无法从 iOS 添加条目到 MySQL

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尽管这里有不同的帖子，但我还没有找到在我的 MySQL 表中添加条目的解决方案。我从 iOS 应用程序发布了一个 json 字典，我想将它输入到数据库中。它由一个带有 4 个字段（名字、姓氏...）的条目组成。下面是服务器端的 php 代码（使用 MAMP 在本地运行）：

<?php

DEFINE('DB_USERNAME', 'root');
DEFINE('DB_PASSWORD', 'root');
DEFINE('DB_HOST', 'localhost');
DEFINE('DB_DATABASE', 'syncList');

$mysqli = new mysqli(DB_HOST, DB_USERNAME, DB_PASSWORD, DB_DATABASE);

if (mysqli_connect_error()) {
    die('Connect Error (' . mysqli_connect_errno(). ') ' . mysqli_connect_error());
}

$body = file_get_contents('php://input');
$jsonArray = json_decode($body);

echo json_encode($jsonArray); // ---> send back for testing.

$firstname = $jsonArray['firstname']; 
$firstname = $mysqli->real_escape_string($firstname);

$lastname = $jsonArray['lastname'];
$lastname = $mysqli->real_escape_string($lastname);

$eds = $jsonArray['eds'];

$dateOfBirth = $jsonArray['dateOfBirth'];

$sql = "INSERT INTO tbl_syncList (firstname, lastname, EDS, dateOfBirth) VALUES ('$firstname', '$lastname', '$eds', '$dateOfBirth')";


if ($mysqli->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $mysqli->error;
}

$mysqli->close();

?>

问题似乎是我的 json_array[key]（如 '$firstname'）无法识别，因为当我将 json_array 写入文件时，我看到密钥在方括号内，就像它是一个数组一个元素：

stdClass Object
(
    [firstname] => Robert
    [eds] => 1234567
    [lastname] => Redford
    [dateOfBirth] => 12.01.1965
)

这是一个正常的发现还是可以解释为什么我不能在我的 sql 语句中提取值？而且我不能返回一个数组对象而不是一个 stdClass 对象，即使我使用 json_decode($body, TRUE)。

我添加了 obj-C 代码，使用 POST 方法发送我的字典。这是调用上面的php脚本。

NSURLSessionConfiguration *defaultConfigObject = [NSURLSessionConfiguration defaultSessionConfiguration];
NSURLSession *defaultSession = [NSURLSession sessionWithConfiguration: defaultConfigObject delegate:nil delegateQueue: [NSOperationQueue mainQueue]];

NSMutableDictionary *dictionary = [[NSMutableDictionary alloc] init];
[dictionary setValue:@"Robert" forKey:@"firstname"];
[dictionary setValue:@"Redford" forKey:@"lastname"];
[dictionary setValue:@"1234567" forKey:@"eds"];
[dictionary setValue:@"12.01.1965" forKey:@"dateOfBirth"];

NSError *error;

//serialize the dictionary data as json
NSData *data = [NSJSONSerialization dataWithJSONObject:dictionary options:0 error:&error];

NSString *urlString = @"http://192.168.1.106:8888/postPerson.php";
NSURL *url = [NSURL URLWithString:urlString];
NSMutableURLRequest *urlRequest = [NSMutableURLRequest requestWithURL:url];

[urlRequest setHTTPMethod:@"POST"];
[urlRequest setHTTPBody:data]; //set the data as the post body
[urlRequest addValue:@"postValues" forHTTPHeaderField:@"METHOD"];
[urlRequest setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[urlRequest setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[urlRequest addValue:[NSString stringWithFormat:@"%d",data.length] forHTTPHeaderField:@"Content-Length"];

NSURLSessionDataTask *dataTask =[defaultSession dataTaskWithRequest:urlRequest completionHandler:^(NSData *dataRaw, NSURLResponse *response, NSError *error) {
    NSDictionary *json = [NSJSONSerialization
                          JSONObjectWithData:dataRaw
                          options:kNilOptions error:&error];

    NSString *result = [NSString stringWithFormat:@"%@", json];

    if (error) {
        UIAlertView * av = [[UIAlertView alloc] initWithTitle:[NSString stringWithFormat:@"Got error %@.\n", error] message:nil delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil, nil];
        [av show];
    }

    UIAlertView *av = [[UIAlertView alloc] initWithTitle:result message:nil delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil, nil];
    [av show]; // --> this is to check the array sent with POST-method which is json_encode() in the php script above.

我希望这可能有用...

Answer 1

由于我们处理的是字符串，因此您的 VALUES 中的变量需要被引用：

VALUES ('$firstname', '$lastname', '$eds', '$dateOfBirth')

有关字符串文字的更多信息，请访问：

• http://dev.mysql.com/doc/refman/5.1/en/string-literals.html