Your current pattern is greedy and matches more than intended judging from the resulting replacement. "Greedy" means the usage of is gobbling up more than you intend, so to make it non-greedy you need to add a? After it so it matches the least amount of characters possible:?
A pattern like is greedy because you intend for it to stop at the first quote encountered, but it actually continues matching content till it hits the final quote in the string, if one exists. The non-greedy solution is to use? Or ^\" Try this pattern instead: string pattern = @"\.
+? Url=(?"". +?"").
+? ContentDatabase=(?"". +?
""). +?"; string replacement = "stsadm. Exe -o deletecontentdb -url ${url} -databasename ${databasename} -databaseserver myfixedservername"; string result = Regex.
Replace(input, pattern, replacement); Console. WriteLine(result).
Your current pattern is greedy and matches more than intended judging from the resulting replacement. "Greedy" means the usage of . * is gobbling up more than you intend, so to make it non-greedy you need to add a?
After it so it matches the least amount of characters possible: . *?. A pattern like \".
*\" is greedy because you intend for it to stop at the first quote encountered, but it actually continues matching content till it hits the final quote in the string, if one exists. The non-greedy solution is to use \". +?
\" or \"^\"+\". Try this pattern instead: string pattern = @"\. +?
Url=(?"". +? "").
+? ContentDatabase=(?"". +?"").
+? "; string replacement = "stsadm. Exe -o deletecontentdb -url ${url} -databasename ${databasename} -databaseserver myfixedservername"; string result = Regex.
Replace(input, pattern, replacement); Console. WriteLine(result).
Can tell regexps in VB are severely impared! Then get the numeric value for the char. Unfortunatley in VB you can not pass submatches to a funtion.
Only work inside the expression. Let me know. Aside from being brief, it works!
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