Question

我正在通过Google表格制作一个签到应用程序，并希望创建一个搜索功能，该功能将运动名称作为HTML表单中的输入，然后从HTML表格中的工作表中返回有关该运动的信息。但是，当我尝试测试Web应用程序时，没有任何反应。我该如何解决这个问题？

这是我的代码：

的index.html

<!DOCTYPE html>
 <html>
  <head>
   <?!= HtmlService.createHtmlOutputFromFile('Stylesheet').getContent(); ?>
  </head>
  <body>

    <fieldset id="tasks-panel">
      <legend>Sports</legend>

      <form name="sport-form" id="sport-form">
        <label for="sport-name">Search a sport by name:</label>
        <input type="text" name="sport-name" id="sport-name" />
        <button onclick='addTable()' id='submit-button'>Press this</button>
      </form>

      <p>List of things:</p>
      <div id="toggle" style="display:none"></div>
    </fieldset>

    <?!= HtmlService.createHtmlOutputFromFile('Javascript').getContent(); ?>

  </body>
</html>

Javascript.html

<script> 
  function addTable() {
  var sportInput = $('sport-name').value();
  var columnNames = ["Names", "Times"];
  var dataArray = google.script.run.getSportData(sportInput);


  var myTable = document.createElement('table');
  $('#divResults').append(myTable);

  var y = document.createElement('tr');
  myTable.appendChild(y);

  for(var i = 0; i < columnNames.length; i++) {
    var th = document.createElement('th'),
        columns = document.createTextNode(columnNames[i]);
    th.appendChild(columns);
    y.appendChild(th);
  }

  for(var i = 0 ; i < dataArray.length ; i++) {
    var row= dataArray[i];
    var y2 = document.createElement('tr');
    for(var j = 0 ; j < row.length ; j++) {
      myTable.appendChild(y2);
      var th2 = document.createElement('td');
      var date2 = document.createTextNode(row[j]);
      th2.appendChild(date2);
      y2.appendChild(th2);
        }
      }
  }
</script>

Code.gs

//Setting up global variables
var ss = SpreadsheetApp.openById("-spreadsheetID-");
var sheet = ss.getSheetByName("Sheet1");

var sportsFromSheet = sheet.getRange("D4:D12");
var namesFromSheet = sheet.getRange("B4:B12").getValues();
var timesFromSheet = sheet.getRange("A4:A12").getValues();
var NAMES = [];
var TIMES = [];
var OUTPUT = [];

//doGet function
function doGet() {
  return HtmlService.createTemplateFromFile('Index').evaluate()
      .setTitle('Check In Data')
      .setSandboxMode(HtmlService.SandboxMode.IFRAME);
}

//Gets both names and times of checked-in people 
function getSportData(input) {
  var sportInput = input;
    getNamesInSport(sportInput);
    getTimesInSport(sportInput);

    OUTPUT = [
      [NAMES],
      [TIMES]
      ];

    Logger.log(OUTPUT);
    return OUTPUT;
}

//Puts the names of every person from an inputted sport into an array.
function getNamesInSport(input) { 
  var data = sportsFromSheet.getValues();

  for (var i = 0; i < data.length; i++) {
    if(data[i] == input){
      NAMES.push(namesFromSheet[i][0]);
    }
  }
}

//Puts the times of every person from an inputted sport into an array.
function getTimesInSport(input){
  var data = sportsFromSheet.getValues();

  for (var i = 0; i < data.length; i ++) {
    if(data[i] == input){
      TIMES.push(timesFromSheet[i][0]);
    }
  }
}

Answer 1

JQuery id选择器必须以＃为前缀，以便从'sport-name'输入中获取值，您需要使用

选择它

var sportInput = $('#sport-name').val();

此外，正如Robin在上面评论的那样，如果您想使用JQuery库，您需要加载它，您显示的代码表明您可能没有这样做？

<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
</head>

虽然如果是这种情况，你可能会立即看到'$ is not defined'错误。

运行Google Web应用程序脚本后出现空白屏幕

1 个答案: