检查包含“http：//”的网址的字符串 [英] Check string containing URL for "http://"

问题描述

我正在尝试检查用户输入的网址，但我打击一些错误和警告。

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if ... else中未使用的变量urlAddress（对于两者）

urlAddress undeclared：在URLRequest中

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

有没有人知道该怎么办？

Does anyone have any idea what to do?

推荐答案

NSString响应 rangeOfString：，而不是 rangeWithString：。

NSString responds to rangeOfString:, not rangeWithString:.

变量urlAddress在 if 语句和 else 语句中声明。这意味着它只生活在那个范围内。一旦你离开if / else语句，变量就会消失。

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.

对于一个URL，最好是以方案开头（如http：//您的代码将很乐意接受apple.http：//.com作为有效的。

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.

您可以使用 hasPrefix：方法，如下：

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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