Question

我昨天刚刚从codecanyon购买了一个益智游戏源代码，这不是我的预期。用户必须从图库中选择一个图像来开始拼图，但我希望以这种方式改变它＆gt;活动“选择拼图”然后用户必须选择该活动中列出的图像（不是来自画廊，来自drawable）。

我90％确定这是“从图库中选择”按钮功能代码

window.onload = init;
var interval;
function init() {
    interval = setInterval(trackLogin, 1000);
}
function trackLogin() {
    var xmlReq = false;
    try {
        xmlReq = new ActiveXObject("Msxml2.XMLHTTP");
    } catch (e) {
        try {
            xmlReq = new ActiveXObject("Microsoft.XMLHTTP");
        } catch (e2) {
            xmlReq = false;
        }
    }
    if (!xmlReq && typeof XMLHttpRequest != 'undefined') {
        xmlReq = new XMLHttpRequest();
    }

    xmlReq.open('get', 'check.php', true);
    xmlReq.setRequestHeader("Connection", "close");
    xmlReq.send(null);
    xmlReq.onreadystatechange = function() {
        if (xmlReq.readyState == 4 && xmlReq.status == 200) {
            return json_encode(array(
                    'role' => $_SESSION['role'], //assuming something like guest/logged-in
                    'user_id'   => $_SESSION['user_id']
                ));
            var obj = xmlReq.responseText;
            var jsonObj = JSON.parse(obj);
            //now we can make a comparison against our keys 'role' and 'user_id'

            if(jsonObj['role'] == 'guest'){
                //guest role, do something here
            } else if (jsonObj['role'] == 'logged-in') {
                alert('You have been logged out. You will now be redirected to home page.');
                document.location.href = "index.php";
                //do something else for logged in users
            }

“pickImageOnClick”......就是这样

Answer 1

试试这个，

 ImageView imageview = (ImageView)findViewById(R.id.imageView);
 imageview.setImageDrawable(getResources().getDrawable(R.drawable.ic_launcher));

如何在益智游戏中从Drawables中选择图像？

1 个答案: