Question

我通过iOS应用程序在我的服务器上运行php脚本，该应用程序检查输入到UITextFields中的详细信息并允许用户登录。从PHP脚本返回的响应只是

   <?php require_once("includes/connection_auth.php"); 
?>
<?php


    $username = $_GET["userName"];
    $userpassword = $_GET["userPassword"];

    //check if users username exists in database
    $selectQuery = mysqli_query($connection, "SELECT * FROM MVUsers WHERE userName = '$username' ");
    if (!$selectQuery) {
        printf("Error: %s\n", mysqli_error($connection));
        exit();
    }


    if ($row = mysqli_fetch_assoc($selectQuery)){
        //user found in database check password matches



         if($row ['userPassword'] == $userpassword){
            //password matches check if confirmed email
                if($row ['userHasConfirmedEmail'] == 1){
                     //allow user to log in
                    echo (0);
                  }
                  else{
                  echo (1);

                 }

          }
          else{
            //password does not match warn user

            echo (2);
               }    

    }



    else{

    echo(3);

   }
    //close connection to database
    mysqli_close($connection);

?>

这个回显的数字作为NSData返回到我的应用程序中，并且在NSURLConnectionDataDelegate回调之一中，我将该数据转换为NSString，然后从该字符串中获取intValue。

- (void)connectionDidFinishLoading:(NSURLConnection *)connection {
NSString *string = [[NSString alloc]initWithData:[self receivedData] encoding:NSUTF8StringEncoding];

NSLog(@"Connection finished loading with string :%@",string);

int returnedInt = [string intValue];
 NSLog(@"%i",returnedInt);

// returnedInt is the compared inside a switch statement. 
}

两个日志语句对于相同的数据具有不同的输出。

2014-02-05 13:20:47.363 MotoVlogger[50944:70b] Connection finished loading with string :






3
2014-02-05 13:20:47.363 MotoVlogger[50944:70b] 0

为什么在一个实例中为3，在另一个实例中为0，为什么在第一个日志语句中都添加了空格？

Answer 1

字符串有一堆换行符;去除那些，你应该得到预期的intValue

Answer 2

试试这个：

-(NSInteger)convert:(NSString *)str
{
    str =[str stringByReplacingOccurrencesOfString:@" " withString:@""];
    NSInteger returnedInt = [str integerValue];
    NSLog(@"%ld",(long)returnedInt);
    return (long)returnedInt;
}


-(int)convert:(NSString *)str
{
    str =[str stringByReplacingOccurrencesOfString:@" " withString:@""];
    int returnedInt = [str intValue];
    NSLog(@"%i",returnedInt);
    return returnedInt;
}

Answer 3

看到我需要的角色就在最后我只是使用

string =[string substringFromIndex:[string length]-1];

有效，但想知道空白的原因

编辑：按照奥斯汀的建议，我不得不删掉很多帖子根据他的回答在评论中这样做的代码。

从字符串中获取错误的int值

3 个答案: