我有一个收藏夹按钮,它调用ajax请求来更新MySQL数据库。
如果有重复的添加或添加的太多,我想发出警报。
如果有重复的添加,有人可以看到一种显示警报的方式吗?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;}
答案 0 :(得分:0)
只需返回文本以提醒您的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javascript,并在其中进行解析:
$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
]);
}
然后在ajax中,确保设置dataType: 'json'
,它将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);
}
}
});
答案 1 :(得分:0)
在大多数情况下,简单为佳。
通过返回messages
,您可以根据消息在前端执行任何操作。
PHP:
<?php
$db = new PDO("mysql:host=localhost;dbname=favourites", 'root', '');
$query1 = "SELECT * FROM " . $email;
$stat1 = $db->prepare($query1);
$stat1->execute();
$count = $stat1->rowCount();
$fave = $count;
if ($fave > 9) {
echo "tooMany"; exit();
} else {
echo "addedFav"; $fave++;
}
JS:
jQuery.post({
url: 'favaddDB.php',
data: jQuery('#addfaveform').serialize()
}).then(function (code) {
switch (code) {
case "addedFav":
alert('Added To Favourites');
break;
case "tooMany":
alert('Too many favourites');
break;
}
}).catch(function (error) {
console.log(error);
});