如何使用php从mysql获取值?
您好,我正在尝试从mysql数据库中获取值,以便在UITextField中显示值。 这里是我的php代码如何使用php从mysql获取值?,php,mysql,ios,Php,Mysql,Ios,您好,我正在尝试从mysql数据库中获取值,以便在UITextField中显示值。 这里是我的php代码 <?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_
<?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();
?>
}
我想在故事板的文本字段中显示地址和电话 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"];
}
}
}
-(iAction)查找:(id)发送方
{
n错误*错误;
NSString*strURL=[NSString stringWithFormat:@]http://192.168.1.6:81/priya/sam.php?"];
NSString*aa=name.text;
NSData*dataURL=[NSData datawithcontentsofull:[NSURL URLWithString:strURL]];
NSDictionary*jsonArray=[NSJSONSerialization JSONObjectWithData:dataURL选项:编织错误:&err];
NSMutableArray*array1=[[NSMutableArray alloc]init];
array1=[jsonArray objectForKey:@“key”];
NSLog(@“%@”,数组1);
对于(int i=0;i是否NSLog状态返回正确的值?那么您需要做的就是将其分配给UITextField
,您可以为其创建一个IBOutlet,只需设置UITextField
outlet的text属性,如self.myTextField.text=strResult
您不应该使用mysql\ucode>其弃用!
- (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"];
}
}
}