Question

对于我的问题的this answer，我可以用一个提交。

我无法处理多个字段。

http：// $ solr_host：8983 / solr / magazines / select＆amp; q = sport + education＆amp; df = title＆amp; q.op = OR - 这适用于一个字段。我怎样才能在多个领域做到这一点。

实现目标的正确方法是什么。提前谢谢。

Answer 1

将Dismax / eDismax与call.enqueue(new Callback<ContactList>() { @Override public void onResponse(Call<ContactList> call, Response<ContactList> response) { if (response.isSuccessful()) { Toast.makeText(getApplicationContext(),"Response Success",Toast.LENGTH_LONG).show(); Log.d("RESPONSE: ", contactList1.toString()); contactList = (List<Map<String,List<EffectList>>>) response.body().getEffectlist(); adapter = new MyContactAdapter1(uploadpage.this, contactList); listView.setAdapter(adapter); } else { } }（查询字段）参数一起使用。

示例：

public class MaxStepsCount {

     /** Dynamic Programming. */
     private static int getMaxWaysDP(int distance) {

           int[] count = new int[distance+1];
           count[0] = 1;
           count[1] = 1;
           count[2] = 2;

           /** Memorize the Sub-problem in bottom up manner*/
           for (int i=3; i<=distance; i++) {
                count[i] = count[i-1] + count[i-2] + count[i-3];               
           }
           return count[distance];
     }


     /** Recursion Approach. */
     private static int getMaxWaysRecur(int distance) {
           if(distance<0) {
                return 0;
           } else if(distance==0) {
                return 1;
           }
           return getMaxWaysRecur(distance-1)+getMaxWaysRecur(distance-2)
                     +getMaxWaysRecur(distance-3);
     }

     public static void main(String[] args) {
           // Steps pf 1, 2 and 3.
           int distance = 10;

           /** Recursion Approach. */
           int ways = getMaxWaysRecur(distance);
           System.out.println(ways);

           /** Dynamic Programming. */
           ways = getMaxWaysDP(distance);
           System.out.println(ways);
     }
}

选中此here

选择具有2个或更多单词的搜索查询，而不对索引中的多个字段使用双引号

1 个答案: