我有一个带有链接的下拉列表,当我选择它会进入页面的选项时,但我想使用一个按钮而不是确定如何更改javascript以实现此目的:
<select name="form" onchange="location = this.options[this.selectedIndex].value;"class="form-control">
<option>Choose a Service</option>
<option value="http://www.google.com">Option 1</option>
<option value="http://www.google.com">Option 2</option>
<option value="http://www.google.com">Option 3</option>
<option value="http://www.google.com">Option 4</option>
<option value="http://www.google.com">Option 5</option>
</select>
答案 0 :(得分:1)
不要将location = this.options[this.selectedIndex].value
放入unChange
,而是将其移至按钮onClick
:
<select id="dropdown" name="form" class="form-control">
<option>Choose a Service</option>
<option value="http://www.google.com">Option-1</option>
<option value="http://www.google.com">Option-2</option>
<option value="http://www.google.com">Option-3</option>
<option value="http://www.google.com">Option-4</option>
<option value="http://www.google.com">Option-5</option>
</select>
<input type="button" value="Go!" onClick='location = document.getElementById("dropdown").options[document.getElementById("dropdown").selectedIndex].value'>
答案 1 :(得分:0)
<select id="dd" name="form">
<option>Choose a Service</option>
<option value="http://www.google.com">Option 1</option>
<option value="http://www.google.com">Option 2</option>
<option value="http://www.google.com">Option 3</option>
<option value="http://www.google.com">Option 4</option>
<option value="http://www.google.com">Option 5</option>
</select>
<button onClick="launchPage" >
<script>
var launchPage = function() {
var e = document.getElementById("dd");
var url = e.options[e.selectedIndex].value;
window.open(url);
};
</script>