是否有任何方法来填写我的表视图取决于用户的输入?例如:用户有一个文本,无论他在该文本中输入什么,如果该文本与mySql中的相同数据匹配,则结果显示在我的表视图中。
在我使用这个php代码的时刻:
<?php
$host="localhost"; // Host name
$username="UserName"; // Mysql username
$password="PassWord"; // Mysql password
$db_name="DataBase"; // Database name
$tbl_name="Notes"; // Table name
// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
// To protect MySQL injection (more detail about MySQL injection)
$mytitle = stripslashes($mytitle);
$mytitle = mysql_real_escape_string($mytitle);
$mytitle = $_POST['title'];
$sql="SELECT * FROM $tbl_name WHERE title = '$mytitle'";
echo "Data found but not loaded";
$result=mysql_query($sql);
// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row
if ($count==1){
echo "Success";
} else {
echo "No data matching found";
}
?>
And mu Xcode code is:
NSInteger success = 0;
@try {
if([[self.lbOne text] isEqualToString:@""] ) {
[self alertStatus:@"lbOne is empty" :@"Failed to Search" :0];
} else {
NSString *post =[[NSString alloc] initWithFormat:@"title=%@",[self.lbOne text]];
NSLog(@"PostData: %@",post);
NSURL *url=[NSURL URLWithString:@"http://MyDomain/ReadNotesByTitle.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);
if([responseData isEqualToString:@"Success"])
{
[self alertStatus:@"Search Success" :@"There are some result" :0];
NSURL *url = [NSURL URLWithString:getDataBySearch];
NSData *data = [NSData dataWithContentsOfURL:url];
jasonArray = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:nil];
listArray = [[NSMutableArray alloc]init];
for (int i = 0; i < jasonArray.count; i++) {
NSString *cUserName = [[jasonArray objectAtIndex:i]objectForKey:@"userName"];
NSString *cTitle = [[jasonArray objectAtIndex:i]objectForKey:@"title"];
NSString *cComments = [[jasonArray objectAtIndex:i]objectForKey:@"comments"];
NSString *cTimeC = [[jasonArray objectAtIndex:i]objectForKey:@"commentsTime"];
NSString *cDateC = [[jasonArray objectAtIndex:i]objectForKey:@"commentsDate"];
[listArray addObject:[[ListOfObjects alloc]initWithUserName:cUserName andTitle:cTitle andComments:cComments andtimeC:cTimeC andDateC:cDateC]];
}
[self.tableView reloadData];
} else {
[self alertStatus:@"No result Found" :@"Search Failed" :0];
}
} else {
//if (error) NSLog(@"Error: %@", error);
[self alertStatus:@"Check Connection" :@"Search Failed" :0];
}
}
}
@catch (NSException * e) {
NSLog(@"Exception: %@", e);
[self alertStatus:@"Search Failed" :@"Error" :0];
}
if (success) {
NSLog(@"Data match , but not loaded");
}
这段代码显示没有匹配的结果,我确定有相同的数据。
有办法做这件事吗?
谢谢人
您尝试过使用switch语句吗?http://www.w3schools.com/php/php_switch.asp
我一直在使用它来改变基于数据的html显示方式。当我使用它时,它是用于下拉菜单的。所以,它很容易有一个案例。然后我根据案件编号回调了html。
另一种可能性是基于查询创建一个变量,然后在变量之间进行比较。http://php.net/manual/en/control-structures.if.php