Javascript 在函数之前 undefined variable

Question

我有一段 javascript，它是从 http://www.developphp.com/view.php?tid=1248 被黑客攻击的。我看到“ undefined variable - 广播”错误。

function cdtd(broadcast) {
    /* expected date format is Month DD, YYYY HH:MM:SS */
    var nextbroadcast = new Date(broadcast);
    var now = new Date();
    var timeDiff = nextbroadcast.getTime() - now.getTime();
    if (timeDiff <= 0) {
        clearTimeout(timer);
        document.getElementById("countdown").innerHTML = "<a href=\"flconlineservices.php\">Internet broadcast in progress<\/a>";
        /* Run any code needed for countdown completion here */
    }
    var seconds = Math.floor(timeDiff / 1000);
    var minutes = Math.floor(seconds / 60);
    var hours = Math.floor(minutes / 60);
    var days = Math.floor(hours / 24);
    hours %= 24;
    minutes %= 60;
    seconds %= 60;
    document.getElementById("daysBox").innerHTML = days + " d";
    document.getElementById("hoursBox").innerHTML = hours + " h";
    document.getElementById("minsBox").innerHTML = minutes + " m";
    // seconds isn't in our html code (javascript error if this isn't commented out)
    /*document.getElementById("secsBox").innerHTML = seconds + " s";*/
    var timer = setTimeout('cdtd(broadcast)',1000);
}

“广播”是从带有此 <script type="text/javascript">cdtd("<?php echo $nextbroadcast; ?>");</script> 的页面传递的。 $nextbroadcast 基于用户查看页面的日期/时间。

我试过var broadcast; , var broadcast = ""; ，和var broadcast = null; 。每当我尝试在函数之前声明变量时，它就会破坏脚本。

我做错了什么吗？该脚本工作正常，但我不想出现错误。

Answer 1

更改以下行:

vartimer = setTimeout('cdtd(广播)',1000);

对此:

var timeout = setTimeout(function() { cdtd(广播); }, 1000);