无法在不同的 div 中填充列表元素
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【中文标题】无法在不同的 div 中填充列表元素【英文标题】:Failing to populate list elements in different divs 【发布时间】:2018-12-09 20:22:50 【问题描述】:无论如何,我有 3 个 divs
,其中包含不同的 list
文本。我有JSON
从中获得列表。但我只得到第一个满的,其他的是空的。当我登录列表时,我得到了所有这些。您将能够在图片中看到这一点。
<div class="pricing">
<div class="container-of-pricing">
<div class="containers">
<div class="container-title">
<p class="cont-title"></p>
</div>
<div class="container-price">
<p class="cont-price"></p>
</div>
<div class="list-div">
<ul>
<li></li>
<li></li>
<li></li>
<li></li>
</ul>
</div>
<div class="button-div">
<button class="select">SELECT</button>
</div>
</div>
<div class="containers">
<div class="container-title">
<p class="cont-title"></p>
</div>
<div class="container-price">
<p class="cont-price"></p>
</div>
<div class="list-div">
<ul>
<li></li>
<li></li>
<li></li>
<li></li>
</ul>
</div>
<div class="button-div">
<button class="select">SELECT</button>
</div>
</div>
<div class="containers last-container">
<div class="container-title">
<p class="cont-title"></p>
</div>
<div class="container-price">
<p class="cont-price"></p>
</div>
<div class="list-div">
<ul>
<li></li>
<li></li>
<li></li>
<li></li>
</ul>
</div>
<div class="button-div">
<button class="select">SELECT</button>
</div>
</div>
</div>
</div>
JSON:
var db = '"titleImage":"source":"./images/trumpet.png","titleText":"Original Trombones","navMenuItems":"menu":["navItem":"Features","navItem":"How it works","navItem":"Pricing"],"masterTitle":"masterText":"Handcrafted, home-made masterpieces","howItWorks":"video":"https://www.youtube.com/embed/tgbNymZ7vqY?controls=0","menu":"menuList":["menuItem":"Privacy","menuItem":"Terms","menuItem":"Contact"],"copyRight":"copyRightText":"Copyright 2016, Original Trombones","premiumMaterials":["offerParagraf":["offerTitle":"Premium Materials","text":"Our trombones use the shiniest brass which is sourced locally. This will increase the longevity of your purchase.","offerTitle":"Fast Shipping","text":"We make sure you recieve your trombone as soon as we have finished making it. We also provide free returns if you are not satisfied.","offerTitle":"Quality Assurance","text":"For every purchase you make, we will ensure there are no damages or faults and we will check and test the pitch of your instrument."]],"containers":["containerTitle":["contTitle":"TENOR TROMBONE","contTitle":"BASS TROMBONE","contTitle":"VALVE TROMBONE"],"containerPrice":["contPrice":"$600","contPrice":"$900","contPrice":"$1200"],"listDiv":["list":["item":["itemText":"Lorem ipsum.","itemText":"Lorem ipsum.","itemText":"Lorem ipsum dolor.","itemText":"Lorem ipsum."],"item":["itemText":"Lorem ipsum.","itemText":"Lorem ipsum.","itemText":"Lorem ipsum dolor.","itemText":"Lorem ipsum."],"item":["itemText":"Plays similar to a Trumpet","itemText":"Great for Jazz Bands","itemText":"Lorem ipsum dolor.","itemText":"Lorem ipsum."]]]]';
还有我获得这些列表的 Java 脚本:
$(document).ready(function()
var obj = JSON.parse(db);
footerData(obj);
pricingData(obj);
offersData(obj);
navigationData(obj);
masterpiecesData(obj);
videoData(obj);
);
function pricingData(obj)
for(i in obj.containers)
for(j in obj.containers[i].containerTitle)
$('.cont-title').eq(j).text(obj.containers[i].containerTitle[j].contTitle);
for(i in obj.containers)
for(j in obj.containers[i].containerPrice)
$('.cont-price').eq(j).text(obj.containers[i].containerPrice[j].contPrice);
//this is the one that is problematic**********************
for(i in obj.containers)
for(j in obj.containers[i].listDiv)
for(k in obj.containers[i].listDiv[j].list)
for(p in obj.containers[i].listDiv[j].list[k].item)
console.log(obj.containers[i].listDiv[j].list[k].item[p].itemText);
$('.list-div ul li').eq(p).text(obj.containers[i].listDiv[j].list[k].item[p].itemText);
;
这是我得到的控制台输出:
如你所见,其他的div是空的,第一个是ok的。
任何帮助或提示?
【问题讨论】:
【参考方案1】:您使用 eq 错误,将匹配项从 $('.list-div ul li') 删除到 index 匹配项。
应该是这样的:
for(i in obj.containers)
for(j in obj.containers[i].listDiv)
for(k in obj.containers[i].listDiv[j].list)
var $ul = $('.list-div ul').eq(k)
for(p in obj.containers[i].listDiv[j].list[k].item)
$ul.find('li').eq(p).text(obj.containers[i].listDiv[j].list[k].item[p].itemText);
【讨论】:
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