我正在构建我的第一个iPhone应用程序,该应用程序从MySQL数据库中获取客户端网络信息并在UITextFields中显示内容。一切正常,将正确的信息删除,并将客户端名称添加到UITableView,然后选择客户端名称时,将执行segue,将其移至第二个视图控制器,在该控制器中,将文本字段的其余部分填充到第二个视图控制器中。客户信息。
这是用于实现此功能的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);
?>
(到目前为止,我使用本教程的目的是http://codewithchris.com/iphone-app-connect-to-mysql-database/。如果需要有关应用程序本身编码的更多信息,可以在上面的链接中找到,我不知道您可能需要查看现有代码的哪些部分来提供帮助,因为我对Php,MySQL和JSON完全陌生。)
现在,我要做的就是编辑从mysql数据库中提取并显示在UITExtFields中的信息,然后将其保存回数据库(仅更新现有记录而不添加新记录)。我环顾四周,但找不到能显示所需PHP以及如何在Xcode中使用PHP文件更新MySQL数据的解决方案。
我几乎已经在PHP方面工作了,我设法创建了一个脚本,可以从Web表单更新数据库,所以我真正需要的是知道如何将UITextField.text的值传递给php文件,而不是网络表单的价值?
如果有帮助,我可以发布更新表格吗?
感谢您的宝贵时间,我们将不胜感激。
更新2:
NSString *url = [NSString stringWithFormat:@"http://server.com/updatedb.php?name=%@&ipaddress=%@&netpass=%@&wifipass=%@&domain=%@", schoolnameField.text, ipaddressField.text, _netpassField.text,_wifipassField.text,_domainField.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);
假设您要更新用户的地址...通过userID识别用户:因此,您可以通过POST Objective-C代码通过Xcode发送要更新的记录:
- (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);
?>
希望它对您有帮助:)
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