2015-04-06 81 views
0

我遇到了问题,我的数据没有通过phpMyAdmin。服务器和数据库连接PHP,但数据未插入到表中。插入的数据不会去phpMyAdmin数据库

这是我到目前为止。

<?php 

    $dbConnect = mysqli_connect("xxxxxxx", "xxxxx","xxxxx"); 
    if (!$dbConnect) 
    die("<p>The database server is not available.</p>"); 
    echo "<p>Successfully connected to the database server.</p>"; 
    $dbSelect = mysqli_select_db($dbConnect,"sxxxxxxx_db"); 
    if (!$dbSelect) 
    die("<p>The database is not available.</p>"); 
    echo "<p>Successfully opened the database.</p>"; 

    $sql = "INSERT INTO customer (Name,Password,Email,Phone) VALUES ('$_POST[namefield]', '$_POST[pwdfield]','$_POST[email]','$_POST[phone]') "; 

    mysqli_query($sql,$dbConnect); 

    mysqli_close($dbConnect); 

    if(isset($_POST['namefield']) && isset($_POST['pwdfield']) && isset($_POST['cpwdfield']) && isset($_POST['email']) && isset($_POST['phone']) && isset($_POST['submit'])) 
     { 
      $name = $_POST['namefield']; 
      $password = $_POST['pwdfield']; 
      $cpassword = $_POST['cpwdfield']; 
      $email = $_POST['email']; 
      $phone = $_POST['phone']; 

      if ($_POST["pwdfield"] == $_POST["cpassword"]) 
    mysqli_query("INSERT INTO user VALUES('','$password')") or die(mysqli_error()); 

     else { 
     echo("Password did not match! Try again. "); 
+0

删除mysqli_close($数据库连接);这一行,并把它放在else块的末尾 – RaMeSh 2015-04-06 07:22:09

回答

0

你已经关闭数据库连接

mysqli_close($dbConnect); 
0

改变SQL略有

$sql = "INSERT INTO customer (Name,Password,Email,Phone) VALUES ('".$_POST['namefield']."', '".$_POST['pwdfield']."','".$_POST['email']."','".$_POST['phone']."') " 
1
  • 关闭连接。
  • 您不应该直接在查询中使用$ _POST,请使用以下方式来防止SQL注入。

//Connect 
 

 
$unsafe_variable = $_POST["user-input"]; 
 
$safe_variable = mysql_real_escape_string($unsafe_variable); 
 

 
mysql_query("INSERT INTO table (column) VALUES ('" . $safe_variable . "')"); 
 

 
//Disconnect