-09-21 at 19:59:06ID: 25389284

I see one other gotcha.  When you are replacing possible end of word characters with a space you might replace a period when it is actually part of the word as in Mr., Mrs., Dr., etc.  Again, knowing what your input data will look like and what is acceptable output will determine just how complicated you need to make your algorithm.

 

by: Cayman21Posted on 2009-09-21 at 20:03:24ID: 25389296

Mine should work for that.  It treats the . as part of the word.  The only way it might not work the way you want it too, is if the . is the 25 char.  In which case it tosses out the whole last word.

20120131-EE-VQP-002

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