java - 我需要一个 SQL 查询来找到你可以用一组字母组成的所有单词，包括最多两个空白 block

Question

我有一个名为字典的数据库表，当前包含所有字典条目的以下字段:

public static final String COLUMN_NAME_UID = "_id_";
public static final String COLUMN_NAME_WORD = "word";
public static final String COLUMN_NAME_WORD = "wordSorted";
public static final String COLUMN_NAME_WORD_LENGTH = "length";
public static final String COLUMN_NAME_COUNT_A = "count_A";
public static final String COLUMN_NAME_COUNT_B = "count_B";
public static final String COLUMN_NAME_COUNT_C = "count_C";
public static final String COLUMN_NAME_COUNT_D = "count_D";
public static final String COLUMN_NAME_COUNT_E = "count_E";
public static final String COLUMN_NAME_COUNT_F = "count_F";
public static final String COLUMN_NAME_COUNT_G = "count_G";
public static final String COLUMN_NAME_COUNT_H = "count_H";
public static final String COLUMN_NAME_COUNT_I = "count_I";
public static final String COLUMN_NAME_COUNT_J = "count_J";
public static final String COLUMN_NAME_COUNT_K = "count_K";
public static final String COLUMN_NAME_COUNT_L = "count_L";
public static final String COLUMN_NAME_COUNT_M = "count_M";
public static final String COLUMN_NAME_COUNT_N = "count_N";
public static final String COLUMN_NAME_COUNT_O = "count_O";
public static final String COLUMN_NAME_COUNT_P = "count_P";
public static final String COLUMN_NAME_COUNT_Q = "count_Q";
public static final String COLUMN_NAME_COUNT_R = "count_R";
public static final String COLUMN_NAME_COUNT_S = "count_S";
public static final String COLUMN_NAME_COUNT_T = "count_T";
public static final String COLUMN_NAME_COUNT_U = "count_U";
public static final String COLUMN_NAME_COUNT_V = "count_V";
public static final String COLUMN_NAME_COUNT_W = "count_W";
public static final String COLUMN_NAME_COUNT_X = "count_X";
public static final String COLUMN_NAME_COUNT_Y = "count_Y";
public static final String COLUMN_NAME_COUNT_Z = "count_Z";

我希望能够搜索实例 test* 并找到所有可以由“t”、“e”、“s”、“t”和通配符组成的单词，例如“tests”(s是通配符)，“setts”(s 是通配符)，“set”，“tet”，“es”，“te”，“best”(b 是通配符)，等等......任何你可以用任何组合制作的东西那些字母。

我试过这样的方法，但是这个例子只找到没有通配符的四个字母的单词:

SELECT * FROM dictionary WHERE 

count_E=1 AND
count_S=1 AND
count_T=2 

SELECT * FROM dictionary WHERE  length <=4

这会产生:

"137075"    "sett"  "estt"
"145808"    "stet"  "estt"
"153675"    "test"  "estt"
"153851"    "tets"  "estt"

现在，我知道，这本质上是一个谨慎的数学问题。

这里是我如何得到所有 5 个字母的单词和一个空格以及最后一个查询中提供的所有字母:

SELECT * FROM dictionary WHERE 

count_E=1 AND
count_S=1 AND
count_T=2 

INTERSECT 
SELECT * FROM dictionary WHERE  length <=5

结果:

"97705"     "netts" "enstt"
"137075"    "sett"  "estt"
"145250"    "state" "aestt"
"145808"    "stet"  "estt"
"152303"    "taste" "aestt"
"152333"    "tates" "aestt"
"152632"    "teats" "aestt"
"153361"    "tents" "enstt"
"153675"    "test"  "estt"
"153676"    "testa" "aestt"
"153733"    "testy" "estty"
"153769"    "teths" "ehstt"
"153851"    "tets"  "estt"
"153874"    "texts" "esttx"
"156575"    "totes" "eostt"
"157952"    "trets" "erstt"
"172060"    "yetts" "estty"

但是，我必须遍历字母组合的所有迭代才能获得所有隐藏的子词...谁能帮我想出一种更优雅的方法来从查询和最多两个通配符？我也知道您可以在 SQL 中使用 REGEXP，所以这可能是一种方式。我现在还不知道，我正在将这个问题带到 hive 中......

是否有一个查询，或一系列查询，或交叉、连接等...可以帮助我解决这个问题？

更新 我想我可能偶然发现了这一点，但不确定它是否正常工作。任何帮助将不胜感激:

SELECT * FROM dictionary WHERE 
(
count_E<=1 AND
count_S<=1 AND
count_T<=1 
)
INTERSECT SELECT * FROM dictionary WHERE length =(count_E+count_S+count_T+1)     ORDER BY length

+1 是占一个空格。对于两个，我正在考虑只做一个 +2，等等...... +0 就是那些字母，以及你可以从中得到的任何东西。

Answer 1

您必须执行以下操作，将表的所有字段串联起来，如下所示:

concatenacion = "(_id||' '||Desc_art||' '||Nom_proveedor||' '||marca) like '"+resultado+"'" +
            "OR (_id||' '||Nom_proveedor||' '||marca||' '||Desc_art) like '"+resultado+"'" +
            "OR (marca||' '||Nom_proveedor||' '||Desc_art||' '||_id) like '"+resultado+"'" +
            "OR (marca||' '||Nom_proveedor||' '||_id||' '||Desc_art) like '"+resultado+"'" +
            "OR (Desc_art||' '||Nom_proveedor||' '||marca||' '||_id) like '"+resultado+"'" +
            "OR (Desc_art||' '||_id||' '||Nom_proveedor||' '||marca) like '"+resultado+"'";

然后提出你的请求，在 WHERE 子句中必须把你的连接例如:

cursor=bd.rawQuery("select _id, Desc_art, cant_art, Desc_bulto, precio"+getDefaultNroLista(codcliente)+", tiene_imagen,marca from listas_precios where "+concatenacion+" ORDER BY Desc_art ASC", null);

我的工作很好，希望有用