你好我想从MySQL中获取值在Xcode显示textfield.but我得到的文本字段（空）值..我的Xcode的...
- (IBAction)find:(id)sender {
NSError *err;
NSString *strURL=[NSString stringWithFormat:@"http://192.168.1.15:81/priya/sam.php?FirstName=%@",name.text];
NSData *dataURL=[NSData dataWithContentsOfURL:[NSURL URLWithString:strURL]];
NSString *strResult=[[NSString alloc] initWithData:dataURL encoding:NSUTF8StringEncoding];
NSLog(@"got response==%@", strResult);
NSDictionary *jsonArray = [NSJSONSerialization JSONObjectWithData:dataURL options: NSJSONReadingMutableContainers error:&err];
NSArray *array=[jsonArray objectForKey:@"key"];
address.text=[NSString stringWithFormat:@"%@", [[array objectAtIndex:0]objectForKey:@"firstName"]];
phone.text=[NSString stringWithFormat:@"%@", [[array objectAtIndex:1]objectForKey:@"lastname"]];
}
和我的PHP代码..
<?php
$host = "localhost";
$user = "xcode";
$pass = "xcode";
$db="xcode";
$r = mysql_connect($host, $user, $pass);
echo mysql_get_server_info() . "\n";
$r2 = mysql_select_db($db);
$result1 = mysql_query("SELECT * FROM Persons WHERE FirstName='{$_GET['FirstName']}'");
$result=array();
while($row = mysql_fetch_array($result1))
array_push($result,array('firstName' => $row[0],'lastname'=>$row[1]));
echo json_encode(array("key" => $result));
mysql_close();
?>
而且我在控制台得到JSON值
{"key":[{"firstName":"Peter","lastname":"Griffin"},{"firstName":"Peter","lastname":"Griffin"}]}
帮我找到这个答案。 提前致谢。
