将数据从IOS应用程序发布到PHP Web服务（托管GODADDY）

我必须将IOS应用程序中的数据发布到基于PHP的Web服务中，发布工作正常，但表中填充了空值。我已经给出了IOS和PHP代码 Am使用GODADDY主机

IOS

PHP代码

$name = $_POST['name'];
$mail = $_POST['mail'];
$servername = "localhost";
$username = "admin";
$password = "admin";
$dbname = "user";
$conn = mysqli_connect($servername, $username, $password, $dbname);
 if (!$conn) {
     die("Connection failed: " . mysqli_connect_error());
 }else{
 }
$sql = "insert into user(name,mail) values ('$name','$mail')";
$result = mysqli_query($conn, $sql);

PHP我的管理员屏幕截图

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可能是您的查询错误

$sql = "insert into user(name,mail) values ('".$name."','".$mail."')";

尝试以下代码，而不是“NSURLConnection*con=[[NSURLConnection alloc]initWithRequest:rq delegate:self]；”

在phpmyadmin上尝试您的查询以检查其是否正确。

尝试或以轻松地发出GET或POST请求，这将在一个方法中处理所有问题

例如：使用SVHttp：您需要在项目中拖放SVHttp文件夹。如果需要，请选中“复制”，然后向您的URL发出如下请求：

SVHTTPClient *request = [SVHTTPClient sharedClient];
NSMutableDictionary *params=[[NSMutableDictionary alloc]init];
[params setObject:@"abc" forKey:@"uname"];
[request setBasicAuthWithUsername:admin password:admin];
[request setSendParametersAsJSON:NO];

[request POST:@"http://your server URL"
   parameters:params
   completion:^(id response, NSHTTPURLResponse *urlResponse, NSError *error) {
NSLog(@"response : %@",response);
}];

---

：）

我修好了，谢谢大家

PHP

---

您的查询已为sql注入打开。它不起作用，是我做的。感谢godaddy主机中的maid是否有任何更改？发布工作正常，但插入了空值int TABLE u plz chek附加屏幕截图？php工作正常，我使用postman插件-chromeSure进行了测试。。如果您面临任何问题，请告诉我。如果你认为这个解决方案正确，请投票！谢谢，我会尝试一下，但它只支持基于json的输入，您可以根据需要进行更改，它会生成简单的API请求。我编辑了答案，请检查。

NSError *err = nil;
NSHTTPURLResponse *response = nil;
NSData   *data = [NSURLConnection sendSynchronousRequest:rq returningResponse:&response error:&err];
NSString *responseString = [[NSString alloc] initWithData:data encoding:NSASCIIStringEncoding];

SVHTTPClient *request = [SVHTTPClient sharedClient];
NSMutableDictionary *params=[[NSMutableDictionary alloc]init];
[params setObject:@"abc" forKey:@"uname"];
[request setBasicAuthWithUsername:admin password:admin];
[request setSendParametersAsJSON:NO];

[request POST:@"http://your server URL"
   parameters:params
   completion:^(id response, NSHTTPURLResponse *urlResponse, NSError *error) {
NSLog(@"response : %@",response);
}];

<?php
$servername = "localhost";
$username = "admin";
$password = "admin";
$dbname = "user";
$conn = mysqli_connect($servername, $username, $password, $dbname);
 if (!$conn) {
     die("Connection failed: " . mysqli_connect_error());
 }else{
 }
$name = $_POST['name'];
$mail = $_POST['mail'];
echo "Name : ". $name;
echo "Mail : ". $mail;
$sql = "insert into user(name,mail) values ('$name','$mail')";
$result = mysqli_query($conn, $sql);
?>

NSString *myRequestString = [NSString stringWithFormat:@"name=%@&mail=%@",_name.text,_mail.text];

NSData *myRequestData = [NSData dataWithBytes: [myRequestString UTF8String] length: [myRequestString length]];

NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL: [NSURL URLWithString: @"http://www.example.com/test/t1.php"]];

[request setHTTPMethod: @"POST"];

[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"content-type"];

[request setHTTPBody: myRequestData];

NSData *returnData = [NSURLConnection sendSynchronousRequest: request returningResponse: nil error: nil];

NSString *response = [[NSString alloc] initWithBytes:[returnData bytes] length:[returnData length] encoding:NSUTF8StringEncoding];
NSLog(@"%@",response);