使用JavaScript将visible属性设置为false时删除空白区域

Question

我有一个DropdownList,当它的Selected Value被更改时(例如:0)我需要将Panel的visible属性设置为True,将另一个Panel的visible属性设置为False.

当选择另一个值时,我需要使用JAVASCRIPT进行反之亦然.

我目前正在实现这一目标,但空间仍然存在.我怎样才能删除空格.

谁能帮我？？

我也附上了代码.

function visible(val) {

    var ddl = document.getElementById("ddl_IDProof");
    var selectedFilterType = drpFilterType.options[ddl.selectedIndex].value;

    if (selectedFilterType == "0") {

        document.getElementById("pnl1").style.visibility = "visible";
        document.getElementById("pnl2").style.visibility = "hidden";
    }

    else {
        document.getElementById("pnl1").style.visibility = "hidden";
        document.getElementById("pnl2").style.visibility = "visible";
    }  
}

Answer 1

用display而不是visibility.这将隐藏整个元素:

    // Show pnl1 (maybe you have to use inline or inline-block insdead of block)
    document.getElementById("pnl1").style.display = "block";
    // Hide pnl2
    document.getElementById("pnl2").style.display = "none";