按 Esc 按钮关闭建议列表

Question

当用户按下Esc按钮时，我希望建议列表将被关闭。

我该怎么做？

这是我的代码：

1. script_suggestion.php：

   <script type="text/javascript">
   
       //document.getElementById("suggestion")
   
       function getSuggestion(q) {
           var ajax;
           if(window.XMLHttpRequest)//for ie7+, FF, Chrome
               ajax = new XMLHttpRequest();//ajax object
           else
               ajax = new ActiveXObject("Microsoft.XMLHTTP");//for ie6 and previous
           ajax.onreadystatechange = function() {
               if(ajax.status === 200 && ajax.readyState === 4) {
                   //if result are not there then don't display them
                   if(ajax.responseText === "")
                       document.getElementById("suggestion").style.visibility = "hidden";
                   else {
                       document.getElementById("suggestion").style.visibility = "visible";
                       document.getElementById("suggestion").innerHTML = ajax.responseText;
                   }
               }
           };
           ajax.open("GET", "suggestion.php?q=" + q, false);
           ajax.send();
       }
   </script>
   

2. PHP 代码：

   <?php include 'script_suggestion.php'; include 'script_close_suggestion_box.php'; ?>

提前致谢。

Answer 1

使用这个：

var forceClose = false;

function getSuggestion(q) {

    if (!forceClose) {

        var ajax;
        if (window.XMLHttpRequest)//for ie7+, FF, Chrome
            ajax = new XMLHttpRequest();//ajax object
        else
            ajax = new ActiveXObject("Microsoft.XMLHTTP");//for ie6 and previous
        ajax.onreadystatechange = function () {
            if (ajax.status === 200 && ajax.readyState === 4) {
                //if result are not there then don't display them
                if (ajax.responseText === "")
                    document.getElementById("suggestion").style.visibility = "hidden";
                else {
                    document.getElementById("suggestion").style.visibility = "visible";
                    document.getElementById("suggestion").innerHTML = ajax.responseText;
                }
            }
        };
        ajax.open("GET", "suggestion.php?q=" + q, false);
        ajax.send();

    }
}

window.document.onkeydown = function (e) {
    if (!e) e = event;
    if (e.keyCode == 27) {
        document.getElementById("suggestion").style.visibility = "hidden";
        forceClose = true;
    }
};

Answer 2

在 getSuggestion() 函数后添加这段代码：

window.document.onkeyup = function (e)
{
if (!e) e = event;
if (e.keyCode == 27)
document.getElementById("suggestion").style.visibility = "hidden";
}