使用URL參數聲明失敗

在我的項目中，我需要向客戶端發送電子郵件，但我面臨斷言失敗無效的參數。我檢查了所有解決方案，但沒有解決問題。使用URL參數聲明失敗

-(void)EmailWebservice 
{ 
    [MBProgressHUD showHUDAddedTo:self.view animated:YES]; 

    AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager]; 
    AFJSONRequestSerializer *serializer = [AFJSONRequestSerializer serializer]; 
    [serializer setValue:@"application/json" forHTTPHeaderField:@"Content-Type"]; 
    [serializer setValue:@"application/json" forHTTPHeaderField:@"Accept"]; 
    manager.requestSerializer = serializer; 

    NSString *getPostLink = [NSString stringWithFormat:@"http://Dummy.com/sendmail/?business_email=%@&_name=%@",TextEmail.text,[[NSUserDefaults standardUserDefaults]valueForKey:@"USERNAME"]]; 

    NSString *encoded = [getPostLink stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]; 

    NSLog(@"Encoding value is %@",encoded); 

    [manager GET:getPostLink parameters:nil success:^(AFHTTPRequestOperation *operation, id responseObject) { 

     NSLog(@"JSON: %@", responseObject); 
     [MBProgressHUD hideHUDForView:self.view animated:YES]; 

}failure:^(AFHTTPRequestOperation *operation, NSError *error) { 

     NSLog(@"the failure is %@", error); 
     [MBProgressHUD hideHUDForView:self.view animated:YES]; 

    }]; 
}

錯誤：

Terminating app due to uncaught exception 'NSInternalInconsistencyException', reason: 'Invalid parameter not satisfying: URLString'

來源

2016-01-05 Arun

它的意圖是'query'分隔符？'是在'path'的最後一個'/'之後出現的嗎？ – vadian

嘗試使用URL編碼您的參數。您的網址中可能有空格或其他非網址字符，因爲您直接替換了該處的電子郵件地址和客戶名稱。請參閱http://stackoverflow.com/questions/718429/creating-url-query-parameters-from-nsdictionary-objects-in-objectivec瞭解如何更優雅地做到這一點。 –

是啊，它是@vadian – Arun

錯誤是告訴你，有什麼不對您的網址。

NSLog(@"Encoding value is %@",encoded);

你可以粘貼到您選擇的瀏覽器，看看是否在頁面加載：我會通過打印網址，你正在與線做啓動？

你可能在你的URL中的參數嘗試硬編碼值：

business_email 
fave_customer_name

編碼您的參數：

-(void)EmailWebservice 
{ 
    [MBProgressHUD showHUDAddedTo:self.view animated:YES]; 

    AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager]; 
    AFJSONRequestSerializer *serializer = [AFJSONRequestSerializer serializer]; 
    [serializer setValue:@"application/json" forHTTPHeaderField:@"Content-Type"]; 
    [serializer setValue:@"application/json" forHTTPHeaderField:@"Accept"]; 
    manager.requestSerializer = serializer; 

    NSString *email = [TextEmail.text stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]; 
    NSString customer = [[NSUserDefaults standardUserDefaults]valueForKey:@"USERNAME"] stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]; 

    NSString *getPostLink = [NSString stringWithFormat:@"http://Dummy.com/csrestapi/sendintroemail/?business_email=%@&fave_customer_name=%@",email,customer]; 

    NSString *encoded = [getPostLink stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]; 

    NSLog(@"Encoding value is %@",encoded); 

    [manager GET:getPostLink parameters:nil success:^(AFHTTPRequestOperation *operation, id responseObject) { 

     NSLog(@"JSON: %@", responseObject); 
     [MBProgressHUD hideHUDForView:self.view animated:YES]; 

    }failure:^(AFHTTPRequestOperation *operation, NSError *error) { 

     NSLog(@"the failure is %@", error); 
     [MBProgressHUD hideHUDForView:self.view animated:YES]; 

    }]; 
}

，因爲它涵蓋了這種類型的問題，您應該閱讀本stackoverflow question。

來源

2016-01-05 13:49:25

使用URL參數聲明失敗

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