Question

我正在尝试获取数据库的结果，即如果存在用户名和密码以便转到另一个页面。我能够创建一个新用户，但它无法获取现有用户的信息以移动到下一页。

这是我的ajax请求：

var $signinForm = $("#sigin-form");
var $errorMessage = $signinForm.find(".error-message");
//$signinForm.submit(function(event){

//get rid of default behavior (submit form)
//event.preventDefault();

$.ajax({
  type: "POST",
  url: "loginProcess.php",
  data: $signinForm.serialize(),
  dataType: "html",
  success: function(data){
    console.log(data);
    //handle any errors
    if (data.success != true) {
      console.log("error");
    }
    //if successful
    else{
      //show no errors

      //redirect user to content page
      window.location = "page1.php";
    }
  }	
});

这是我的php mysqli请求，它应该像以前使用它们一样工作，但它不是取结果：

$myDB = new mysqli($url, $username, $password, $dbName);

/* test database connection */
if ($myDB->connect_errno) {
echo "Failed to connect to MySQL: (" . $myDB->connect_errno . ") " . $myDB->connect_error;
}

if(!($getUser = $myDB->prepare("SELECT id FROM users WHERE username = ? AND password = ?"))) {
echo "Prepare failed: " . $getUser->errno . " " . $getUser->error;
}
if(!($getUser->bind_param("ss", $_POST['username'], $_POST['password']))) {
echo "bind_param failed: " . $getUser->errno . " " . $getUser->error;					
}
if(!($getUser->execute())) {
echo "Execute failed1: " . $getUser->errno . " " . $getUser->error;
}
if(!($getUser->bind_result($userID))) {
echo "bind_result failed: " . $getUser->errno . " " . $getUser->error;
}

if(!($getUser->fetch())) {
echo "fetch failed: " . $getUser->errno . " " . $getUser->error;
}
$_SESSION['userID'] = $userID;
$getUser->close();
?>

使用ajax获取mysql内容

0 个答案: