#1
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Replacing A Line
I'm trying to figure out how to replace a line in a resource file.
I have a line like: VALUE "SpecialBuild", "1/24/2007 - 12:02:02 PM" I want to replace the date and time part with the current date and time. The problem is the date and time currently there will be different on each build. I'm doing a find on VALUE "SpecialBuild", and I don't really care about the rest of it. But I don't know how to do that. |
#2
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See attached sample.
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#3
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Thanks for the quick response, that worked great.
Could you please explain the regular expression line that you are using here? What does this tell me? (VALUE\s*"SpecialBuild"\s*,\s*")[[^"]]+" Then the replacement line ($1)[vbld_FormatDateEx(Now, "m/d/yyyy - H:MM:SS AP")]" Thanks Joe |
#4
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#5
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I'm not an expert in Regular expression could you advice how i can replace a text eg. Testing 1234 with Testing 5678
The 1234 will change so i cannot fix the value. Please advice. Thanks, angpy |
#6
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got it resolved.
All i need is Testing [[^*]] and $1Testing 5678 |
#7
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Huh... somehow some of my content is being removed.
I have Test1 and Test2 being removed after i did a regular expression Testing 1234 to Testing 5678. Any idea? |
#8
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Please ZIP and send or post:
1) The info from Help | About | Install Info 2) The .bld file 3) The file being searched/replaced Thanks. |
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