Question

我想让用户选择访问他的网站的选项，我有以下代码，任何人都可以帮我纠正它;

<script type="text/javascript">
    function onSubmitForm() {
      document.selectDisplay.action = document.getElementById("selectwebiste").value;
      return true;
    }
</script>

<form method="get" id="uesfulwebsite">
    <select id="selectwebiste">
        <option value="http://www.google.com">Google</option>
        <option value="http://www.yahoo.com">Yahoo</option>
        <option value="http://www.facebook.com">facebook</option>
        <option value="http://stackoverflow.com/">Stack Over Flow</option>
    </select>
    <input type="submit" value="Go!" />
</form>

Answer 1

您的表单需要事件onsubmit =＆＃34; onSubmitForm（）;

<script type="text/javascript">

function onSubmitForm() 
{
    document.forms[0].action = document.getElementById("selectwebiste").value;

    return true;
}
</script>

<form method="get" id="uesfulwebsite" onsubmit="onSubmitForm();">
<select id="selectwebiste">
    <option value="http://www.google.com">Google</option>
    <option value="http://www.yahoo.com">Yahoo</option>
    <option value="http://www.facebook.com">facebook</option>
    <option value="http://stackoverflow.com/">Stack Over Flow</option>
</select>
<input type="submit"  value="Go!" />

表单操作通过用户选项选择更改javascript

1 个答案: