我正在尝试使用Xcode（个人项目）为iPhone应用程序实现基本登录功能，所以此时我不需要考虑安全性。
我开始使用static functionallity，我将数据发布到我本地主机上的URL（使用NAMP），以检查用户名＆amp;密码参数与我的php文件中指定的密码参数匹配。
这很好 - 所以我试图编辑php文件，以便它连接到NAMP上托管的MySQL数据库，并在'Customer'表中检查与发布的用户名和密码匹配的用户名和密码。它返回一个计数，如果计数为1，我希望登录成功。
这是我的“登录”按钮中的代码：
- (IBAction)login:(id)sender {
NSInteger success = 0;
@try {
if([[self.txtEmail text] isEqualToString:@""] || [[self.txtPassword text] isEqualToString:@""] ) {
[self alertStatus:@"Please enter Email and Password" :@"Sign in Failed!" :0];
} else {
NSString *post =[[NSString alloc] initWithFormat:@"username=%@&password=%@",[self.txtEmail text],[self.txtPassword text]];
NSLog(@"PostData: %@",post);
NSURL *url=[NSURL URLWithString:@"http://localhost:8888/jsonlogin2.php"];
NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];
NSString *postLength = [NSString stringWithFormat:@"%lu", (unsigned long)[postData length]];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
[request setURL:url];
[request setHTTPMethod:@"POST"];
[request setValue:postLength forHTTPHeaderField:@"Content-Length"];
[request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
[request setHTTPBody:postData];
//[NSURLRequest setAllowsAnyHTTPSCertificate:YES forHost:[url host]];
NSError *error = [[NSError alloc] init];
NSHTTPURLResponse *response = nil;
NSData *urlData=[NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];
NSLog(@"Response code: %ld", (long)[response statusCode]);
if ([response statusCode] >= 200 && [response statusCode] < 300)
{
NSString *responseData = [[NSString alloc]initWithData:urlData encoding:NSUTF8StringEncoding];
NSLog(@"Response ==> %@", responseData);
NSError *error = nil;
NSDictionary *jsonData = [NSJSONSerialization
JSONObjectWithData:urlData
options:NSJSONReadingMutableContainers
error:&error];
success = [jsonData[@"success"] integerValue];
NSLog(@"Success: %ld",(long)success);
if(success == 1)
{
NSLog(@"Login SUCCESS");
} else {
NSString *error_msg = (NSString *) jsonData[@"error_message"];
[self alertStatus:error_msg :@"Sign in Failed!" :0];
}
} else {
//if (error) NSLog(@"Error: %@", error);
[self alertStatus:@"Connection Failed" :@"Sign in Failed!" :0];
}
}
}
@catch (NSException * e) {
NSLog(@"Exception: %@", e);
[self alertStatus:@"Sign in Failed." :@"Error!" :0];
}
if (success) {
[self performSegueWithIdentifier:@"login_success" sender:self];
}
}
- (void) alertStatus:(NSString *)msg :(NSString *)title :(int) tag
{
UIAlertView *alertView = [[UIAlertView alloc] initWithTitle:title
message:msg
delegate:self
cancelButtonTitle:@"Ok"
otherButtonTitles:nil, nil];
alertView.tag = tag;
[alertView show];
}
这是我的php文件：
<?php
ob_start();
$host="localhost"; // Host name
$username="root"; // Mysql username
$password="root"; // Mysql password
$db_name="coffee"; // Database name
$tbl_name="Customer"; // Table name
// Connect to server and select database.
