Question

脚本成功检索数据但我无法使用underscore.js显示“行”的结果。失败的具体点是“var = resultContentTemplate”。无法理解这一点。

$(document).ready(function(){
var symbols = symbols || ['GOOG','A','AA','AAN'];
var yqlUrl = "http://query.yahooapis.com/v1/public/yql";
var historicalUrl = 'http://finance.yahoo.com/d/quotes.csv';

var queryTemplate = _.template("select * from csv where url='" + historicalUrl + "?s=<%= symbol %>&f=n0s0l1' and columns='name,symbol,LastTradePriceOnly'");

var resultPlaceholderTemplate = _.template(
  '<li id="<%= id %>">Please wait, Loading quotes...</li>');


var resultContentTemplate = _.template(
  '<ul>'
  + '<li><% _.each(results, function(row) { %>'
  + '<%=row.name %>'
  + '<%=row.symbol %>'
  + '<%=row.LastTradePriceOnly %>'
  + '<% }); %>'
  + '</li>'
  + '</ul>');

// display the results of the query, replacing the 'loading' placeholder
function displayResult(id, queryResult, symbol) {
  var resultsAsHtml = resultContentTemplate({results: queryResult.row, symbol: symbol});
  $('#' + id).html(resultsAsHtml);
}

_.each(symbols, function(symbol) {

    var resultId = _.uniqueId();

    // lay down a placeholder
    $('#resultContainer').append(resultPlaceholderTemplate({id:resultId, symbol:symbol}));

    $.ajax({
      url: yqlUrl,
      data: {q: queryTemplate({symbol:symbol}), format: 'json'},
      context: $('#resultContainer')
    }).done(function(output) {
      console.log(output);
      var response = _.isString(output) ? JSON.parse(output) : output;
      displayResult(resultId, response.query.results, symbol);
    }).fail(function(err) {
      console.log('the thing failed with an error');
      console.log(err.responseText);
    });

});

});

Answer 1

请参阅http://jsfiddle.net/mpBMK/

您的代码每个符号发出一个ajax请求。因此，为每个符号调用displayResult，而queryResult参数只有一个子节点，即该符号的行。

resultContentTemplate已经获得单行，因此您不应该迭代结果。代替：

var resultContentTemplate = _.template(
'<ul><li><%=result.name %><%=result.symbol %><%=result.LastTradePriceOnly %></li></ul>');

...
//in displayResult(...) which is called once per symbol
var resultsAsHtml = resultContentTemplate({result: queryResult.row, symbol: symbol});

小提琴应该清楚问题是什么以及如何修复它。

使用underscore.js显示yahoo finance的ajax结果

1 个答案: