If I have this string: "wwww:xxxx:yyyy:zzzzzz:0"
How can I return only the "zzzzzz:0" value? Knowing that zzzzz will not always be the same length.
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2 Answers
Use Patterns for this:
string.match("wwww:xxxx:yyyy:zzzzzz:0", "[^:]+:[^:]+$") --> 'zzzzzz:0'
This is naïve and focuses entirely on expressing this particular intent. [^:]+ matches a sequence of one or more characters that are not :, : after that matches itself, and $ matches the end of the string.
Based on the format of your string, you might want to look into splitting the string using : as a separator. This can be accomplished with e.g., string.gmatch and the [^:]+ pattern from above. See Split string in Lua? for a detailed explanation of this problem.
In a complex situation where the components are expected to e.g., support quoting or escape sequences, and if you are highly concerned with time or memory optimisation you should look to use a csv-like parser.
Try also this somewhat simpler pattern:
string.match("wwww:xxxx:yyyy:zzzzzz:0", ".+:(.+:.+)$")
It looks for the last item of the form foo:bar.
1 Comment
Aki
Note that in this case the string must have at least three component (which is a fair assumption in situations like this). Otherwise, one can e.g., prepend delimiting character to the string (in that case you only need to ensure two components, and that's even a better assumption since we're looking for the last two).