Question

我试图从laravel中的不同表中选择不同的表。在oracle中，我已经成功实现了查询并可以正常工作。现在，我正在尝试将其“翻译”成laravel。我该怎么办？

在oraclesql中，查询如下。 TEMP表以及此查询中使用的所有其他表都存在。是否可以使用DB :: raw做到这一点？你能给我我的建议吗？

INSERT INTO TEMP (OBJECT_TYPE, OBLECT_ID)
SELECT DISTINCT HRP1001_CG.OBJECT_TYPE, HRP1001_CG.OBJECT_ID FROM HRP1001_SC, HRP1001_CG, CONFIG
WHERE
(HRP1001_SC.OBJECT_TYPE = 'CG')  
AND
(HRP1001_SC.REL_OBJ_TYPE = 'SC')   
AND
(HRP1001_SC.REL_OBJ_ID = CONFIG.SC)
AND
((HRP1001_SC.ST_DATE < CONFIG.DES_DATE) AND (HRP1001_SC.END_DATE > CONFIG.DES_DATE)) 
AND
(HRP1001_CG.REL_OBJ_ID = HRP1001_SC.OBJECT_ID)
AND
((HRP1001_CG.OBJECT_TYPE ='CG') OR (HRP1001_CG.OBJECT_TYPE ='SM'))
ORDER BY HRP1001_CG.OBJECT_ID;

更新：我也尝试了这段代码，但也没有收到结果:(。

$data = DB::table('hrp1001_sc')
            ->join('config', 'config.sc', '=', 'hrp1001_sc.rel_obj_id')
            ->join('config', 'config.des_date', '>', 'hrp1001_sc.st_date')
            ->join('config', 'config.des_date', '<', 'hrp1001_sc.end_date')
            ->join('hrp1001_cg', 'hrp1001_cg.rel_obj_id', '=', 
             'hrp1001_sc.object_id')
            ->where('hrp1001_sc.object_type', '=', 'cg')
            ->where('hrp1001_sc.rel_obj_type', '=', 'sc')
            ->select('hrp1001_sc.object_id')
            ->distinct()
            ->get();

Answer 1

找到解决方案了，就是这样

$q = "insert into temp(object_type, object_id)";
    $q = $q."select distinct hrp1001_cg.object_type, hrp1001_cg.object_id from 
    hrp1001_cg, hrp1001_sc, config where";
    $q =$q." hrp1001_sc.object_type = 'CG'";
    $q = $q." and hrp1001_sc.rel_obj_type = 'SC'";
    $q = $q." and hrp1001_sc.rel_obj_id = config.sc";
    $q = $q." and hrp1001_sc.st_date < config.des_date";
    $q = $q." and hrp1001_sc.end_date > config.des_date";
    $q = $q." and hrp1001_cg.rel_obj_id = hrp1001_sc.object_id";
    $q = $q." and";
    $q = $q." ((hrp1001_cg.object_type = 'CG') or (hrp1001_cg.object_type = 'SM'))";

    $xx = DB::insert($q);

如何实现oracle查询以从laravel中的不同表中选择

1 个答案: