我有一个 NSString，例如：
"Had a #great time at the #party last night."
我想把它分成一个数组，如下所示：
"Had a "
"#great"
" time at the "
"#party"
" last night."
我怎么能这样做？
问问题
924 次
9 回答
4
NSString *str = @"Had a #great time at the #party last night.";
NSMutableArray *arr = [[NSMutableArray alloc] init];
NSArray *array = [str componentsSeparatedByString:@"#"];
NSMutableString *retStr= [[NSMutableString alloc] initWithString:[array objectAtIndex:0]];
[arr addObject:retStr];
for(int i=1 ; i<[array count];i++)
{
NSArray *array1 = [[array objectAtIndex:i] componentsSeparatedByString:@" "];
{
NSMutableString *retStr= [[NSMutableString alloc]init];
for (int i = 0;i< [array1 count]; i++)
{
if(i==0)
{
[retStr appendFormat:@" #%@ ",[array1 objectAtIndex:i]];
[arr addObject:retStr];
retStr= [[NSMutableString alloc]init];
}
else
{
[retStr appendFormat:@"%@ ",[array1 objectAtIndex:i]];
}
}
[arr addObject:retStr];
}
}
NSLog(@"%@",arr);
您将根据需要获得正确的输出
于 2013-04-29T12:29:21.253 回答
3
像这样尝试它会帮助你，
NSString *str=@"how are #you #friend";
NSArray *arr=[str componentsSeparatedByString:@" "];
NSPredicate *p = [NSPredicate predicateWithFormat:@"not SELF contains '#'"];
NSArray *b = [arr filteredArrayUsingPredicate:p];
NSLog(@"%@",b);
以上谓词将返回不包含'#'符号的单词
NSPredicate *p = [NSPredicate predicateWithFormat:@"not SELF like '#*'"];
它会返回不以字母“#”开头的单词
运单：-
(
how,
are
)
EDIT:-
NSString *str=@"how are #you #friend";
NSArray *arr=[str componentsSeparatedByString:@"#"];
NSMutableArray *result=[[NSMutableArray alloc]initWithObjects:[arr objectAtIndex:0], nil];
for(int i=1;i<[arr count];i++){
[result addObject:[NSString stringWithFormat:@"#%@",[arr objectAtIndex:i]]];
}
NSLog(@"%@",result);
运单：-
(
"how are ",
"#you ",
"#friend"
)
于 2013-04-29T12:17:29.700 回答
1
尝试使用这个正则表达式：(#.+?\\b)|(.+?(?=#|$))
它找到以主题标签开头的单词和以主题标签结尾的子序列
NSString * string = @"Had a #great time at the #party last night.";
NSError * error = nil;
NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:@"((#.+?\\b)|(.+?(?=#|$)))"
options:0
error:&error];
NSArray * matches = [regex matchesInString:string options:0 range:NSMakeRange(0, [string length])];
for (NSTextCheckingResult* match in matches ) {
NSLog(@"%@", [string substringWithRange:[match range]]);
}
输出：
2013-04-29 16:57:51.688 Had a
2013-04-29 16:57:51.689 #great
2013-04-29 16:57:51.690 time at the
2013-04-29 16:57:51.691 #party
2013-04-29 16:57:51.692 last night.
