使用JavaScript中的循环仅显示最新的3个结果-Showonlythelatest3resultsusingaloopinJavaScript

0  

I think your are looking for this. if our query always returns top three items then why we explicitly goes to Javascript?

我想你正在寻找这个。如果我们的查询总是返回前三项,那么为什么我们明确转到Javascript?

$pnp.sp.web.lists.getByTitle("O365RoadMap").items.top(3).orderBy("Created", false).get().then(function(items) {
console.log(items);
});

2  

The following is specific to arrays. However, it appears there's a $pnp-specific answer which it seems to me should be more efficient.

以下内容特定于数组。但是,似乎有一个特定于$ pnp的答案,在我看来应该更有效率。

Assuming your current code works (showing the full list), you can grab just the first three entries in items like this:

假设您当前的代码有效(显示完整列表),您可以只抓住以下项目中的前三个条目:

items = items.slice(0, 3);

...or the last three like this:

......或者最后三个像这样:

items = items.slice(-3);

1  

I am guessing that $pnp is the PnP Javascript Library With jQuery In SharePoint based on a quick search online.

我猜这个$ pnp是基于在线快速搜索的jQuery In SharePoint的PnP Javascript库。

I don't know too much about PnP; however, I noticed this documentation in https://github.com/SharePoint/PnP-JS-Core/wiki/Working-With:-Items:

我不太了解PnP;但是,我在https://github.com/SharePoint/PnP-JS-Core/wiki/Working-With:-Items中注意到了这个文档:

// use odata operators for more efficient queries

//使用odata运算符进行更有效的查询

$pnp.sp.web.lists.getByTitle("My List").items.select("Title", "Description").top(5).orderBy("Modified", true).get().then((items: any[]) => { console.log(items); });

$ pnp.sp.web.lists.getByTitle(“My List”)。items.select(“Title”,“Description”)。top(5)。orderBy(“Modified”,true).get()。then( (items:any [])=> {console.log(items);});

As you can see they are using "top(5)"; therefore, you could change the order, so you get first the last elements, then use top(3) to obtain those three records you wish to display.

正如你所看到的,他们正在使用“top(5)”;因此,您可以更改顺序,因此首先获得最后一个元素,然后使用top(3)获取要显示的三个记录。

Since $pnp seems to follow a chain pattern, I wonder if you could apply again the order method after obtaining the top(3).

由于$ pnp似乎遵循链式模式,我想知道你是否可以在获得top(3)后再次申请订单方法。

1  

To keep two arrays:

要保留两个数组:

var items = [1,2,3,4,5,6]
var n = 3
var remainingArray = items.splice(0, items.length-n);

console.log(remainingArray)//<--first n elements
 
console.log(items)//<--last n elements

-1  

Got the three latest results by updating this part with a slice attribute. Works perfectly now.

通过使用切片属性更新此部件获得了三个最新结果。现在工作得很好。

items.slice(0, 3).forEach(item => {