Question

使用独立的Google Apps脚本和Google电子表格。我有这个脚本作为JSON返回一个空数组和三个对象，但我希望只得到三个对象。它是一个搜索，当搜索一个zipcode时，脚本将返回任何匹配。问题是，它成功返回匹配，但它也为每个不匹配的行返回一个null，按行google sheet上显示的顺序。要使其正常运行，应运行function testDoGetWithZipcode()。

我不知道我是否应该获得这些空值，如果它们重要，或者我如何解决它。到目前为止，我似乎没有学到任何关于JSON的知识，但在此之前，我甚至在此之前做了一个半小时的关于Javascript和JSON的Lynda.com课程并阅读了JSON.org网站并阅读了有关JSON的Mozilla文档。我已经调整了所有函数中的变量，因为起初我认为它在function formatOrganization()中，但现在我完全被难倒了。

s = SpreadsheetApp.openById("1280aUAvFoUDP2rtpCFS2JYR7TuQNYcd5gm8QudukiGc");
var sheet = s.getSheetByName("RAP - Data");
var data = sheet.getDataRange().getValues();
var headings = data[0];

function zipcodeQuery(zipcode) {
 zipcodeArray = [];
 for (var i = 1; i < data.length; i++){
  if (zipcode === data[i][4].toString()){
    zipcodeArray.push(data[i]);
  }
 }
  return zipcodeArray
}

function formatOrganization(rowData){
 var organization = {}
 for (var i = 0; i < headings.length; i++){
   Logger.log('Headings: ' + headings[i]);
   organization[headings[i].toString()] = rowData[i];
 }
 return  organization
}

function executeZipcodeQuery(request) {

     zipcodes = request.parameters.zipcode;

      // The object to be returned as JSON
  response = {
        organizations : []
      }

      // Fill the organzations dictionary with requested organizations
      for (var i = 0; i < zipcodes.length; i++) { 
        sheetData = zipcodeQuery(zipcodes[i]);
        if(sheetData !== undefined) {
          for (var orgIndex = 0; orgIndex < sheetData.length; orgIndex++) { 
            var org = formatOrganization(sheetData[orgIndex]);
            if(org !== undefined) {
               Logger.log('Org object: ' + org);
               if(typeof org === 'object') {
                  //FIXME
                 var orgId = parseInt(org.Id);
                 Logger.log('Org Id: ' + orgId);
                 response.organizations[orgId] = org
                 //response.organizations.push({orgId : org});
               }
            }
          }  
        }
      }
      if (response.organizations.length > 0)
      {
        return ContentService.createTextOutput(JSON.stringify(response.organizations));
      } 
      else 
      {
        return ContentService.createTextOutput('Invalid Request. zipcode(s) not found.');
      } 
}

function testDoGetWithZipcode() {
  var testRequest = {"parameter":{"zipcode":"19132"},"contextPath":"","contentLength":-1,"queryString":"zipcode=19132","parameters":{"zipcode":["19132"]}};

  var textResult = doGet(testRequest);
  textResult.setMimeType(ContentService.MimeType.JSON);

  Logger.log('Mime Type: ' + textResult.getMimeType());
  Logger.log('Result content: ' + textResult.getContent());
}

我得到的回报是这样的（删节，因为电子表格中有超过180行，并且它们都在null或对象的返回中表示）：

[
   null,
   ....
   null,
   {
      "Id":61,
      "Category":"Day / Drop in Centers",
      "Organization Name":"Philadelphia Recovery Community Center (PRCC)",
      "Address":"1701 W Lehigh Ave, Philadelphia, PA 19132",
      "Zip Code":19132,
      "Days":"Mon, Tues, Thurs, Fri: 12-8pm, Wed: 9-5pm, Sat: 9-1pm",
      "Time: Open":"",
      "Time: Close":"",
      "People Served":"Women, Men, Families",
      "Description":"Case management, outpatient treatment, youth programs, training programs",
      "Phone Number":"215-223-7700"
   },
....
   null,
   {
      "Id":81,
      "Category":"Emergency Shelter",
      "Organization Name":"Station House",
      "Address":"2601 N Broad St, Philadelphia, PA 19132",
      "Zip Code":19132,
      "Days":"",
      "Time: Open":"",
      "Time: Close":"",
      "People Served":"Men",
      "Description":"After hours reception for single men\n 2601 N. Broad Street\n After 4 pm",
      "Phone Number":"215-225-9230"
   },
   null,
...
]

Answer 1

您的原始对象是：

response = {
    organizations : []
  }

organizations的键/值对的值是一个数组。但是你使用的符号就好像organizations是一个对象。

response.organizations[orgId] = org

您可以使用以下命令将值推送到数组中：

response.organizations.push(org);

我可能会尝试这样的事情：

var tempObject = {}; //Reset every time
tempObject[orgId] = org;
response.organizations.push(tempObject);

JSON包含带有三个对象的空数组 - 预计只有三个对象

1 个答案: