【发布时间】:2015-06-27 09:23:04
【问题描述】:
我正在编写一种池应用程序 - 我有大约 50 个 div - 显示 1 个,隐藏 49 个。在每个 div 中都有一个带有单选按钮的表单,在用户单击单选按钮后,JS 调用 AJAX - PHP 请求并将答案插入 MySQL。成功后显示下一个div,以此类推...
一切正常,直到我回答得更快 - 大约 3 个问题/秒。
然后一些答案将停止存储。例如答案 1,2,4,6,7,10 被存储。 3,5,8,9 不见了……
我怎么能避免呢?答案的一致性非常重要。
这是我的 JS
$('input[type=radio]').click(function () {
$.ajax({
url: 'assets/ajax/save_answer.php',
data: {action: 'save_answer', question: currentQuestion, answer: $('input[name=answer-' + currentQuestion + ']:checked').val(), answer_pattern: 'agree-disagree', seconds: currentTime},
type: 'post',
dataType: 'json',
success: function (output) {
if (output.success) {
$('.loading img').css("display", "none");
$('#' + currentQuestion).remove();
currentQuestion += currentQuestion;
$('#' + currentQuestion).show();
} else {
alert("Answer wasn't stored");
$('.loading img').css("display", "none");
}
},
error: function () {
alert("Answer wasn't stored");
$('.loading img').css("display", "none");
}
});
PHP函数
if (isset($_POST['action']) && $_POST['action'] == "save_answer") {
$resp = new stdClass();
$config = HTMLPurifier_Config::createDefault();
$purifier = new HTMLPurifier($config);
$user = $purifier->purify($_SESSION['user_id']);
$projekt = $purifier->purify($_SESSION['projekt_id']);
$otazka = $purifier->purify($_POST['question']);
$odpoved = $purifier->purify($_POST['answer']);
$answer_pattern = $purifier->purify($_POST['answer_pattern']);
$seconds = $purifier->purify($_POST['seconds']);
if(check_answer($answer_pattern, $odpoved)){
$stmt = $db->prepare("SELECT * FROM odpovedi WHERE uzivatel = :uzivatel AND projekt = :projekt AND otazka = :otazka");
$stmt->bindParam(':uzivatel', $user);
$stmt->bindParam(':projekt', $projekt);
$stmt->bindParam(':otazka', $otazka);
$stmt->execute();
$answers_count = count($stmt->fetchAll());
if ($answers_count != 1) {
$stmt = $db->prepare("INSERT INTO odpovedi (uzivatel, projekt, otazka, odpoved, seconds) VALUES (:uzivatel, :projekt, :otazka, :odpoved, :seconds) ");
$stmt->bindParam(':uzivatel', $user);
$stmt->bindParam(':projekt', $projekt);
$stmt->bindParam(':otazka', $otazka);
$stmt->bindParam(':odpoved', $odpoved);
$stmt->bindParam(':seconds', $seconds);
$stmt->execute();
$stmt = $db->prepare("SELECT * FROM odpovedi WHERE uzivatel = :uzivatel AND projekt = :projekt AND otazka = :otazka");
$stmt->bindParam(':uzivatel', $user);
$stmt->bindParam(':projekt', $projekt);
$stmt->bindParam(':otazka', $otazka);
$stmt->execute();
$answer_inserted = count($stmt->fetchAll());
if ($answer_inserted < 1) {
$resp->success = false;
} else {
$resp->success = true;
}
}else{
$resp->success = true;
}
}else{
$resp->success = false;
}
}
打印 json_encode($resp);
至少应该显示警报,对吗?但它不是... 谢谢,
托马什
【问题讨论】:
-
用php代码更新你的问题
-
注意:使用
alert("Answer wasn't stored"); -
你应该在alert函数中转义单引号字符:
alert('Answer wasn\'t stored'); -
对,我在页面上编辑了警报,因为我来自捷克共和国并且警报在我的语言中,所以我这样做只是为了清晰。
-
首先你必须看到它失败的地方,所以在每一步添加错误报告: if ($stmt = $db->prepare()) { if ($stmt->bindparam( 你可以绑定所有这些都在同一时间)) { if ($stmt->execute()) { echo 'Success!' } else { echo $db->error, E_USER_ERROR; } 回声 $db-> 错误,E_USER_ERROR; } else { echo $db->error, E_USER_ERROR; }
标签: javascript php jquery mysql ajax