Question

我想创建一个寄存器php，它将检查表用户是否存在，如果没有则创建它，如果它确实存在，那么它会将POST数据放在表中，然后将其显示在表中的页面上。但是我得到了这个错误

<?php
$host   =   "[RETRACTED]"  ;;
$pass   =   "[RETRACTED]"  ;

$link  =  mysql_connect ( $host ,  $user ,  $pass );
if (! $link ) {
    die( 'Could not connect: '  .  mysql_error ());
}

$db_selected  =  mysql_select_db ( $db ,  $link );
if (! $db_selected ) {
    die ( 'Can\'t use $db : '  .  mysql_error ());
}

// do stuff





$query = "SELECT ID FROM `users` ";
$result = mysql_query($link, $query);

if(empty($result)) {
                $query = "CREATE TABLE `users`(
                          `username` varchar(255) NOT NULL,
                          `password` varchar(255) NOT NULL,
                          PERMISSION_LEVEL int,
                          APPLICATION_COMPLETED int,
                          APPLICATION_IN_PROGRESS int,
                          PRIMARY KEY  (ID)
                          )";
                $result = mysql_query($link, $query);
}


mysql_query("insert into `users` (username, password) 
values
  ('$_POST[username]','$_POST[password]')")
or die(mysql_error());
echo "Done!!!!";

$result = mysqli_query($con,"SELECT * FROM `users`");


echo "<table border='1'>
<tr>
<th>Username</th>
<th>Password</th>
</tr>";

while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['username'] . "</td>";
echo "<td>" . $row['password'] . "</td>";
echo "</tr>";
}
echo "</table>";



$filename = "users/$userName.txt";
$dirname = dirname($filename);
if (!is_dir($dirname))
{
    mkdir($dirname, 0755, true);
}
 $myfile = fopen("$filename", "w");
 fclose($myfile);
mysql_close ( $link );
?>

Answer 1

而不是

$query = "SELECT ID FROM `users` ";
$result = mysql_query($link, $query);

if(empty($result)) {
                $query = "CREATE TABLE `users`(
                          `username` varchar(255) NOT NULL,
                          `password` varchar(255) NOT NULL,
                          PERMISSION_LEVEL int,
                          APPLICATION_COMPLETED int,
                          APPLICATION_IN_PROGRESS int,
                          PRIMARY KEY  (ID)
                          )";
                $result = mysql_query($link, $query);
}

用于检查表是否存在并创建表

$query = "CREATE TABLE IF NOT EXISTS  `users`(
                          `username` varchar(255) NOT NULL,
                          `password` varchar(255) NOT NULL,
                          PERMISSION_LEVEL int,
                          APPLICATION_COMPLETED int,
                          APPLICATION_IN_PROGRESS int,
                          PRIMARY KEY  (ID)
                          )";
$result = mysql_query($link, $query);

表＆＃39; sk43＆＃39;不存在

1 个答案: