php if语句(如何查看用户输入重复记录)

Question

如何检查用户和数据库之间的输入是否重复？这是用户向数据库输入数据的表单并且无论保存相同或不同的记录，此代码的结果都将有效。

    $con = mysql_connect("localhost","root","usbw");
    if (!$con)
      {
      die('Could not connect: ' . mysql_error());
      }
    mysql_select_db("booking", $con);

    $year = $_POST['year'];
    $month = $_POST['month'];
    $day = $_POST['day'];
    $time = $_POST['time'];
    $classroom = $_POST['classroom']; 
    $subject = $_POST['subject'];
    $s=implode(" ",$subject);
    $sql1 = "SELECT * FROM `test` WHERE `year` LIMIT 0, 30 ";
    $result=mysql_query("$sql1");
    $row =mysql_fetch_row($result);




        $sql = "INSERT INTO `test`(`year`, `month`, `day`, `time`, `classroom`,`subject`) VALUES  ('$year','$month','$day','$time','$classroom','$s')";



    ?>

我写了一段代码，我认为可以确定相同的记录如果相同，则用户无法输入。如果不是同一个用户可以输入。代码表示如下:

‖

if($row[1]==$year){ 
　   echo'the data was repeated';

任何人都可以帮助解决我的 if 语句问题吗？

更新:

if($row[1]==$year){ 
　   echo'the data was repeated';} else{
    $sql = "INSERT INTO `test`(`year`, `month`, `day`, `time`, `classroom`,`subject`) VALUES  ('$year','$month','$day','$time','$classroom','$s')";}

I wrote this syntax error....(Parse error: syntax error, unexpected '@' in )我应该使用 While 循环来重写吗？你能告诉我如何更正代码吗？

Answer 1

$con = mysql_connect("localhost","root","usbw");
if (!$con) {
    die('Could not connect: ' . mysql_error());
}
mysql_select_db("booking", $con);

$year = $_POST['year'];
$month = $_POST['month'];
$day = $_POST['day'];
$time = $_POST['time'];
$classroom = $_POST['classroom']; 
$subject = $_POST['subject'];
$s=implode(" ",$subject);

$sql = "INSERT INTO `test`(`year`, `month`, `day`, `time`, `classroom`,`subject`) VALUES  ('$year','$month','$day','$time','$classroom','$s')";
$qr = mysql_query($sql);

if (!$qr) {
    echo mysql_error();
}

还有 mysql....你需要改变 typs...

CREATE TABLE `test` (
  `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `year` int(11) DEFAULT NULL,
  `month` int(11) DEFAULT NULL,
  `day` int(11) DEFAULT NULL,
  `time` int(11) DEFAULT NULL,
  `classroom` int(11) DEFAULT NULL,
  `subject` int(11) DEFAULT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `year` (`year`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;