从Xcode中获取MySQL的特定用户数据

Question

我的代码工作正常，我得到了我需要的所有信息，但是我必须从用户那里写下我想要的信息在我的php文件中。我想从用户那里获得我在文本字段中写下他的电子邮件的信息。我怎么能这样做？

这是我的php文件：

<?php

$servername = "*****";
$username = "*****";
$password = "*****";
$dbname = "*****";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

//I want to get this email information from my xcode text field   
    $sql = "SELECT id, nome, email, cpf, senha FROM usuario_log where email = 'albino@peggou.com.br'";
    $result = $conn->query($sql);

    if ($result->num_rows > 0) {
        // output data of each row
        while($row = $result->fetch_assoc()) {
            $resultado = array("status" => "0", "id" => $row["id"], "nome" => $row["nome"], "email" => $row["email"], "cpf" => $row["cpf"], "senha" => $row["senha"]);
        }
    } else {
         echo "0 results";
    }
    echo json_encode($resultado);

        $conn->close();
?>

这是我的.m文件，所有信息都显示在我的日志中。

NSString *urlString = [NSString stringWithFormat: @"http://peggou.com.br/dados.php"];
    NSURL *url = [NSURL URLWithString:urlString];
    NSData *data = [NSData dataWithContentsOfURL:url];
    NSDictionary *json = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:nil];
    NSString *statusUser = [json objectForKey:@"status"];
    NSString *idUser = [json objectForKey:@"id"];
    NSString *nomeUser = [json objectForKey:@"nome"];
    NSString *emailUser = [json objectForKey:@"email"];
    NSString *cpfUser = [json objectForKey:@"cpf"];
    NSString *senhaUser = [json objectForKey:@"senha"];
    NSLog(@"\n Status: %@,\n ID: %@,\n Nome: %@,\n E-mail: %@,\n CPF: %@,\n Senha: %@ \n", statusUser, idUser, nomeUser, emailUser, cpfUser, senhaUser);