无法解析函数内的变量

发布于 2024-12-03 01:02:29 字数 1027 浏览 2 评论 0原文

我正在使用 Jay Salvats vegas 插件 (http://vegas.jaysalvat.com) 生成全屏背景。

我不想静态设置背景图像 src，而是想生成 6 张随机图像。

下面的 img 变量正在工作（即使效率有点低）；但是，我似乎无法将变量输出到 src。

请帮忙。

$( function() {
    var img1 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img2 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img3 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img4 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img5 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img6 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");


$.vegas( 'slideshow', {
    delay: 8000,
    backgrounds: [
        { src: '+img1+', fade: 4000 },
        { src: '+img2+', fade: 4000 },
        { src: '+img3+', fade: 4000 },
        { src: '+img4+', fade: 4000 },
        { src: '+img5+', fade: 4000 },
        { src: '+img6+', fade: 4000 }
    ]
} )( 'overlay' );

} );

原文

I'm using Jay Salvats vegas plugin (http://vegas.jaysalvat.com) to generate a fullscreen background.

Rather than statically setting the background image src, I would like to generate 6 random images.

The img variables below are working (even if a bit inefficient); however, I can't seem to output the variable to the src.

Please help.

$( function() {
    var img1 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img2 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img3 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img4 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img5 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img6 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");


$.vegas( 'slideshow', {
    delay: 8000,
    backgrounds: [
        { src: '+img1+', fade: 4000 },
        { src: '+img2+', fade: 4000 },
        { src: '+img3+', fade: 4000 },
        { src: '+img4+', fade: 4000 },
        { src: '+img5+', fade: 4000 },
        { src: '+img6+', fade: 4000 }
    ]
} )( 'overlay' );

} );

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趁年轻赶紧闹 2024-12-10 01:02:29

为什么将变量名称放在字符串中？它会导致图像的来源为+img1+，而不是变量的实际值。

$( function() {
    var img1 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img2 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img3 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img4 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img5 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img6 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");


$.vegas( 'slideshow', {
    delay: 8000,
    backgrounds: [
        { src: img1, fade: 4000 },
        { src: img2, fade: 4000 },
        { src: img3, fade: 4000 },
        { src: img4, fade: 4000 },
        { src: img5, fade: 4000 },
        { src: img6, fade: 4000 }
    ]
} )( 'overlay' );

} );

Why did you put the variable names in a string? It will cause the source of the image to be +img1+ instead of the actual value of the variable.

$( function() {
    var img1 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img2 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img3 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img4 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img5 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");
    var img6 = new String("/images/bg/"+Math.floor(Math.random()*101) + ".png");


$.vegas( 'slideshow', {
    delay: 8000,
    backgrounds: [
        { src: img1, fade: 4000 },
        { src: img2, fade: 4000 },
        { src: img3, fade: 4000 },
        { src: img4, fade: 4000 },
        { src: img5, fade: 4000 },
        { src: img6, fade: 4000 }
    ]
} )( 'overlay' );

} );

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