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我需要从 xcode 打开我的 php url,并且我在 url 中传递了一个变量。我正在执行以下代码,但它不起作用,我已经检查过在 Web 浏览器中输入此 url 是否有效。

NSString *urlformatted = [NSString stringWithFormat:@"http://arcamm.uc3m.es/notifications/saveID.php?id=%@",deviceToken];

    NSLog(@"web service URL: %@", urlformatted);

    [urlformatted stringByAppendingString:[urlformatted stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];

    NSURLRequest *request = [NSURLRequest requestWithURL:[NSURL URLWithString: urlformatted] cachePolicy:NSURLRequestUseProtocolCachePolicy timeoutInterval:60.0];

    NSURLConnection *connection = [[NSURLConnection alloc] initWithRequest: request delegate:self startImmediately:YES];
    NSLog(@"connection: %@", [connection debugDescription]);
    if (connection) {
        NSLog(@"Connection succeeded");
    } else {
        NSLog(@"Connection failed");
    }

它告诉我连接成功,但变量没有出现在我的数据库中。而且我也得到了 deviceToken 的正确值。

编辑:发现我的错误,我收到了格式化的 deviceToken,所以我做了以下操作来转义 <> 和“”

NSString *deviceTokenString = [[deviceToken description] stringByTrimmingCharactersInSet:[NSCharacterSet characterSetWithCharactersInString:@"<>"]];
    deviceTokenString = [deviceTokenString stringByReplacingOccurrencesOfString:@" " withString:@""];
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1 回答 1

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编辑:发现我的错误,我收到了格式化的 deviceToken,所以我做了以下操作来转义 <> 和“”

NSString *deviceTokenString = [[deviceToken description] stringByTrimmingCharactersInSet:[NSCharacterSet characterSetWithCharactersInString:@"<>"]];
    deviceTokenString = [deviceTokenString stringByReplacingOccurrencesOfString:@" " withString:@""];
于 2013-06-07T09:45:24.577 回答