Javascript jQuery徽标函数中没有重复项

我试图做一个脚本，以改变我的网页底部的8个标志的标志和链接

然而，我目前的脚本创建了大量的副本，我只想在同一时间显示同一公司的1个徽标

我应该如何删除重复项

HTML:

<div class="bbe-container-wrap sponsorlogo" data-wrapper-id="TEYYD">
<div class="bbe-container container " data-container-id="IDXBz">
    <div class="row row-eq-height">
        <div class="col-md-3 img1" data-column-id="ZfWwM">
            <div class="bbe-col-content-wrap">
                <div class="bbe-component-block" data-block-id="RSyVB">
                    <a href="https://www.atvise.nl"><img src="/wp-content/uploads/2017/09/huis-ter-duin-logo-hotel1-1024x768.jpg" alt=""></a>
                </div>
            </div>
        </div>
        <div class="col-md-3 img2" data-column-id="tLyPm">
            <div class="bbe-col-content-wrap">
                <div class="bbe-component-block" data-block-id="UbKko">
                    <a href="https://www.atvise.nl"><img src="/wp-content/uploads/2017/09/Dental4U-300x109.jpg" alt=""></a>
                </div>
            </div>
        </div>
        <div class="col-md-3 img3" data-column-id="lIvMg">
            <div class="bbe-col-content-wrap">
                <div class="bbe-component-block" data-block-id="ILkzc">
                    <a href="https://www.atvise.nl"><img src="/wp-content/uploads/2017/09/van-der-Wiel-logo.jpg" alt=""></a>
                </div>
            </div>
        </div>
        <div class="col-md-3 img4" data-column-id="AUEmT">
            <div class="bbe-col-content-wrap">
                <div class="bbe-component-block" data-block-id="WgxcB">
                    <a href="https://www.atvise.nl"><img src="/wp-content/uploads/2017/08/Booming-logo.jpg" alt=""></a>
                </div>
            </div>
        </div>
    </div>
    <div class="row row-eq-height">
        <div class="col-md-3 img5" data-column-id="LMXqR">
            <div class="bbe-col-content-wrap">
                <div class="bbe-component-block" data-block-id="bleWv">
                    <a href="https://www.atvise.nl"><img src="/wp-content/uploads/2017/09/Horesca-logo.png" alt=""></a>
                </div>
            </div>
        </div>
        <div class="col-md-3 img6" data-column-id="laEKc">
            <div class="bbe-col-content-wrap">
                <div class="bbe-component-block" data-block-id="cNGGz">
                    <a href="https://www.atvise.nl"><img src="/wp-content/uploads/2017/09/home.png" alt=""></a>
                </div>
            </div>
        </div>
        <div class="col-md-3 img7" data-column-id="oLIWV">
            <div class="bbe-col-content-wrap">
                <div class="bbe-component-block" data-block-id="YCAcz">
                    <a href="https://www.atvise.nl"><img src="/wp-content/uploads/2017/09/happazz_logo_facebook-01.png" alt=""></a>
                </div>
            </div>
        </div>

        <div class="col-md-3 img8" data-column-id="mIwTu">
            <div class="bbe-col-content-wrap">
                <div class="bbe-component-block" data-block-id="tjCGs">
                    <a href="https://www.atvise.nl"><img src="/wp-content/uploads/2017/09/ricoh.png" alt=""></a>
                </div>
            </div>
        </div>
    </div>
</div>

一些快速提示：

用相同的值填充数组1乘1的值，每个值填充一行代码=来吧，让计算机来做繁琐的工作，这就是编码的全部内容；）

---

for（var i=0，i您可能可以通过以下代码实现整个过程：

imgs= ['08/Booming-logo.jpg','08/Langenegger-logo.jpg','08/effect.png','08/Van-Duyn-van-der-Geer-Logo.png',
       '08/VanEgmondtotaal-logo.jpg','09/ricoh.png','08/advocaten.png','09/home.png','09/ABS-Brouwer-LOGO-TBV-WEBSITE-GALA-02.jpeg',
       '09/Bjverhuur-logo.png','09/Brouwer-logo.jpg','09/Dental4U-300x109.jpg','09/happazz_logo_facebook-01.png',
       '09/Horesca-logo.png','09/huis-ter-duin-logo-hotel1-1024x768.jpg','09/van-der-Wiel-logo.jpg',
       '09/VP-Group-BMW-MINI_staand.jpg'];
lnks= ['#lnk1','#lnk2','#third','#forth','#5','#and','#so','#on','#','#','#','#','#','#','#','#','#'];

