Javascript:条件比较的麻烦 – 胡闹

Question

有时这段代码可以正常工作，但这取决于我将要执行的部分放在哪里，这实际上是说它根本不起作用 - 但我真的不知道为什么不这样做。也许你们中的一个人能够提供一些见解:

function displayBanner(currentDate) {
    var munf = currentDate.getMonth();

    var imageSrc = "defaultLogo.gif";   

    imageSrc = (munf == 9) ? ("fallLogo.gif") : ("defaultLogo.gif");
    imageSrc = (munf == 8) ? ("fallLogo.gif") : ("defaultLogo.gif");
    imageSrc = (munf == 10) ? ("fallLogo.gif") : ("defaultLogo.gif");

    imageSrc = (munf == 11) ? ("winterLogo.gif") : ("defaultLogo.gif");
    imageSrc = (munf == 0) ? ("winterLogo.gif") : ("defaultLogo.gif");
    imageSrc = (munf == 1) ? ("winterLogo.gif") : ("defaultLogo.gif");

    imageSrc = (munf == 2) ? ("springLogo.gif") : ("defaultLogo.gif");
    imageSrc = (munf == 3) ? ("springLogo.gif") : ("defaultLogo.gif");
    imageSrc = (munf == 4) ? ("springLogo.gif") : ("defaultLogo.gif");

    imageSrc = (munf == 7) ? ("summerLogo.gif") : ("defaultLogo.gif");
    imageSrc = (munf == 5) ? ("summerLogo.gif") : ("defaultLogo.gif");
    imageSrc = (munf == 6) ? ("summerLogo.gif") : ("defaultLogo.gif");

    return imageSrc;
}

Answer 1

你有逻辑问题。所发生的情况是，即使找到了正确的月份，所有比较都会发生。你应该尝试 switch...case 方法:

switch(numf) {
    case 9:
    case 8:
    case 10:
        imageSrc = 'fallLogo.gif';
        break;
    case 11:
    case 0:
    case 1:
        imageSrc = 'winterLogo.gif';
        break;
    case 2:
    case 3:
    case 4:
        imageSrc = 'springLogo.gif';
        break;
    case 7:
    case 5:
    case 6:
        imageSrc = 'summerLogo.gif';
        break;
    default:
        imageSrc = 'defaultLogo.gif';
}