我正在为我为大学开发的ctf游戏网站制作排行榜。我开始玩一些简单的动画,并发现一些我无法弄清楚的错误:1。你必须在登录按钮上点击两次或添加按钮才能在第一时间实际工作。 2.当提交错误的答案时,动画只能运行一次,然后对于任何后续的错误答案都没有任何反应。
这是页面的链接:https://nerdts-4ed61.firebaseapp.com/leaderboards.html
<footer>
<h6 class="right-ans">
Correct! Your progress will be added to your account.
</h6>
</br>
<div id="addFlagInput" class="input-group">
<div class="input-group-prepend">
<span
class="input-group-text"
style="background: transparent; color: white; border-top: none; border-left: none; border-bottom: none; border-width: 2px; border-color: rgba(0,0,0,0.09);"
id=""
>Name and flag</span
>
</div>
<input
id="inputbox1"
type="text"
class="form-control transparent-input"
/>
<input
id="inputbox2"
type="text"
class="form-control transparent-input"
/>
<div class="input-group-append">
<button
id="submitBtn"
onclick="submit()"
class="btn btn-light"
style="background: transparent; color: white; border-color: rgba(0,0,0,0.09);"
type="button"
>
Submit
</button>
</div>
</div>
</footer>
</div>
<!--ANIMATION LOGIN BTN-->
<script>
function addFlag() {
const element = document.querySelector("#addFlagBtn");
element.classList.add("animated", "fadeInDown");
const display = element.style.display;
if (display == "none") {
element.style.display = "block";
} else {
element.style.display = "none";
}
}
</script>
<!--check user state for flag add-->
<script>
function isUser() {
firebase.auth().onAuthStateChanged(function(user) {
if (user) {
//show input form
const inElement = document.querySelector("#addFlagInput");
inElement.classList.add("animated", "fadeInUp");
const visibility = inElement.style.visibility;
if (visibility == "hidden") {
inElement.style.visibility = "visible";
} else {
inElement.style.visibility = "hidden";
}
} else {
//tooltip: you must register to add a flag
const element = document.querySelector("#addFlagBtn");
element.setAttribute("title", "You must register to add a flag.");
}
});
}
</script>
<!--Submit button functionality-->
<script>
function hideText() {
document.querySelector(".right-ans").style.display = "none";
}
function submit() {
const submitElement = document.querySelector("#inputbox2");
const answer = submitElement.value;
console.log("Your answer is: ", answer);
var database = firebase.database();
var dbRef = database.ref("flags/");
dbRef.orderByKey().on("value", snapshot => {
if (snapshot.exists()) {
snapshot.forEach(function(data) {
var val = data.val();
if (val.id == answer) {
document
.querySelector(".right-ans")
.classList.add("animated", "heartBeat");
document.querySelector(".right-ans").style.display = "inline";
console.log("correct!");
setTimeout(hideText, 5000);
}
else {
const inputdiv = document.querySelector("#addFlagInput");
inputdiv.classList.remove("fadeInUp");
inputdiv.classList.add("animated","swing");
}
});
}
});
}
</script>
是什么
dbRef.orderByKey().on("value", snapshot => {
if (snapshot.exists()) {
snapshot.forEach(function(data) {
做?看起来您拥有实际添加或删除firebase事件侦听器中的类的代码。每次运行函数时,都会再次创建内部的变量或函数,并且每当到达该函数的末尾或在作用域之外时,如果变量是在该作用域内创建的,而不是分配给外部的任何其他变量,它被摧毁了。
因此,每次提交函数运行时,都会再次创建看起来像firebase的事件。
该事件仅在您第二次单击该按钮时运行,因为该事件尚未存在,第一次单击该按钮时,因为它仅在第一次提交()发生后才存在。