NetSuite RESTlet 错误代码：INVALID_RETURN_DATA_FORMAT

Question

我尝试通过 RESTlet API 调用已保存的搜索，并且已部署 RESTlet，但在尝试运行它时，出现以下错误：

错误代码：INVALID_RETURN_DATA_FORMAT

错误消息：数据格式无效。您应该返回 TEXT。

查看我的脚本，我无法确定错误在哪里。下面是我尝试运行的脚本

function GetSearchResult(){
    //array container for search results
    var output = new Array();
    
    //get search results
    var results = nlapiSearchRecord('transaction','customsearchid',null,null);
    var columns = results[0].getAllColumns();
    
    //loop through the search results
    for(var i in results){
        //create placeholder object place holder
        var obj = new searchRow(
          //set the values of the object with the values of the appropriate columns
          results[i].getValue(columns[0]),
          results[i].getValue(columns[1]),
          results[i].getValue(columns[2]),
          results[i].getValue(columns[3]),
          results[i].getValue(columns[4]),
          results[i].getValue(columns[5]),
          results[i].getValue(columns[6]),
          results[i].getValue(columns[7]),
          results[i].getValue(columns[8]),
          results[i].getValue(columns[9]),
          results[i].getValue(columns[10]),
          results[i].getValue(columns[11]),
          results[i].getValue(columns[12])
          );
        
        //add the object to the array of results
        output.push(obj);
    }
    
    //return the array of search objects
    return output;
}

//Object to serve a place holder for each search row
function searchRow(internalid,lineid,subsidiaryid,locationid,departmentid,accountid,date,name,memo,amount,uniqueid,product,period){
    this.internalid = internalid;
    this.lineid = lineid;
    this.subsidiaryid = subsidiaryid;
    this.locationid = locationid;
    this.departmentid = departmentid;
    this.accountid = accountid;
    this.date = date;
    this.name = name;
    this.memo = memo;
    this.amount = amount;
    this.uniqueid = uniqueid;
    this.product = product;
    this.period = period;

}

谢谢！

Answer 1

确保您发送的请求Content-Type头包含。application/json我相信您还需要而return JSON.stringify(output);不仅仅是return output;

您也不想将 JavaScriptfor..in与数组一起使用；它仅适用于对象迭代。相反，您应该使用标准for循环。