dtcrw26206 2014-10-19 19:52
浏览 61
已采纳

POST从iOS到MySQL数据库

I'm trying to post data from my UITextFields to a MySQL database. The console says my connection is successful, but for some reason the data isn't posting to the database? Any idea as to why? Here's my code:

ViewController.h

-(IBAction)addParty:(id)sender;

@property (strong, nonatomic) IBOutlet UITextField *firstname;
@property (strong, nonatomic) IBOutlet UITextField *lastname;
@property (strong, nonatomic) IBOutlet UITextField *phone;
@property (strong, nonatomic) IBOutlet UITextField *email;
@property (strong, nonatomic) IBOutlet UITextField *guests;

@property (nonatomic, copy) NSDictionary *venueDetail;
@end

ViewController.m

- (IBAction)addParty:(id)sender
{

NSString *strURL = [NSString stringWithFormat:@"guestfirst=%@&guestlast=%@", firstname.text, lastname.text];

NSData *postData = [strURL dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];
NSString *postLength = [NSString stringWithFormat:@"%d",[postData length]];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
[request setURL:[NSURL URLWithString:@"http://myurl.com/phpfile.php"]];
[request setHTTPMethod:@"POST"];
[request setValue:postLength forHTTPHeaderField:@"Content-Length"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Current-Type"];
[request setHTTPBody:postData];


NSURLConnection *conn = [[NSURLConnection alloc]initWithRequest:request delegate:self];

if(conn)
{
    NSLog(@"Connection Successful");
}
else
{
    NSLog(@"Connection could not be made");
}

}

Post.php

<?Php
include('connect.php');

if (isset ($_POST["guestfirst"]) && isset ($_POST["guestlast"])){
    $guestfirst = $_POST["guestfirst"];
    $guestlast = $_POST["guestlast"];
} else {
    $guestfirst = "none";
    $guestlast = "none";
}

// Insert value into DB
$sql = "INSERT INTO events (guestfirst, guestlast) VALUES ('$guestfirst', '$guestlast');";
$res = mysql_query($sql,$conn) or die(mysql_error());

mysql_close($conn);

if($res) {          
$response = array('status' => '1');                 
} else {
die("Query failed");
}

echo json_encode($res);
exit();

?>
  • 写回答

1条回答 默认 最新

  • dongshao8566 2014-10-19 20:05
    关注

    There are a few problems I see with your current code.

    1. When you're fetching existing articles, you are assigning the query to $get_events, but fetching from $events.
    2. You're fetching before your insert. (maybe that's your intent?)
    3. You output more than one JSON object
    4. Your query uses $events rather than `events`.
    5. You call die() if a MySQL error occurs, so you will never receive a JSON response saying there was an error.
    6. You are using an old MySQL library AND your script is vulnerable to SQL injection.

    Below is a version that should fix all of these except for the old library:

    connect.php

    <?php
    $conn = mysql_connect('localhost', 'root', 'root');
    mysql_select_db('testing');
    ?>
    

    post.php

    <?php
    require_once("connect.php");
    
    if (isset ($_POST["guestfirst"]) && isset ($_POST["guestlast"]))
    {
        $guestfirst = mysql_real_escape_string($_POST["guestfirst"]);
        $guestlast = mysql_real_escape_string($_POST["guestlast"]);
    }
    else
    {
        $guestfirst = "No Entry";
        $guestlast = "none";
    }
    
    $sql = "INSERT INTO `events` (guestfirst, guestlast) VALUES ('$guestfirst', '$guestlast');";
    $res = mysql_query($sql, $conn) or $error = mysql_error();
    
    if($res)
    {
        $ret = array(
            'inserted' => true
        );
    }
    else
    {
        $ret = array(
            'inserted' => false,
            'error' => $error
        );
    }
    
    mysql_close($conn);
    echo json_encode($ret);
    exit();
    ?>
    

    list.php (or whatever else you called it)

    <?php
    require_once("connect.php");
    
    $get_events = mysql_query("SELECT * FROM `events`", $conn);
    $articles = array();
    
    while ($row = mysql_fetch_assoc($get_events))
    {
        $articles[] = $row;
    }
    
    mysql_close($conn);
    echo json_encode($articles);
    exit();
    ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥15 javaweb项目无法正常跳转
  • ¥15 VMBox虚拟机无法访问
  • ¥15 skd显示找不到头文件
  • ¥15 机器视觉中图片中长度与真实长度的关系
  • ¥15 fastreport table 怎么只让每页的最下面和最顶部有横线
  • ¥15 R语言卸载之后无法重装,显示电脑存在下载某些较大二进制文件行为,怎么办
  • ¥15 java 的protected权限 ,问题在注释里
  • ¥15 这个是哪里有问题啊?
  • ¥15 关于#vue.js#的问题:修改用户信息功能图片无法回显,数据库中只存了一张图片(相关搜索:字符串)
  • ¥15 texstudio的问题,