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ios 判断包含字符串

假设要查询的字符串味: NSString *searchStr = @"a string with substr"; 查询(是否)被包含的字符串为@"substr",则通过下面的判断: if ([searchStr rangeOfString:@"substr"].location != NSNotFound) { //条件为真,表示字符串searchStr包...

假设要查询的字符串味: NSString *searchStr = @"a string with substr"; 查询(是否)被包含的字符串为@"substr",则通过下面的判断: if ([searchStr rangeOfString:@"substr"].location != NSNotFound) { //条件为真,表示字符串searchStr包...

以下是自己封装的一个方法,可以根据返回的bool值进行判断 实例代码 -(BOOL)isEmpty:(NSString *) str { NSRange range = [str rangeOfString:@" "]; if (range.location != NSNotFound) { return YES; //yes代表包含空格 }else { return NO; //...

正则表达式不好返回下标,字符串直接用indexOf不就行了 public static void main(String[] args){ String msg = "MESSAGE=rpd[1810]: %DAEMON-4: bgp_listen_accept: Connection attempt from unconfigured neighbor: 123.255.91.29+62490"; Str...

//判断是否有中文 -(BOOL)IsChinese:(NSString *)str { for(int i=0; i< [str length];i++){ int a = [str characterAtIndex:i]; if( a > 0x4e00 && a < 0x9fff) { return YES; } } return NO; } 用法 if ([self PanDuan:@"aa!@#$%^&*(我)"]) { ...

设选定复选框时为区分字母大小写,不选定是为不区分字母大小写private sub command1_click()if check1.value=0 then m1=text1.text m2=text2.textelse m1=lcase(text1.text) m2=lcase(text2.text)endifa=len(m1)b=len(m2)if ab then msgbox "两...

假设要查询的字符串味: NSString *searchStr = @"a string with substr"; 查询(是否)被包含的字符串为@"substr",则通过下面的判断: if ([searchStr rangeOfString:@"substr"].location != NSNotFound) { //条件为真,表示字符串searchStr包...

同意一楼,要么都转成大写,要么都转成小写进行比较,contains是关键

rangeOfString 如果不够用的话,可以查询一下range开头的函数 NSString* str = @"1232343453453"; NSRange range = [str rangeOfString:@"a"]; if (range.length > 0) { }

假设要查询的字符串味: NSString *searchStr = @"a string with substr"; 查询(是否)被包含的字符串为@"substr",则通过下面的判断: if ([searchStr rangeOfString:@"substr"].location != NSNotFound) { //条件为真,表示字符串searchStr包...

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