在 Oracle 中将 varchar 拆分为单独的列

发布于 2024-10-20 15:55:05 字数 291 浏览 2 评论 0原文

我有点困惑：我被要求接受以数据库中的特定字符串开头的注释，并将结果分成单独的列。

例如，如果返回值是这样的：

COLUMN_ONE
--------------------
'D7ERROR username'

返回值必须是：

COL_ONE    COL_TWO
--------------------
D7ERROR   username

一旦结果集被结构化，是否可以定义列，只是为了将字符串分成两部分？

原文

I'm in a bit of a pickle: I've been asked to take in comments starting with a specific string from a database, and separate the result into separate columns.

For example -- if a returned value is this:

COLUMN_ONE
--------------------
'D7ERROR username'

The return needs to be:

COL_ONE    COL_TWO
--------------------
D7ERROR   username

Is it even possible to define columns once the result set has been structured just for the sake of splitting a string into two?

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旧夏天 2024-10-27 15:55:05

取决于数据的一致性 - 假设单个空格是您想要出现在第一列和第二列中的内容之间的分隔符：

SELECT SUBSTR(t.column_one, 1, INSTR(t.column_one, ' ')-1) AS col_one,
       SUBSTR(t.column_one, INSTR(t.column_one, ' ')+1) AS col_two
  FROM YOUR_TABLE t

Oracle 10g+ 具有正则表达式支持，可以根据您需要解决的情况提供更大的灵活性。它还有一个正则表达式子字符串方法...

参考：

Depends on the consistency of the data - assuming a single space is the separator between what you want to appear in column one vs two:

SELECT SUBSTR(t.column_one, 1, INSTR(t.column_one, ' ')-1) AS col_one,
       SUBSTR(t.column_one, INSTR(t.column_one, ' ')+1) AS col_two
  FROM YOUR_TABLE t

Oracle 10g+ has regex support, allowing more flexibility depending on the situation you need to solve. It also has a regex substring method...

Reference:

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じее 2024-10-27 15:55:05

使用 REGEXP_SUBSTR 非常简单：

SELECT REGEXP_SUBSTR(t.column_one, '[^ ]+', 1, 1) col_one,
       REGEXP_SUBSTR(t.column_one, '[^ ]+', 1, 2) col_two
FROM YOUR_TABLE t;

With REGEXP_SUBSTR is as simple as:

SELECT REGEXP_SUBSTR(t.column_one, '[^ ]+', 1, 1) col_one,
       REGEXP_SUBSTR(t.column_one, '[^ ]+', 1, 2) col_two
FROM YOUR_TABLE t;

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我只土不豪 2024-10-27 15:55:05

简单的方法就是转换成列

SELECT COLUMN_VALUE FROM TABLE (SPLIT ('19869,19572,19223,18898,10155,'))

CREATE TYPE split_tbl as TABLE OF VARCHAR2(32767);

CREATE OR REPLACE FUNCTION split (p_list VARCHAR2, p_del VARCHAR2 := ',')
   RETURN split_tbl
   PIPELINED IS
   l_idx PLS_INTEGER;
   l_list VARCHAR2 (32767) := p_list;
   l_value VARCHAR2 (32767);
BEGIN
   LOOP
      l_idx := INSTR (l_list, p_del);

      IF l_idx > 0 THEN
         PIPE ROW (SUBSTR (l_list, 1, l_idx - 1));
         l_list := SUBSTR (l_list, l_idx + LENGTH (p_del));
      ELSE
         PIPE ROW (l_list);
         EXIT;
      END IF;
   END LOOP;

   RETURN;
END split;

Simple way is to convert into column

SELECT COLUMN_VALUE FROM TABLE (SPLIT ('19869,19572,19223,18898,10155,'))

CREATE TYPE split_tbl as TABLE OF VARCHAR2(32767);

CREATE OR REPLACE FUNCTION split (p_list VARCHAR2, p_del VARCHAR2 := ',')
   RETURN split_tbl
   PIPELINED IS
   l_idx PLS_INTEGER;
   l_list VARCHAR2 (32767) := p_list;
   l_value VARCHAR2 (32767);
BEGIN
   LOOP
      l_idx := INSTR (l_list, p_del);

      IF l_idx > 0 THEN
         PIPE ROW (SUBSTR (l_list, 1, l_idx - 1));
         l_list := SUBSTR (l_list, l_idx + LENGTH (p_del));
      ELSE
         PIPE ROW (l_list);
         EXIT;
      END IF;
   END LOOP;

   RETURN;
END split;

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