Adobe Campaign：Javascript XML E4X Switch Case

如何解决Adobe Campaign：Javascript XML E4X Switch Case

鉴于 Adob​​e Campaign 中的下表数据模型

我开发了以下脚本，它混合了 javascript、xml、e4x（由 adobe 活动使用）。该脚本基本上遍历每一行并为每个开关执行案例内的代码。我正在寻找一种方法来简化开关/外壳，因为它们有点多余？有人可以提出更好的方法吗？

  var query = xtk.queryDef.create(
               <queryDef schema="temp:enrich" operation="select">
                  <select>                                       
                    <node expr="@id"/>
                    <node expr="@fun"/>
                    <node expr="@news"/>
                    <node expr="@events"/>
                    <node expr="@student"/>                       
                  </select> 
               </queryDef>)    
  var result = query.ExecuteQuery();  

for each(var i in result.enrich)
{
    //Debug: logInfo(i.@id+ " "+i.@fun+" " +i.@news+" " +i.@student);
    var recipient = <recipient _key = "@id" id = {i.@id} />;
    
    var fun     = parseInt(i.@fun);
    var news    = parseInt(i.@news);
    var events  = parseInt(i.@events);
    var student = parseInt(i.@student);    

    switch(fun) {
      case 0:
        nms.subscription.Unsubscribe("uosFunStuff",recipient);
        break;
      case 1:
        nms.subscription.Subscribe("uosFunStuff",recipient,false);
        break;
      default:
        // donothing
    }        
    switch(news) {
      case 0:
        nms.subscription.Unsubscribe("uosUniversityNews",recipient);
        break;
      case 1:
        nms.subscription.Subscribe("uosUniversityNews",false);
        break;
      default:
        // donothing
    }         
    switch(events) {
      case 0:
        nms.subscription.Unsubscribe("uosEvents",recipient);
        break;
      case 1:
        nms.subscription.Subscribe("uosEvents",false);
        break;
      default:
        // donothing
    }     
    switch(student) {
      case 0:
        nms.subscription.Unsubscribe("uosStudentLife",recipient);
        break;
      case 1:
        nms.subscription.Subscribe("uosStudentLife",false);
        break;
      default:
        // donothing
    }      

}

解决方法

大量重复的代码和一些替换让我想到使用带有地图的单独函数。

function updateSubscription(val,label,recipient) {
    if (val) {
        nms.subscription.Unsubscribe(label,recipient);
    } else {
        nms.subscription.Subscribe(label,recipient,false);
    }
}

for each(var i in results.enrich()) {
    //Debug: logInfo(i.@id+ " "+i.@fun+" " +i.@news+" " +i.@student);
    var recipient = <recipient _key = "@id" id = {i.@id} />;

    updateSubscription(parseInt(i.@fun),'uosFunStuff',recipient);    
    updateSubscription(parseInt(i.@news),'uosUniversityNews',recipient);    
    updateSubscription(parseInt(i.@events),'uosEvents',recipient);    
    updateSubscription(parseInt(i.@student),'uosStudentLife',recipient);
}

作为旁注，通常最好在 switch 上使用 if-else，直到您有大约 5 个选项。对此有很多注意事项，但对于零或一，if-else 是一个更好的主意。

,

介绍

可以通过 3 种方式优化代码。由于我仅在 javascript 中进行了测试，因此您必须了解哪些对您有用。

标签和函数数组

var labels = ["uosFunStuff","uosUniversityNews","uosEvents","uosStudentLife"];
var functionArray = [
    function(input){return parseInt(input.@fun)  },function(input){return parseInt(input.@news)  },function(input){return parseInt(input.@events)  },function(input){return parseInt(input.@student)  }
];

for each(var i in results.enrich()) {

    var recipient = <recipient _key = "@id" id = {i.@id} />;

    var  functionIndex = 0;
    for each(var label in labels) { 
        if(functionArray[functionIndex++](i) == 1){
              nms.subscription.Subscribe(label,false);
        }else{          
             nms.subscription.Unsubscribe(label,recipient);
        }           
    });

}

标签数组，键数组

var labels = ["uosFunStuff","uosStudentLife"];
var keys = [ "@fun","@news","@events","@student"];

for each(var i in results.enrich()) {
    var recipient = <recipient _key = "@id" id = {i.@id} />;

    var  functionIndex = 0;
    for each(var label in labels) { 
        if(parseInt(i[keys[functionIndex++]])== 1){
              nms.subscription.Subscribe(label,recipient);
        }           
    });
}

对象（这个可能不起作用）

var labelsAndKeys = {
uosFunStuff:"@fun",uosUniversityNews:"@news",uosEvents:"@events",uosStudentLife:"@student"
};

for each(var i in results.enrich()) {
    var recipient = <recipient _key = "@id" id = {i.@id} />;

    for each(var label in labels) { 
        if(parseInt(i[labelsAndKeys[labels])== 1){
              nms.subscription.Subscribe(label,recipient);
        }           
    });
}