使用带有Key和Value的数组的jQuery Text Complete

Question

我使用的是带有以下代码的jQuery.textcomplete插件：

$('#textarea3').textcomplete([
    { // html
        mentions: {
            'test': '[e-mail]', 
            'othertest': '[e-mail]'
        },
        match: /B@(w*)$/,
        search: function (term, callback) {
            callback($.map(this.mentions, function (mention) {
                return mention.indexOf(term) === 0 ? mention : null;
            }));
        },
        index: 1,
        replace: function (mention) {
            return '@' + mention + ' ';
        }
    }
]);

当用户键入@te时，我想显示选项，但只显示mention.[value]，我不想像一样返回replace: function

replace: function (mention) {
    return '@' + mention.[key] + ' ';
}

那么，我该怎么办呢？

Answer 1

将提到dict的内容转换为数组，您就可以完成。

var mentions = {
  'test': '[e-mail]', 
  'othertest': '[e-mail]'
};
var arrayOfMentions = Object.keys(mentions).map(function(key) {
  var val = {};
  val[key] = mentions[key];
  return val;
});
$('textarea').textcomplete([
  {
    mentions: arrayOfMentions,
    match: /B@(w*)$/,
    search: function (term, callback) {
      callback($.map(this.mentions, function (mention) {
        var value = mention[Object.keys(mention)[0]];
        return value.indexOf(term) === 0 ? mention : null;
      }));
    },
    template: function (mention) {
      return mention[Object.keys(mention)[0]];
    },
    index: 1,
    replace: function (mention) {
      var key = Object.keys(mention)[0];
      return '@' + key + ' ';
    }
  }
]);

请参阅jsbinhttp://jsbin.com/farewa/edit?js，输出

Answer 2

类似的东西。。。我在拉小提琴。。。

        callback($.map(this.mentions, function (mention) {
            for (property in this.mentions) {
                  return mention.indexOf(property) === 0 ? mention : null;
            }
        }));

Answer 3

我的情况与此相反：我需要在keys中搜索，但替换为value:

$('#maintext').textcomplete([{
  mentions: arrayOfMentions,
  match: /B@(w*)$/,
  search: function(term, callback) {
    callback($.map(this.mentions, function(mention) {
      var key = Object.keys(mention)[0];
      return key.indexOf(term) === 0 ? mention : null;
    }));
  },
  template: function(mention) {
    return '@' + Object.keys(mention)[0];
  },
  index: 1,
  replace: function(mention) {
    var key = Object.keys(mention)[0];
    if(mention[key]){
        return mention[key];
    } else {
        return '@' + key + ' ';
    }
  }
}]);

请参见此处：https://jsfiddle.net/FabioBeneditto/od5jumvh/