连接到mysql数据库的iOS不显示新行

Question

我有这个代码，它从MySQL数据库中检索用户的名字。但是，它不会加载我刚刚放入数据库的新条目。

-(void) fetchUsers {
    @try
    {
        NSURL * url = [NSURL URLWithString:@"http://malaebna.com/usersMalaebna.php"];
        NSData *data = [NSData dataWithContentsOfURL:url];

        NSMutableArray *jsonArray2 = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:nil];

        preCalcArray = [[NSMutableArray alloc] init];

        for (int i = 0; i < jsonArray2.count; i++) {
            NSString *mEmail = [[jsonArray2 objectAtIndex:i] objectForKey:@"email"];
            NSString *mPass = [[jsonArray2 objectAtIndex:i] objectForKey:@"password"];
            NSString *mNumber = [[jsonArray2 objectAtIndex:i] objectForKey:@"number"];
            NSString *mReservations = [[jsonArray2 objectAtIndex:i] objectForKey:@"reservations"];

            NSLog(mEmail);
            NSLog(mPass);

            [preCalcArray addObject:[[users alloc] initWithEmail:mEmail andPassword:mPass andNumber:mNumber andReservations:mReservations]];

            //[self.tableView reloadData];
        }

    }
    @catch (NSException *ex) {
    }
}

例如，我有一个创建帐户页面，我将创建一个帐户exampleemail@gmail.com。

它将成功插入数据库。

但是，当我返回登录页面并获取用户时，不会获取新条目。

我必须等待或刷新我的Chrome浏览器上的PHP页面才能显示在应用程序中。

这是我的php fetch脚本：

<?php

// Create connection
$con = new mysqli("localhost","","","");
 $con ->set_charset("utf8");

// Check connection
if (mysqli_connect_errno())
{
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

// This SQL statement selects ALL from the table 'Locations'
$sql = "SELECT * FROM users";

// Check if there are results
if ($result = mysqli_query($con, $sql))
{
    // If so, then create a results array and a temporary one
    // to hold the data
    $resultArray = array();
    $tempArray = array();

    // Loop through each row in the result set
    while($row = $result->fetch_object())
    {
        // Add each row into our results array
        $tempArray = $row;
        array_push($resultArray, $tempArray);
    }

    // Finally, encode the array to JSON and output the results
    echo json_encode($resultArray, JSON_UNESCAPED_UNICODE);
}


// Close connections
mysqli_close($con);
?>

我非常需要帮助。