iPhone rangeOfString
Hello I have a string like this:
Results for: 123D12
2010
2009
2008
2007
2006
YEAR: 2006 WEEK: 2 PRODIDNUM: 37911
ACCESSKEY: FA3540B52F
2005
2004
YEAR: 2004 WEEK: 22 PRODIDNUM: 46178
ACCESSKEY: 58B2509373
and I want to spot the ACCESSKEY to help m开发者_如何学Ce get the hexadecimal(in this case are FA3540B52F and 58B2509373).
The issue is that when I use rangeOfString to get the accesskey it stops only to the first one! This is my code:
if ([strippedString rangeOfString:@"ACCESSKEY"].location != NSNotFound ) {
NSUInteger int1=[strippedString rangeOfString:@"ACCESSKEY"].location;
NSString *finssid = [strippedString substringWithRange:NSMakeRange(int1+11,10)];
NSLog(@"Output = Found it %d \n",int1);
NSLog(@"Output =%@ \n \n",finssid);
}
What I am doing wrong here?
Try this:
NSRange searchRange = NSMakeRange(0, [strippedString length]);
BOOL keepGoing = YES;
// Find all ssid
while (keepGoing) {
NSRange accessWord = [strippedString rangeOfString:@"ACCESSKEY" options:NSCaseInsensitiveSearch range:searchRange];
if (accessWord.location != NSNotFound) {
// since we have found the access key, we can assume somethings
// ACCESSKEY: FA3540B52F
int pos = accessWord.location + 11;
NSString *ssid = [strippedString substringWithRange:NSMakeRange(pos, pos + 10)];
NSLog(@"SSID: %@", ssid);
// reset our search up a little for our next loop around
searchRange = NSMakeRange(pos, [strippedString length] - pos);
} else {
keepGoing = NO;
}
}
`
Use rangeOfString:options:range:
instead of rangeOfString:
. Save the index you found a match at, then using the range:
argument start on the index after that. This way you loop through and previous matches fall outside the range you are looking in.
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