Swift httppost数据未插入MySQL数据库

Question

我正在尝试将数据发送到php并将其插入到mysql数据库，但它似乎不起作用。我尝试将数据发送到php只是为了将其编码为json并将其回显为swift并返回结果，因此这意味着php文件已接收到数据。但是，插入数据无效。

swift2 httppost

func post() {

    let myUrl = NSURL(string: "http://localhost:8080/json/json.php");
    let request = NSMutableURLRequest(URL:myUrl!);
    request.HTTPMethod = "POST"
    // Compose a query string
    let postString = "firstName=James&lastName=Bond";

    request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding);

    let task = NSURLSession.sharedSession().dataTaskWithRequest(request) {
        data, response, error in

        if error != nil
        {
            print("error=\(error)")
            return
        }

        // You can print out response object
        print("response = \(response)")

        // Print out response body
        let responseString = NSString(data: data!, encoding: NSUTF8StringEncoding)
        print("responseString = \(responseString)")

        //Let’s convert response sent from a server side script to a NSDictionary object:

        do{

            let myJSON = try NSJSONSerialization.JSONObjectWithData(data!, options: .MutableLeaves) as? NSDictionary

            if let parseJSON = myJSON {
                // Now we can access value of First Name by its key
                let firstNameValue = parseJSON["firstName"] as? String
                print("firstNameValue: \(firstNameValue)")
            }


        }catch let error as NSError {
            print("JSON Error: \(error.localizedDescription)")
        }

    }

    task.resume()

}

json.php

<?php
// Read request parameters
$firstName= $_REQUEST["firstName"];
$lastName = $_REQUEST["lastName"];// Store values in an array

$conn = mysqli("localhost", "root", "root", "notify");

$query = mysqli_query($conn, "INSERT INTO user values('', '$firstName',   
'$lastName')");

 ?>

Answer 1

如果让服务器只是响应请求，那么问题就出在服务器内部，而不是客户端代码。我建议在PHP代码中添加一些错误处理：

<?php

// specify that this will return JSON

header('Content-type: application/json');

// open database

$con = mysqli_connect("localhost","user","password","notify");

// Check connection

if (mysqli_connect_errno()) {
    echo json_encode(array("success" => false, "message" => mysqli_connect_error(), "sqlerrno" => mysqli_connect_errno()));
    exit;
}

// get the parameters

$field1 = mysqli_real_escape_string($con, $_REQUEST["firstName"]);
$field2 = mysqli_real_escape_string($con, $_REQUEST["lastName"]);

// perform the insert

$sql = "INSERT INTO user (first_name, last_name) VALUES ('{$field1}', '{$field2}')";

if (!mysqli_query($con, $sql)) {
    $response = array("success" => false, "message" => mysqli_error($con), "sqlerrno" => mysqli_errno($con), "sqlstate" => mysqli_sqlstate($con));
} else {
    $response = array("success" => true);
}

echo json_encode($response);

mysqli_close($con);

?>

笔记：

1. 我不建议以身份登录root。

2. 确保mysqli_real_escape_string用于保护自己免受SQL注入攻击（请参阅第1点）。

3. 我不知道您的user表中是否包含其他字段，但是如果这样，您可能要在insert语句中指定列名。即使您只有这两列，这也是“确保未来”代码的好方法。

4. 注意，我已经更改了它以生成JSON响应。我这样做是因为它使客户端代码更容易解​​析和处理响应。我留给NSJSONSerialization你。