检查包含& quot; http://& quot;的URL的字符串 [英] Check string containing URL for "http://"

问题描述

我正在尝试检查用户输入的URL，但是我正在与一些错误和警告进行斗争.

I am trying to check the URL entered by the user, but I am fighting against some errors and warnings.

-(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 可能不响应 -rangeWithString
条件为"if…else"(两个都适用)的未使用变量 urlAddress (两者)
urlAddress 未声明:在 URLRequest

NSString may not respond to -rangeWithString
Unused variable urlAddress in the condition "if … else" (for both)
urlAddress undeclared : in the URLRequest

有人知道该怎么做吗?

推荐答案

NSString响应 rangeOfString:，而不是 rangeWithString:.

NSString responds to rangeOfString:, not rangeWithString:.

在 if 语句和 else 语句中都声明了变量urlAddress.这意味着它仅存在于该范围内.离开if/else语句后，变量就消失了.

The variable urlAddress is declared both in the if statement, and in the else statement. That means it only lives in that scope. Once you leave the if/else statement, the variable is gone.

对于URL，最好以方案开头(例如"http://")，并且您的代码将很高兴接受apple.http://.com为有效.

For a URL it's best if it begins with the scheme (like "http://"), and your code will gladly accept apple.http://.com as being valid.

您可以改为使用 hasPrefix:方法，如下所示:

You can use the hasPrefix: method instead, like this:

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];

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