php - 在 AJAX 调用中显示来自 PHP 的消息

Question

我有一个收藏夹按钮，调用 ajax 请求来更新 MySQL 数据库。

如果有重复添加或添加过多，我希望收到警报。

任何人都可以找到一种方法，如果有重复添加，我可以显示警报吗？我的代码如下:

AJAX 请求

$.ajax({
    type: 'post',
    url: 'favaddDB.php',
    data: $('#addfaveform').serialize(),
    success: function () {
      alert('Added To Favourites');
    }
});

PHP

$db = new PDO("mysql:host=localhost;dbname=favourites", 'root', '');                                                                  
$query1="SELECT * FROM `$email` ";
$stat1=$db->prepare($query1);
$stat1->execute();// IMPORTANT add PDO variables here to make safe

//Check if fave adds >9 
$count = $stat1->rowCount();
$fave=$count;
if ($fave>9) {die(); exit();} // HERE I WISH TO RUN AN ALERT OR SEND BACK A MESSAGE TO DISPLAY
else {$fave=$fave+1;}

Answer 1

只需返回文本即可提醒您的 JavaScript:

$db = new PDO("mysql:host=localhost;dbname=favourites", 'root', '');                                                                  
$query1="Query here ($email/similar should NOT BE HERE! Add them via execute/prepare.";
$stat1=$db->prepare($query1);
$stat1->execute();// IMPORTANT add PDO variables here to make safe

//Check if fave adds >9 
$count = $stat1->rowCount();
$fave=$count;
if ($fave>9) {die("Here is a message");} // HERE I WISH TO RUN AN ALERT OR SEND BACK A MESSAGE TO DISPLAY
else {$fave=$fave+1; die("Here is another message"); }

Ajax 请求:

$.ajax({
  type: 'post',
  url: 'favaddDB.php',
  data: $('#addfaveform').serialize(),
  success: function (message) {
    alert(message);
  }
});

此外，您应该考虑使用 JSON，将整个对象传递回您的 JavaScript，并在那里解析它:

$db = new PDO("mysql:host=localhost;dbname=favourites", 'root', '');                                                                  
$query1 = "Query here ($email/similar should NOT BE HERE! Add them via execute/prepare.";
$stat1 = $db->prepare($query1);
$result = $stat1->execute();// IMPORTANT add PDO variables here to make safe

// Tell javascript we're giving json.
header('Content-Type: application/json');

if (!$result) {
    echo json_encode(['error' => true, 'message' => 'A database error has occurred. Please try again later']);
    exit;
}

//Check if fave adds >9 
$count = $stat1->rowCount();
$fave = $count;
if ($fave > 9) {
    echo json_encode(['error' => false, 'fave' => $fave, 'message' => 'Fave > 9!']);
} // HERE I WISH TO RUN AN ALERT OR SEND BACK A MESSAGE TO DISPLAY
else {
   $fave = $fave+1;
   echo json_encode([
       'error' => false,
       'fave' => $fave,
       'message' => 'Current fave count: ' . $fave
   ]);
}

在你的ajax中，确保设置dataType: 'json'，它会自动将其解析为一个对象:

$.ajax({
    type: 'post',
    url: 'favaddDB.php',
    data: $('#addfaveform').serialize(),
    dataType: 'JSON',
    success: function (res) {
        if (res.error) {
            //Display an alert or edit a div with an error message
            alert(res.message);
        } else {
            //Maybe update a div with the fave count
            document.getElementById('#favcount').value = res.fave;
            alert(res.message);
        }
    }
});