Adobe Campaign：Javascript XML E4X 开关盒

Question

给定以下 table Adob​​e Campaign 中的数据模型

我开发了以下脚本，它是 javascript、xml、e4x（由 adobe campaign 使用）的混合体。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 switches/cases，因为它们有点多余？谁能提出更好的方法？

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

}

Answer 1

大量重复代码和少量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，在您有大约 5 个选项之前，通常最好使用 if-else 而不是 switch。对此有很多注意事项，但对于零或一个，if-else 是更好的主意。

Answer 2

简介

代码可以通过 3 种方式进行优化。由于我只在 javascript 中进行了测试，因此您必须了解适合您的方法。

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, recipient,false);
        }else{          
             nms.subscription.Unsubscribe(label, recipient);
        }           
    });

}

标签数组，键数组

var labels = ["uosFunStuff", "uosUniversityNews", "uosEvents","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,false);
        }else{          
             nms.subscription.Unsubscribe(label, recipient);
        }           
    });
}

对象（这个可能不行）

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,false);
        }else{          
             nms.subscription.Unsubscribe(label, recipient);
        }           
    });
}