于 2013-04-29T12:58:57.500 回答
1
如果您想在字符串上一次通过有效地做到这一点，您可以尝试类似（临时代码 - 测试错误/边界情况等......）：
int main (int argc, const char * argv[])
{
NSString *msg = @"Had a #great time at the #party last night.";
Boolean inTag = NO;
NSMutableArray *segments = [[NSMutableArray alloc] init];
NSUInteger idx = 0;
NSUInteger i=0;
for (; i < [msg length]; i++)
{
unichar ch = [msg characterAtIndex:i];
if (inTag && ch == ' ')
{
[segments addObject:[msg substringWithRange:NSMakeRange(idx, i - idx)]];
idx = i;
inTag = NO;
}
if (ch == '#')
{
[segments addObject:[msg substringWithRange:NSMakeRange(idx, i - idx)]];
idx = i;
inTag = YES;
}
}
if (i > idx)
{
[segments addObject:[msg substringWithRange:NSMakeRange(idx, i - idx - 1)]];
}
for(NSString *seg in segments)
{
NSLog(@"%@", seg);
}
}
这输出：
2013-04-29 08:34:34.984 Craplet[95591:707] Had a
2013-04-29 08:34:34.986 Craplet[95591:707] #great
2013-04-29 08:34:34.986 Craplet[95591:707] time at the
2013-04-29 08:34:34.987 Craplet[95591:707] #party
2013-04-29 08:34:34.987 Craplet[95591:707] last night
于 2013-04-29T12:37:32.423 回答
0
尝试这个：
for (int i=0;i<[YourArray count];i++) {
NSString * mystr=[YourArray objectAtIndex:i];
NSString *temp=[mystr substringToIndex:1];
if (![temp isEqualToString:@"#"]) {
//add your string in new array and use this arry.....
}
}
于 2013-04-29T12:02:58.847 回答
0
试试这个代码
for (int i=0;i<[YourArray count];i++) {
NSString * str=[YourArray objectAtIndex:i];
NSString *myString=[str substringToIndex:1];
NSString *stringfinal = [myString
stringByReplacingOccurrencesOfString:@"#" withString:@""];
}
于 2013-04-29T12:12:44.617 回答
0
尝试
NSString *string = @"Had a #great time at the #party last night.";
NSArray *components = [string componentsSeparatedByString:@" "];
NSMutableArray *formattedArray = [NSMutableArray array];
NSMutableString *mutableString = [NSMutableString string];
for (NSString *string in components)
{
if (![string hasPrefix:@"#"]){
if (!mutableString){
mutableString = [NSMutableString string];
}
[mutableString appendFormat:@" %@",string];
}else{
if (mutableString) {
[formattedArray addObject:mutableString];
mutableString = nil;
}
[formattedArray addObject:string];
}
}
if (mutableString) {
[formattedArray addObject:mutableString];
}
NSLog(@"%@",formattedArray);
编辑 ：
(
" Had a",
"#great",
" time at the",
"#party",
" last night."
)
于 2013-04-29T12:22:37.173 回答
0
NSScanner的一次性解决方案
NSString *string = @"Had a #great time at the #party last night.";
NSMutableArray *array = [@[] mutableCopy];
NSScanner *scanner = [NSScanner scannerWithString:string];
while (![scanner isAtEnd]) {
NSString *s;
[scanner scanUpToString:@"#" intoString:&s];
if(s) [array addObject:s];
s = nil;
[scanner scanUpToString:@" " intoString:&s];
if(s) [array addObject:s];
}
结果：
(
"Had a ",
"#great",
"time at the ",
"#party",
"last night."
)
如果要保留前导空格，请将其稍微更改为
NSString *string = @"Had a #great time at the #party last night.";
NSMutableArray *array = [@[] mutableCopy];
NSScanner *scanner = [NSScanner scannerWithString:string];
BOOL firstSegment = YES;
while (![scanner isAtEnd]) {
NSString *s;
[scanner scanUpToString:@"#" intoString:&s];
if(s) [array addObject: (!firstSegment) ? [@" " stringByAppendingString:s] : s];
s = nil;
[scanner scanUpToString:@" " intoString:&s];
if(s) [array addObject:s];
firstSegment = NO;
}
结果：
(
"Had a ",
"#great",
" time at the ",
"#party",
" last night."
)
于 2013-05-01T23:32:35.303 回答
0
尝试使用正则表达式。使用此代码，您将能够提取主题标签：
NSString * string = @"Had a #great time at the #party last night.";
NSError * error = nil;
NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:@"((?:#){1}[\\w\\d]{1,140})" options:0 error:&error];
NSArray * matches = [regex matchesInString:string options:0 range:NSMakeRange(0, [string length])];
for ( NSTextCheckingResult* match in matches )
{
NSString * hashtag = [string substringWithRange:[match range]];
NSLog(@"match: %@", hashtag);
}
有了这个，您将能够建立您正在寻找的数组结果。
于 2013-04-29T12:26:05.267 回答