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&abc で始まり & で終わる特定の文字列を取得する方法を知りたいです。had prefix と sufix で試しました。しかし、これは改行ではありません。

&xyz;123:183:184:142&
&abc;134:534:435:432&
&qwe;323:535:234:532&

私のコード:

NSMutableArray *substrings = [NSMutableArray new];
NSScanner *scanner = [NSScanner scannerWithString:s];
[scanner scanUpToString:@"&abc" intoString:nil]; //
NSString *substring = nil;
[scanner scanString:@"&abc" intoString:nil]; // Scan the # character
if([scanner scanUpToString:@"&" intoString:&substring]) {
   // If the space immediately followed the &, this will be skipped
   [substrings addObject:substring];
   NSLog(@"substring is :%@",substring);
}
// do something with substrings
[substrings release];

"scanner scanUpToString:@"&abc" を作成し、":"==3 を "#" までカウントする方法????

4

2 に答える 2

3 
    NSArray *arr = [NSArray arrayWithObjects:@"&xyz;123:183:184:142&",
                    @"&abc;134:534:435:432&",
                    @"&qwe;323:535:234:532&",
                    @"& I am not in it",
                    @"&abc I am out &" ,nil];

    NSPredicate *predicate = [NSPredicate predicateWithFormat:@"self BEGINSWITH[cd] %@ AND self ENDSWITH[cd] %@",@"&abc",@"&"];

    NSLog(@"Sorted Array %@",[arr filteredArrayUsingPredicate:predicate]);
    NSArray *sortedArray = [arr filteredArrayUsingPredicate:predicate];

    NSMutableArray *finalResult = [NSMutableArray arrayWithCapacity:0];
    for(NSString *string in sortedArray)
    {
        NSString *content = string;
        NSRange range1 = [content rangeOfString:@"&abc"];
        if(range1.length > 0)
            content = [content stringByReplacingCharactersInRange:range1 withString:@""];

        NSRange range2 = [content rangeOfString:@"&"];
        if(range2.length > 0)
            content = [content stringByReplacingCharactersInRange:range2 withString:@""];

        [finalResult addObject:content];
    }

    NSLog(@"%@",finalResult);
于 2013-04-11T15:45:28.040 に答える
0

NSRegularExpressionを使用してみてください:

- (BOOL)isValidString:(NSString *)string
{
    NSRegularExpression *regularExpression = [NSRegularExpression regularExpressionWithPattern:@"^&abc.*&$" options:0 error:NULL];
    NSTextCheckingResult *result = [regularExpression firstMatchInString:string options:0 range:NSMakeRange(0, [string length])];
    return (result != nil);
}
于 2013-04-11T15:52:38.850 に答える