回答編集履歴
3
修正
answer
CHANGED
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@@ -108,7 +108,7 @@
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108
108
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src = data[0],
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109
109
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name = data[1],
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110
110
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$li = $('<li />'),
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111
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-
$img = $('<img />').attr('src', src)
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111
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+
$img = $('<img />').attr('src', src),
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112
112
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$text = $('<p />').text(name);
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113
113
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114
114
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$li.append($img, $text);
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2
追記
answer
CHANGED
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@@ -141,4 +141,19 @@
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141
141
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$(function() {
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142
142
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new filtering();
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143
143
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});
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144
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+
```
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145
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+
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146
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+
追記ですが、groupVal, placeVal, genreValを求める部分を共通の関数で処理するよう
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147
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+
下記の形に書き換えることも出来ます。
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148
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+
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149
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+
```js
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150
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+
getVal: function(key, value){
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151
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+
return DB[key][value] || [];
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152
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+
},
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153
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+
changeList: function() {
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154
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+
var groupVal = this.getVal('group', this.$group.val()),
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155
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+
placeVal = this.getVal('place', this.$place.val()),
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156
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+
genreVal = this.getVal('genre', this.$genre.val()),
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157
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+
// 以下同様
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158
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+
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144
159
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```
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1
修正
answer
CHANGED
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@@ -108,11 +108,9 @@
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108
108
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src = data[0],
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109
109
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name = data[1],
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110
110
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$li = $('<li />'),
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111
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-
$img = $('<img />'),
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111
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+
$img = $('<img />').attr('src', src);
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112
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-
$text = $('<p />');
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112
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+
$text = $('<p />').text(name);
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113
113
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114
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-
$img.attr('src', src);
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115
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-
$text.text(name);
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116
114
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$li.append($img, $text);
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117
115
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118
116
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return $li;
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@@ -123,22 +121,19 @@
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123
121
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this.$searchcontainer.append($list);
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124
122
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},
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125
123
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changeList: function() {
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126
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-
this.$searchcontainer.empty();
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127
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-
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128
124
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var groupVal = DB.group[this.$group.val()] || [],
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129
125
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placeVal = DB.place[this.$place.val()] || [],
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130
126
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genreVal = DB.genre[this.$genre.val()] || [],
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131
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-
result;
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127
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+
result = $.unique(groupVal.concat(placeVal, genreVal));
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132
128
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133
129
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console.log(groupVal, placeVal, genreVal);
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134
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-
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135
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-
result = $.unique(groupVal.concat(placeVal, genreVal));
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136
130
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console.log('result', result);
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137
131
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|
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138
132
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if(result.length === 0){
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139
133
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result = DB.all;
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140
134
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}
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135
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+
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141
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-
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136
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+
this.$searchcontainer.empty();
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142
137
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this.showResult(result);
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143
138
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}
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144
139
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};
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