1

我正在创建下面的地图函数 chouchdb 1.3.0。如果我将它作为临时视图运行它可以正常工作，但是一旦我保存它，它就无法编译。

function(doc) {
    var datestamp = convert_date(doc["report"]["period"]);
    var report_suite = doc["report"]["reportSuite"]["id"];
    var segment = doc["report"]["segment_id"];
    emit([report_suite, segment, datestamp], [doc["report"]["elements"][0]["id"]]);
  }

function convert_date(date){
    var months = {"Jan": "01", "Feb": "02", "Mar": "03", "Apr": "04", "May": "05",    "Jun": "06", "Jul": "07", "Aug": "08", "Sep": "09", "Oct": "10", "Nov": "11", "Dec": "12" };
    var year = date.substring(13, 17);
    var month = months[date.substring(8,11)];
    var day = date.substring(5,7).replace(" ", "0");
    var datestamp = year + month + day;
    return datestamp;
}

错误消息是：“表达式不评估函数”

我用谷歌搜索并尝试失败：

  • 在函数周围放置括号
  • 内联 convert_date 函数
4

2 回答 2

2

如果将 convert_date 函数放在 map 函数中会怎样？

  function(doc) {
    var convert_date = function(date) {
      var months = {"Jan": "01", "Feb": "02", "Mar": "03", "Apr": "04", "May": "05",    "Jun": "06", "Jul": "07", "Aug": "08", "Sep": "09", "Oct": "10", "Nov": "11", "Dec": "12" };
      var year = date.substring(13, 17);
      var month = months[date.substring(8,11)];
      var day = date.substring(5,7).replace(" ", "0");
      var datestamp = year + month + day;
      return datestamp;
    }
    var datestamp = convert_date(doc["report"]["period"]);
    var report_suite = doc["report"]["reportSuite"]["id"];
    var segment = doc["report"]["segment_id"];
    emit([report_suite, segment, datestamp], [doc["report"]["elements"][0]["id"]]);
  }
于 2013-05-16T10:03:27.080 回答
0

好吧，这很奇怪。我现在已经打开了实际的设计文档，复制了它并删除了所有其他功能，现在它又可以工作了。

于 2013-05-16T10:49:37.687 回答