显示来自数据库的多个记录的Ajax调用

I have an ajax call that looks like so;

我有一个ajax调用，看起来是这样的;

success: function(data) {

        //Return the results of the campaign data and populate the page.

         $(data).find('campaign').each(function(i) {

            campaignTitle = $(data).find('campaignTitle').text(),
            campaignDesc = $(data).find('campaignDesc').text(),
            campaignType = $(data).find('campaignType').text(),
            campaignStatus = $(data).find('campaignStatus').text(),
            campaignDuration = $(data).find('duration').text(),
            campaignAuthor = $(data).find('whoCreated').text(),
            campaignCreated = $(data).find('dateCreated').text(),
            campaignTypeText = $(data).find('type').text(),
            campaignDataID = $(data).find('typeID').text(),
            campaignTypeName = $(data).find('typeName').text();

         });

There are 2 records being returned but the only difference in the 2 is the Data ID which I get from a relational join in my table.

返回了两个记录，但这两个记录中唯一的区别是我从表中的关系连接中获得的数据ID。

My page is a single "record" with campaign details on it.

我的页面是一个带有竞选细节的“记录”。

For example, there is a title of the campaign.

例如，有一个运动的名称。

$('[name=campaignTitle]').empty().append(campaignTitle);

Now because its returning multiple records, its duplicating the campaign title which I understand that it will do that.

因为它返回了多个记录，它复制了竞选标题我知道它会这么做。

Since all of the data is the same in the rows except the data ID's, I only need to show everything one time and then I would keep the multiple ID's that it returns and do something with them.

由于除了数据ID外，行中的所有数据都是相同的，所以我只需要一次显示所有数据，然后保留它返回的多个ID并对它们进行处理。

I am trying to do something like so :

我试着这样做:

campaignTitle = $(data).find('campaignTitle')[0].text(),

This returns undefined.

这将返回未定义。

In short, I only want the campaign title to show one time, even if there are 5 records and I would do this by specifying which it as [0] since I know it would be the same as 1-4.

简而言之，我只希望活动标题显示一次，即使有5条记录，我将指定哪个为[0]，因为我知道它将等于1-4。

1 个解决方案

#1

Why don't you return a DataSet or DataTable in your return and break down the value-pairs?

为什么不返回一个数据集或数据表并分解值对呢?