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。这一切都可以正常工作,将正确的信息删除并添加到UITAITHVIEW中,然后选择客户端名称时,执行了移动到第二 View Controller 的SEGUE,其中文本字段与其余部分填充了文本字段客户信息。
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);
?>
(我使用本教程达到了目前的水平 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才能提供帮助,因为我是 Php 的新手、MySQL 和 JSON。)
现在,我要做的是让从 mysql 数据库中提取并显示在 UITExtFields 中的信息可以被编辑,然后保存回数据库(仅更新现有记录而不是添加新记录记录)。我环顾四周,但找不到显示所需 PHP 以及如何在 Xcode 中使用 PHP 文件更新 MySQL 数据的解决方案。
我几乎已经让 PHP 端工作了,我已经设法创建了一个脚本来从 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 Web 表单的值?
如果有帮助,我可以发布我的更新表格吗?
感谢您抽出宝贵时间,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);
最佳答案
假设您想更新用户的地址... 您通过 userID 识别用户:因此您使用 POST 通过 Xcode 发送要更新的记录 Objective-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);
?>
希望它对你有帮助:)
关于php - 如何在 Xcode 5 中使用 UITextField 内容更新 MySQL 数据库?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23523645/