Question

嗨我想从mysql数据库中获取值以在UITextField中显示值。这里我的PHP编码

<?php

$host = "localhost"; 
$user = "xcode"; 
$pass = "xcode"; 
$db="xcode";

$r = mysql_connect($host, $user, $pass);

if (!$r) {
    echo "Could not connect to server\n";
trigger_error(mysql_error(), E_USER_ERROR);
} else {
    echo "Connection established\n"; 
}

echo mysql_get_server_info() . "\n"; 
$r2 = mysql_select_db($db);

if (!$r2) {
    echo "Cannot select database\n";
 trigger_error(mysql_error(), E_USER_ERROR);
 } else {
echo "database selected\n"; 
 }
$sql="select address,phone from login where name='{$_GET['name']}'";
if(!mysql_query($sql))
{
trigger_error(mysql_error(), E_USER_ERROR);
}
else
{
   echo"1 record added";
}
mysql_close();

?>

这里是我的Xcode

- (IBAction)find:(id)sender {
NSString *strURL=[NSString stringWithFormat:@"http://192.168.1.12:81/priya/sam.php?name=%@",name.text];



NSData *dataURL=[NSData dataWithContentsOfURL:[NSURL URLWithString:strURL]];

NSString *strResult=[[NSString alloc] initWithData:dataURL encoding:NSUTF8StringEncoding];
NSLog(@"%@",strResult);

}

我想在故事板中的文本字段中显示地址和电话。

Answer 1

回答我的问题..

 - (IBAction)find:(id)sender
   {


   NSError *err;
   NSString *strURL=[NSString stringWithFormat:@"http://192.168.1.6:81/priya/sam.php?"];
   NSString *aa=name.text;
   NSData *dataURL=[NSData dataWithContentsOfURL:[NSURL URLWithString:strURL]];

   NSDictionary *jsonArray = [NSJSONSerialization JSONObjectWithData:dataURL options: kNilOptions error:&err];


   NSMutableArray *array1=[[NSMutableArray alloc]init];
   array1=[jsonArray objectForKey:@"key"];
   NSLog(@"%@",array1);


   for(int i=0;i<(array1.count);i++)
   {
    NSString *bb=[[array1 objectAtIndex:i]objectForKey:@"firstName"];
   if([bb isEqualToString:aa])
   {
       address.text=[[array1 objectAtIndex:i]objectForKey:@"firstName"];
       phone.text=[[array1 objectAtIndex:i]objectForKey:@"lastname"];
    }
   }

   }

感谢所有

如何使用PHP从mysql获取值？

1 个答案: