preg_replace() help: trim right space of match before replace

[jeff00seattle]

Hi

 

I having beating my head over the following, I want to trim all right spaces from match before replace, and then with replacement append a value with a single space, using preg_replace().

 

For example, I have a text that is filled with dates with Month/Day and Month/Day/Year. I want to replace all Month/Day(s) and append with Year. To elaborate:

Match 6/13 => Replace 6/13/2009

6/13/2009 => Do nothing

 

So far, the following Match regex works: '[0-9]{1,2}\/[0-9]{1,2}\s{1,}'

 

4/28 hfdkkdi sielsl fidijg. 5/14      fiisollei sikedi jdsildh 6/3 diel dkisl hgiels dhpwld, 6/13/2009 yeus spwiv wojje rtyu

 

But replace '\0/2009 ' does the following:

 

4/28 /2009 hfdkkdi sielsl fidijg. 5/14    /2009 fiisollei sikedi jdsildh 6/3 /2009 diel dkisl hgiels dhpwld, 6/13/2009 yeus spwiv wojje rtyu

 

Notice that it does not replace the space(s) after the Month/Day

However, when using this re

 

So, how can I create a replace that removes all spaces after the match and then append replace pattern so it would look like this?:

 

4/28/2009 hfdkkdi sielsl fidijg. 5/14/2009 fiisollei sikedi jdsildh 6/3/2009 diel dkisl hgiels dhpwld, 6/13/2009 yeus spwiv wojje rtyu

 

Thanks

 

Jeff in Seattle