Php 如何在Xcode 5中使用UITextField内容更新MySQL数据库？

我正在构建我的第一个iPhone应用程序，它从MySQL数据库中提取客户端网络信息，并在UITextFields中显示内容。这一切正常，正确的信息被下拉，客户端名称被添加到UITableView中，然后当选择客户端名称时，执行一个序列，移动到第二个视图控制器，其中文本字段填充了其余的客户端信息

这是用于使其工作的PHP代码

<?php

$ipaddress = $_GET['ipaddress'];
$netpass = $_GET['netpass'];
$wifipass = $_GET['wifipass'];
$domain = $_GET['domain'];
$name = $_GET['name'];


 $con=mysqli_connect("localhost","drift","Drift","drift"    );
 // Check connection
 if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$sql = "UPDATE networks SET Ip='$ipaddress', Domain='$domain', Wifi='$wifipass',     Password='$netpass$  WHERE Name = '$name'";

mysqli_query($con,$sql);


$result = mysqli_query ($con, "SELECT * FROM networks");

mysqli_close($con);

?>

---

假设您要更新用户的地址。。。 您通过userID识别用户：因此您可以通过Xcode和POST发送要更新的记录 目标C代码：

- (void)updateDataBase
{

NSString *name = @"name";// this is the name for to find the correct record.

NSString *url = [NSString stringWithFormat:@"http://www.domain.com/file.php?name=%@&ipaddress=%@&schoolname=%@&netpass=%@&wifipass=%@&domain=%@&server=%@", name, ipaddressField.text, schoolnameField.text, netpassField.text,wifipassField.text,domainField.text, serverField.text];

// build the request
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
[request setURL:[NSURL URLWithString:url]];
[request setHTTPMethod:@"POST"];
NSMutableData *body= [NSMutableData data];
[request setHTTPBody:body];

// getting answer from the server.
// you can echo message from the server let's say :"Update finish" or something like that...
NSData *returnData = [NSURLConnection sendSynchronousRequest:request returningResponse:NULL error:nil];

NSString *returnString = [[NSString alloc] initWithData:returnData encoding:NSUTF8StringEncoding];
NSLog(@"returned: %@", returnString);

}

PHP代码：

<?php

$ipaddress = $_GET['ipaddress'];
$schoolname = $_GET['schoolname'];
$netpass = $_GET['netpass'];
$wifipass = $_GET['wifipass'];
$domain = $_GET['domain'];
$server = $_GET['server'];
$name = $_GET['name'];


$con=mysqli_connect($mysql_host,$mysql_user,$mysql_pass, $mysql_db);
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$sql = "UPDATE networks SET Ip='$ipaddress', Domain='$domain', Server='$server', Wifi='$wifipass',  WHERE name = '$name'";

mysqli_query($con,$sql);


$result = mysqli_query ($con, "SELECT * FROM users");

}

mysqli_close($con);

?>

---

您可能需要使用一些凭据（用户名和密码）对用户进行身份验证，并且您可以使用在iOS中发布。@MarcusAdams我已经看过NSMutableURLRequest文档，它看起来确实很有希望。我还发现这篇文章非常接近我要找的@user3613006，请检查它是否对你有帮助为什么？奇怪的是，我在我的应用程序中使用了相同的代码，而且它工作得很好。让我再检查一遍，也许我会成功的simpler@user3613006对不起，我犯了错误的php脚本，现在它的修复，检查它现在请。。。昨天我凭记忆写的，现在我检查了我的php脚本，不管怎样都可以，如果成功请告诉我！！：）这是我运行的PHP或MySQL版本导致的问题，我安装了MAMP并设置了一个新表，它可以使用完全相同的代码正常工作！酷：）很高兴这对你有帮助：）如果这对你有帮助，也请投赞成票；）
<?php

$ipaddress = $_GET['ipaddress'];
$schoolname = $_GET['schoolname'];
$netpass = $_GET['netpass'];
$wifipass = $_GET['wifipass'];
$domain = $_GET['domain'];
$server = $_GET['server'];
$name = $_GET['name'];


$con=mysqli_connect($mysql_host,$mysql_user,$mysql_pass, $mysql_db);
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$sql = "UPDATE networks SET Ip='$ipaddress', Domain='$domain', Server='$server', Wifi='$wifipass',  WHERE name = '$name'";

mysqli_query($con,$sql);


$result = mysqli_query ($con, "SELECT * FROM users");

}

mysqli_close($con);

?>