Question

我正在尝试检查用户输入的网址，但我正在对抗一些错误和警告。

-(BOOL) textFieldShouldReturn:(UITextField *)textField {
    //check "http://"
    NSString *check = textField.text;
    NSString *searchString = @"http://";
    NSRange resultRange = [check rangeWithString:searchString];
    BOOL result = resultRange.location != NSNotFound;
    if (result) {
        NSURL *urlAddress = [NSURL URLWithString: textField.text];
    } else {
        NSString *good = [NSString stringWithFormat:@"http://%@", [textField text]];
        NSURL *urlAddress = [NSURL URLWithString: good];
    }
    // open url
    NSURLRequest *requestObject = [NSURLRequest requestWithURL:urlAddress];
}

他们说：

NSString可能无法回复-rangeWithString   条件“if ... else”（对于两者）的未使用变量urlAddress
  urlAddress未声明的：URLRequest

有谁知道该怎么做？

Answer 1

NSString响应rangeOfString:，而不是rangeWithString:。

变量urlAddress在if语句和else语句中都被声明。这意味着它只存在于该范围内。离开if / else语句后，变量就消失了。

对于URL，最好是以方案开头（如“http：//”），您的代码很乐意接受apple.http：//.com为有效。

您可以使用hasPrefix:方法，如下所示：

BOOL result = [[check lowercaseString] hasPrefix:@"http://"];
NSURL *urlAddress = nil;

if (result) {  
    urlAddress = [NSURL URLWithString: textField.text];
}
else {
    NSString *good = [NSString stringWithFormat:@"http://%@", [textField text]];
    urlAddress = [NSURL URLWithString: good];
}

NSURLRequest *requestObject = [NSURLRequest requestWithURL:urlAddress];

Answer 2

if ( [[url lowercaseString] hasPrefix:@"http://"] )
    return url;
else
    return [NSString stringWithFormat:@"http://%@", url];

Answer 3

我相信你想要rangeOfString方法。

Answer 4

要删除urlAddress警告，您应在NSURL *urlAddress上方if..else声明：

NSURL *urlAddress = nil;
if (result) {  
    urlAddress = [NSURL URLWithString: textField.text];    
} else {   
    NSString *good = [NSString stringWithFormat:@"http://%@", [textField text]];  
    urlAddress = [NSURL URLWithString: good];     
}

Answer 5

使用此

NSString *stringUrl = url.absoluteString;

if ([stringUrl localizedCaseInsensitiveContainsString:@"your string"]) {
    //do something
}

它首先将url转换为字符串，然后检查此url是否包含您的字符串

Answer 6

如果你有一个NSURL用

将它转换为NSString

   NSString stringURL = url.absoluteString

然后检查此NSString是否包含带有此代码的http：//

 [stringURL containsString: @"http://"]


    - (BOOL)containsString:(NSString *)string {
    return [self containsString:string caseSensitive:NO];
}

- (BOOL)containsString:(NSString*)string caseSensitive:(BOOL)caseSensitive {
    BOOL contains = NO;
    if (![NSString isNilOrEmpty:self] && ![NSString isNilOrEmpty:string]) {
        NSRange range;
        if (!caseSensitive) {
            range =  [self rangeOfString:string options:NSCaseInsensitiveSearch];
        } else {
            range =  [self rangeOfString:string];
        }
        contains = (range.location != NSNotFound);
    }

    return contains;
}

检查包含“http：//”URL的字符串

6 个答案: