我正在努力将其插入到 mySQL 数据库中。连接不必保证任何安全,因为它只是项目概念的证明。我正在使用以下 php 文件。我们将非常感谢您的帮助,因为这真的让我很沮丧。
<?php
$host = "localhost"; //Your database host server
$db = "CrossfitPotch"; //Your database name
$user = "root"; //Your database user
$pass = "root"; //Your password
$deadlift = $_GET['deadlift'];
$username = $_GET['username'];
$con = mysql_connect($host,$user,$pass) or die(mysql_error());
mysql_select_db($db,$con) or die(mysql_error());
$sql="INSERT INTO CFP (Deadlift) VALUES ('$_GET[deadlift]') WHERE Email = '$username'";
//$sql="INSERT INTO CFP (Deadlift) VALUES ('$_GET[deadlift]')";
$res = mysql_query($sql,$con) or die(mysql_error());
mysql_close($con);
if ($res) {
echo "success";
}else{
echo "faild";
}// end else
?>
我的 Xcode 插入方法如下所示:
-(IBAction)insert:(id)sender
{
// create string contains url address for php file, the file name is phpFile.php, it receives parameter :name
NSString *url= [NSString stringWithFormat:@"http://localhost/Justin_Test/Deadlift.php?username=%@", @"j@cfp.com"];
// to execute php code
NSData *dataURL = [NSData dataWithContentsOfURL:[NSURL URLWithString: url]];
NSString *AddDeadlift = [NSString stringWithFormat:@"http://localhost/Justin_Test/Deadlift.php?username=%@&username=%@&deadlift=", _NameOfUser.text, _UserDeadlift.text];
[NSData dataWithContentsOfURL:[NSURL URLWithString:AddDeadlift]];
}
最佳答案
我执行了以下操作来发布到 mySQL。只是想添加一个答案,以防其他人遇到这个问题。首先将包含数据的文本框分配给您的头文件。创建分配给按钮的单击事件。我的 PHP 文件如下所示:
<?php
$host = "localhost"; //Your database host server
$db = "CrossfitPotch"; //Your database name
$user = "root"; //Your database user
$pass = "root"; //Your password
$deadlift = $_GET['deadlift'];
$username = $_GET['username'];
//$deadlift = 10;
//$username = 'j@cfp.com';
$con = mysql_connect($host,$user,$pass) or die(mysql_error());
mysql_select_db($db,$con) or die(mysql_error());
$sql = "UPDATE CFP SET Deadlift = '$deadlift' WHERE Email = '$username';";
//$sql="INSERT INTO CFP (Deadlift) VALUES ('$_GET[deadlift]') WHERE Email = '$username'";
//$sql="INSERT INTO CFP (Deadlift) VALUES ('$_GET[deadlift]')";
$res = mysql_query($sql,$con) or die(mysql_error());
mysql_close($con);
if ($res) {
echo "success";
}else{
echo "faild";
}// end else
?>
我的事件“插入”如下所示,并带有空白字段的警报消息:
-(IBAction)insert:(id)sender
{
if([[self.txtName text] isEqualToString:@""] || [[self.txtWeight text] isEqualToString:@""])
{
UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"No Weight Entered"
message:@"Please Enter A Weight To Add"
delegate:nil
cancelButtonTitle:@"Dismiss"
otherButtonTitles:nil];
[alert show];
} else
{
NSString *AddDeadlift = [NSString stringWithFormat:@"http://localhost/Justin_Test/Deadlift.php?username=%@&deadlift=%@", txtName.text, txtWeight.text];
[NSData dataWithContentsOfURL:[NSURL URLWithString:AddDeadlift]];
}
}
关于php - iPhone xCode 插入 mySQL,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25823288/