how can i replace a string with some pattern matching

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Hi
Can anybody help me to get the solution using regular expression in c#
basically i have to add some text before *p0x0Y pattern . In this pattern p,x,y will be fix value and 0 can be replace by any number .

i have to edit a

C#

string str= "l8c1E*p0x0Y"

In this string *p0x0Y can be *p1x1Y or *p2x2Y .

and after editing output should be according *p0x0Y format

C#

string output = "18c1EXXX*p0x0Y"

Thanks in advance .

Posted 8-Oct-13 23:55pm

lalit.mca2006

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OriginalGriff · Accepted Answer · 2013-10-09T00:03:00

Try:

C#

string str = "l8c1E*p0x0Y";
string output = Regex.Replace(str, @"(\*p\dx\dY)", @"XXX$1");

"Hi OriginalGriff , Your solution worked for me , now i want to ask one more thing ... i have to replace left side string into right side string in below text .. In this text "left arrow sign" is one symbol which i could not copy paste so i wrote it in text ,its just one symbol. So please give me regular expression code to replace it . "left arrow sign"*p0x0Y --> "left arrow sign"xxx "left arrow sign"*p0x0Y Am i clear about my requirement , Original ?"

Without knowing the exact character, I can't be sure, but...the dot character matches anything, so it might be that what you want is:

C#

string output = Regex.Replace(str, @"(.)(\*p\dx\dY)", @"$1XXX$1$2");

Give it a try. In the replacement code "$n" indicates a numbered match group from the match string, so "$1XXX$1" should give you "your-character XXX your-character" in the output.

If you are going to use regexes much, I'd recommend you get a copy of Expresso [^] - it's free, and it examines and generates Regular expressions.