oracle中怎么很好的-------按字符匹配程度排序显示在目标列表中？有SQL 语句就可以了

发布网友发布时间：2022-04-07 17:22

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懂视网时间：2022-04-07 21:44

The method to do the map match

1 The algorithm procedures

First ,we should find the most nearest road to the point from all roads in haerbin;

Second, we should find the most nearest point to the point from the road.

For example, as figure 1 ,we will do the map match for the point O.

First ,we choose the most nearest road A by compare the distance between O and A with the distance between O and B.

Second, we choose the most nearest point A3 by compare the distance between O and A1,

the distance between O and A2, the distance between O and A3 with the distance between O and A4.

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Figure 1

2 The algorithm code – oracle

select g.id into abcid from h_gl g where

sdo_nn(g.geom,SDO_GEOMETRY(2001,8307,SDO_POINT_TYPE(In_X,In_Y, null),NULL,NULL),‘SDO_NUM_RES=1‘)=‘TRUE‘;

Select sdo_lrs.convert_to_lrs_geom(sdo_lrs.project_pt(g.geom,m.DIMINFO, sdo_geometry(3001, 8307,

null, sdo_elem_info_array(1, 1,1), sdo_ordinate_array(In_X, In_Y, null))))

v_sql :=‘update table t set t.road_id=‘||abcid||‘,t.longtitude_newgps =‘||v_taxi_pt.SDO_ORDINATES(1) ||‘,

t.latitude_newgps=‘||v_taxi_pt.SDO_ORDINATES(2)||‘ where t.gpsdata_id =‘||IN_GPSDATA_ID;

[Oracle]在数据库中实现地图匹配

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热心网友时间：2022-04-07 18:52

如果是oracle 10g以上版本，可以用regexp_substr函数进行正则表达式匹配，这样处理字符匹配就更容易了！相关方法可以在百度中搜索oracle regexp_substr！
例：select regexp_substr(’I love oracle very much’,'o[[:alpha:]]{5}’) from al;
结果出来是oracle

热心网友时间：2022-04-07 20:10

这种匹配程度在关系数据库里面很难做到的，匹配度建议还是拿出结果后在程序里面处理。。。单sql处理匹配度很难

热心网友时间：2022-04-07 21:44

select * from tab_name where col_1 like ‘%aaa’ order by col_1;

热心网友时间：2022-04-07 23:36

现出查询结果,再想着排列吧...

oracle中怎么很好的-------按字符匹配程度排序显示在目标列表中？ 有SQL 语句就可以了

The method to do the map match

1 The algorithm procedures

2 The algorithm code – oracle

oracle中怎么很好的-------按字符匹配程度排序显示在目标列表中？有SQL 语句就可以了