var rnd,i,vec=[]; 
for (i=imgs.length;i--;) vec[i]=i; // create a vector of same length as imgs
for (rnd=[]; rnd.push(vec.splice(Math.floor(vec.length*Math.random()),1)[0])<8;) {} 
                                   // generate 8 unique random numbers

// replace images and links in your document:
$('.bbe-component-block').each(function(i){
   $('img',this)[0].src= '/wp-content/uploads/2017/'+imgs[rnd[i]];
   $('a',  this)[0].href=lnks[rnd[i]];
})

imgs=['08/blooming logo.jpg'、'08/Langenegger logo.jpg'、'08/effect.png'、'08/Van Duyn Van der Geer logo.png'，
“08/VanegmondTotal logo.jpg”、“09/ricoh.png”、“08/advocaten.png”、“09/home.png”、“09/ABS-Brouwer-logo-TBV-WEBSITE-GALA-02.jpeg”，
‘09/Bjverhuur logo.png’、‘09/Brouwer logo.jpg’、‘09/Dental4U-300x109.jpg’、‘09/happaz_logo_facebook-01.png’，
“09/Horesca logo.png”、“09/huis-ter-duin-logo-hotel1-1024x768.jpg”、“09/van der Wiel logo.jpg”，
'09/VP-Group-BMW-MINI_staand.jpg'；
lnks=['#lnk1'、'#lnk2'、'#第三'、'#第四'、'#第五'、'#和'、'所以'、'#上'、'#'、'#'、'#'、'#'、'#'、'、'#'、'和'；
var rnd，i，vec=[]；
for（i=imgs.length；i--；）vec[i]=i；//创建与imgs长度相同的向量
对于（rnd=[]；rnd.push（vec.splice（Math.floor（vec.length*Math.random（）），1）[0]），当然，要减少随机性。先随机选择，然后迭代过去。或者简单地洗牌数组，然后从1开始迭代。谢谢，效果非常好。我只想更改2个徽标，但我意识到这正是脚本如此复杂的原因。完美的解决方案。
for (var i = 0, i <= 16; i++){
    link[i] = '#';
}

imgs= ['08/Booming-logo.jpg','08/Langenegger-logo.jpg','08/effect.png','08/Van-Duyn-van-der-Geer-Logo.png',
       '08/VanEgmondtotaal-logo.jpg','09/ricoh.png','08/advocaten.png','09/home.png','09/ABS-Brouwer-LOGO-TBV-WEBSITE-GALA-02.jpeg',
       '09/Bjverhuur-logo.png','09/Brouwer-logo.jpg','09/Dental4U-300x109.jpg','09/happazz_logo_facebook-01.png',
       '09/Horesca-logo.png','09/huis-ter-duin-logo-hotel1-1024x768.jpg','09/van-der-Wiel-logo.jpg',
       '09/VP-Group-BMW-MINI_staand.jpg'];
lnks= ['#lnk1','#lnk2','#third','#forth','#5','#and','#so','#on','#','#','#','#','#','#','#','#','#'];

var rnd,i,vec=[]; 
for (i=imgs.length;i--;) vec[i]=i; // create a vector of same length as imgs
for (rnd=[]; rnd.push(vec.splice(Math.floor(vec.length*Math.random()),1)[0])<8;) {} 
                                   // generate 8 unique random numbers

// replace images and links in your document:
$('.bbe-component-block').each(function(i){
   $('img',this)[0].src= '/wp-content/uploads/2017/'+imgs[rnd[i]];
   $('a',  this)[0].href=lnks[rnd[i]];
})

rnd.push(vec.splice(Math.floor(vec.length*Math.random()),1)[0])<8