mysql_connect("$host", "$username", "$password") or die(mysql_error());
echo "Connected to MySQL";
mysql_select_db("$db_name") or die(mysql_error());
echo "Connected to Database";
// Define $username and $password
$username=$_POST['username'];
$password=md5($_POST['password']);
// To protect MySQL injection (more detail about MySQL injection)
//$username = stripslashes($username);
//$password = stripslashes($password);
//$username = mysql_real_escape_string($username);
//$password = mysql_real_escape_string($password);
$sql="SELECT * FROM $tbl_name WHERE username='$username' and password='$password'";
$result=mysql_query($sql);
echo "returned $result";
// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
// If result matched $username and $password, table row must be 1 row
if ($count=="1") {
echo '{"success":1}';
} else {
echo "Unsuccessful! $count";
}
ob_end_flush();
?>
这是我在Xcode调试器中获得的错误代码，无论输入的用户凭据是否与db中的记录匹配。
[38992:a0b] PostData: username=francis&password=pass
[38992:a0b] Response code: 200
TheCoffeeHouse[38992:a0b] Response ==> Connected to MySQLConnected to Databasereturned Resource id #2Unsuccessful! 0
[38992:a0b] Success: 0
正如你在php文件中看到的那样，我已经使用了：'echo'返回了$ result“;”试着找出发生了什么。不知道它返回的'资源ID＃2'是什么。
我还登录了PhpMyAdmin，打开了db并对记录执行了一个简单的SELECT查询，工作正常。所以我推断数据库和应用程序没有正确通信。
最后我应该指出，我对Objective-C很新。
提前感谢您的帮助。
亲切的问候，
Francis
答案 0 :(得分：0)
这是我的代码，它正在处理我的应用程序。谢谢你的PHP代码。它帮助我看到登录成功。我希望你也能解决问题。
<?php
$username = "root";
$database = "testdb";
mysql_connect("localhost", $username);
@mysql_select_db($database) or die("Unable to find databse");
$name = $_POST["name"];
$password=$_POST['password'];
$query = "SELECT * FROM test WHERE Name = '$name' and Password ='$password' ";
$result = mysql_query($query);
$rows = array();
while ($r = mysql_fetch_assoc($result))
{
$rows[] = $r;
}
$count=mysql_num_rows($result);
mysql_close();
if ($count=="1") {
echo '{"success":1}';
} else {
echo "Unsuccessful! ";
}
?>
- (IBAction)signinTapped:(id)sender {
NSInteger success = 0;
@try {
if ([[self.userTextname text]isEqualToString:@""]||[[self.userTextpassword text] isEqualToString:@""]) {
[self alertStatus:@"please enter Email and password":@"sign in Failed!":0];
}else{
NSString *post = [[NSString alloc ]initWithFormat:@"name=%@&password=%@",[self.userTextname text],[self.userTextpassword text]];
NSLog(@"PostData:%@",post);
NSURL *url = [NSURL URLWithString:@"http://localhost/json.php"];
NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];
NSString *postLength = [NSString stringWithFormat:@"%lu",(unsigned long)[postData length]];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc]init];
[request setURL:url];
[request setHTTPMethod:@"POST"];
[request setValue:postLength forHTTPHeaderField:@"Content-Length"];
[request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
[request setHTTPBody:postData];
NSError *error = [[NSError alloc]init];
NSHTTPURLResponse *response =nil;
NSData *urlData=[NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];
NSLog(@"Response code: %ld",(long)[response statusCode]);
if ([response statusCode] >=200 && [response statusCode]<300) {
NSString *responseData = [[NSString alloc]initWithData:urlData encoding:NSUTF8StringEncoding];
NSLog(@"Response ==> %@",responseData);
NSError *error =nil;
NSDictionary *jsonData =[NSJSONSerialization
JSONObjectWithData:urlData
options:NSJSONReadingMutableContainers
error:&error];
success = [jsonData[@"success"]integerValue];
NSLog(@"Success: %ld",(long)success);
if (success == 1) {
NSLog(@"Login SUCCESS" );
}else{
NSString *error_msg =(NSString *)jsonData[@"error_message"];
[self alertStatus:error_msg :@"Sign in Failed!" :0];
}
}else{
[self alertStatus:@"Connnection Failed" :@"sign in failed!":0];
}
}
}
@catch (NSException *exception) {
NSLog(@"Exception: %@",exception);
[self alertStatus:@"Sign in Failed." :@"Error!":0];
} if (success) {
[self performSegueWithIdentifier:@"loginSuccess" sender:self];
}
}