我的php代码没有向mySQL输入数据,也没有给我任何反馈


My php code is not inputting data to mySQL and giving me no feedback

我不确定为什么我的数据库没有从我的网站接收信息。此外,没有弹出错误消息,所以我根本不确定数据是否在任何地方。它似乎允许我连接到数据库,但当我单击"添加员工"时,我只得到一个空白页面。有什么建议吗?编辑:我已经将代码更改为只输入一个变量,但仍然只返回一个空白页。

<html>
   <head>
      <title>Add New Record in MySQL Database</title>
   </head>
   <body>
      <?php
         if(isset($_POST['add'])) {
            $dbhost = '';
            $dbuser = 'j';
            $dbpass = 'os';
            $conn = mysql_connect($dbhost, $dbuser, $dbpass);
            if(! $conn ) {
               die('Could not connect: ' . mysql_error());
            }
            if(! get_magic_quotes_gpc() ) {
               $emp_name = addslashes ($_POST['employee_name']);
            } else {
               $emp_name = $_POST['employee_name'];
            }
            $sql = "INSERT INTO employee ". "(employee_name) ". "VALUES('$emp_name')";
            mysql_select_db('test_db');
            $retval = mysql_query( $sql, $conn );
            if(! $retval ) {
               die('Could not enter data: ' . mysql_error());
            }
            echo "Entered data successfully'n";
            mysql_close($conn);
         } else {
      ?>
            <form method = "post" action = "<?php $_PHP_SELF ?>">
               <table width = "400" border = "0" cellspacing = "1" cellpadding = "2">
                  <tr>
                     <td width = "100">
                        Employee Name
                     </td>
                     <td>
                        <input name = "employee_name" type = "text" id = "employee_name">
                     </td>
                  </tr>
                  <tr>
                     <td width = "100"> 
                     </td>
                     <td> 
                     </td>
                  </tr>
                  <tr>
                     <td width = "100"> 
                     </td>
                     <td>
                        <input name = "add" type = "submit" id = "add" value = "Add Employee">
                     </td>
                  </tr>
               </table>
            </form>
            <?php
         }
            ?>
   </body>
</html>
$sql = "INSERT INTO employee ". "(employee_name) ". "VALUES('$emp_name', NOW())";

您在sql中传递了两个值,但只指定了一列。employee_name

您可能需要删除NOW()值或添加另一列(employee_name,date_aded)

$sql = "INSERT INTO employee ". "(employee_name) ". "VALUES('$emp_name')";
$sql = "INSERT INTO employee ". "(employee_name, date_added) ". "VALUES('$emp_name', NOW())";