我在 couchdb http://guide.couchdb.org/draft/cookbook.html#unique上遵循本指南,以便distinct
从视图中返回列表。
我的地图功能如下所示:
function(doc) {
if(doc.PartnerName !=null) {
emit(doc.PartnerName, null);
}
}
而且,我有一个减少功能:
function(keys, values) {
return true;
}
当我通过点击运行它时:
/dbName/_design/Partners/_view/my-view-name
我得到了这个:
{"rows":[
{"key":null,"value":true}
]}
如果我添加?reduce=false
到最后,我会得到想要的结果:
{
"total_rows":11,"offset":0,
"rows":[
{"id":"a","key":"PARTNER_ONE","value":null},
{"id":"b","key":"PARTNER_ONE","value":null},
{"id":"c","key":"PARTNER_ONE","value":null},
{"id":"d","key":"PARTNER_ONE","value":null},
{"id":"e","key":"PARTNER_ONE","value":null},
{"id":"f","key":"PARTNER_ONE","value":null},
{"id":"g","key":"PARTNER_TWO","value":null},
{"id":"h","key":"PARTNER_TWO","value":null},
{"id":"i","key":"PARTNER_TWO","value":null},
{"id":"j","key":"PARTNER_THREE","value":null},
{"id":"k","key":"PARTNER_FOUR","value":null}
]}
但是,理想情况下,我会尝试获得一个不同的列表,因此在上面的示例中,它将是 PARTNER_ONE、PARTNER_TWO、PARTNER_THREE、PARTNER_FOUR