根据提交时的多个选中复选框显示结果答案

【问题标题】：display result according to multiple checked checkboxes on submit根据提交时的多个选中复选框显示结果
【发布时间】：2018-04-26 10:17:55
【问题描述】：

我已经创建了 php 页面（从数据库中获取数据）并再次检查了每个复选框。每个复选框都与不同的页面链接。我希望当我选中一个或多个复选框并单击提交按钮时。它显示链接到这些选中复选框的 php 页面这是我的代码，但它没有根据选中的复选框显示页面

<ul id="Names" class="stateNames">
    <li>Alabama
        <ul class="airports">
            <li><input type="checkbox" class="destination"/> Birmingham, AL</li>
            <li><input type="checkbox" class="destination"/> Huntsville, AL</li>
        </ul>
    <li>Alaska
        <ul class="airports">
            <li><input type="checkbox" class="destination"/> Anchorage, AK</li>
            <li><input type="checkbox" class="destination"/> Fairbanks, AK</li>
            <li><input type="checkbox" class="destination"/> Juneau, AK</li>
        </ul>
    </li>
</ul>
<input type="button" onclick="clickHandler()" value="Submit" />
<script>
function clickHandler() {
endLocDest1 = "";
endLocDest2 = "";
endLocDest3 = "";
endLocDest4 = "";
endLocDest5 = "";

for(i = 0; i < document.getElementsByClassName('destination').length; i++) {
    if (document.getElementsByClassName('destination')[i].checked) {
    
   endLocDest1=         window.open('http://localhost/project/conct5.php');
   endLocDest2=         window.open('http://localhost/project/conct7.php');
   endLocDest3=         window.open('http://localhost/project/conct8.php');
endLocDest4=         window.open('http://localhost/project/conct7.php');
   endLocDest5=         window.open('http://localhost/project/conct8.php');


   }
};
}
</script>

【问题讨论】：

标签： php html ajax jscript

【解决方案1】：

这将是您要查找的 JavaScript：

function clickHandler() {
endLocDest1 = "http://localhost/project/conct5.php";
endLocDest2 = "http://localhost/project/conct7.php";
endLocDest3 = "http://localhost/project/conct8.php";
endLocDest4 = "http://localhost/project/conct7.php";
endLocDest5 = "'http://localhost/project/conct8.php";

for(i = 0; i < document.getElementsByClassName('destination').length; i++) {
    if (document.getElementsByClassName('destination')[i].checked) {
        switch(i) {
            case 0:
                window.open(endLocDest1);
                break;
            case 1:
                window.open(endLocDest2);
                break;
            case 2:
                window.open(endLocDest3);
                break;
            case 3:
                window.open(endLocDest4);
                break;
            case 4:
                window.open(endLocDest5);
                break;
       }
   }
};
}

我建议不要使用 i 为每个元素设置 ids。

【讨论】：