Javascript函数返回到原来的位置
Javascript函数返回到原来的位置
提问于 2018-04-06 03:24:26
我实际上不是一个程序员，但我必须做好这个网站的工作。为此我需要你的帮助。
我正在处理一些javascript，我设法做到了这一点：
<script>
function funcaosabores1() {
document.getElementById("testeagora1").innerHTML = "";
document.getElementById('contento1').style.visibility="visible";
document.getElementById('contento2').style.visibility="hidden";
document.getElementById('contento3').style.visibility="hidden";
document.getElementById('contento4').style.visibility="hidden";
document.getElementById('contento5').style.visibility="hidden";
}
function funcaosabores2() {
document.getElementById("testeagora2").innerHTML = "";
document.getElementById('contento2').style.visibility="visible";
document.getElementById('contento1').style.visibility="hidden";
document.getElementById('contento3').style.visibility="hidden";
document.getElementById('contento4').style.visibility="hidden";
document.getElementById('contento5').style.visibility="hidden";
}
function funcaosabores3() {
document.getElementById("testeagora3").innerHTML = "";
document.getElementById('contento3').style.visibility="visible";
document.getElementById('contento1').style.visibility="hidden";
document.getElementById('contento2').style.visibility="hidden";
document.getElementById('contento4').style.visibility="hidden";
document.getElementById('contento5').style.visibility="hidden";
}
function funcaosabores4() {
document.getElementById("testeagora4").innerHTML = "";
document.getElementById('contento4').style.visibility="visible";
document.getElementById('contento1').style.visibility="hidden";
document.getElementById('contento2').style.visibility="hidden";
document.getElementById('contento3').style.visibility="hidden";
document.getElementById('contento5').style.visibility="hidden";
}
function funcaosabores5() {
document.getElementById("testeagora5").innerHTML = "";
document.getElementById('contento5').style.visibility="visible";
document.getElementById('contento1').style.visibility="hidden";
document.getElementById('contento2').style.visibility="hidden";
document.getElementById('contento3').style.visibility="hidden";
document.getElementById('contento4').style.visibility="hidden";
}
</script>例如，我不知道如何制作:当我单击funcaosabores2时，funcaosabores1被单击并显示，第一个是隐藏的，第二个是显示的。但我不能再点击第一个，因为它已经被点击了。(如果它被称为return，则为Idk)
这是脚本中调用的div：
<div class="animacao_saborgingerale" id="contento2" style="visibility:hidden;"></div>
<div class="animacao_saboruvasyrah" id="contento3" style="visibility:hidden;"></div>
<div class="animacao_sabortangerina" id="contento4" style="visibility:hidden;"></div>
<div class="animacao_saboruvabranca" id="contento5" style="visibility:hidden;"></div>
<div class="sabor-melancia"><p onclick="funcaosabores1()" id="testeagora1">MELANCIA</p> </div>
<div class="sabor-gingerale"><p onclick="funcaosabores2()" id="testeagora2">GINGER ALE</p></div>
<div class="sabor-uvasyrah"><p onclick="funcaosabores3()" id="testeagora3">UVA SYRAH</p></div>
<div class="sabor-tangerina"><p onclick="funcaosabores4()" id="testeagora4">TANGERINA</p></div>
<div class="sabor-uvabranca"><p onclick="funcaosabores5()" id="testeagora5">UVA BRANCA</p></div> 这看起来很乱，但如果你们能帮我的话我就在这里！谢谢。
它现在是如何的CodePen。@nielsdebruin
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/49680034
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