如何在 Oracle 中查询 CLOB 列

我正在尝试运行一个查询，其中有几列是 CLOB 数据类型.如果我像往常一样运行查询，所有这些字段都只有 (CLOB) 作为值.

我尝试使用 DBMS_LOB.substr(column) 但我收到错误

ORA-06502:PL/SQL:数字或值错误:字符串缓冲区太小

如何查询 CLOB 列?

解决方案

当获取 CLOB 列的子字符串并使用具有大小/缓冲区限制的查询工具时，有时您需要将 BUFFER 设置为更大的大小.例如，在使用 SQL Plus 时，使用 SET BUFFER 10000 将其设置为 10000，因为默认值为 4000.

运行 DBMS_LOB.substr 命令，您还可以指定要返回的字符数量和偏移量.因此，使用 DBMS_LOB.substr(column, 3000) 可能会将其限制为足够小的缓冲区数量.

请参阅 oracle 文档 有关 substr 命令的更多信息

<前>DBMS_LOB.SUBSTR (LOB CHARACTER SET ANY_CS 中的 lob_loc，整数数量:= 32767，整数中的偏移量 := 1)返回 VARCHAR2 字符集 lob_loc%CHARSET;

I'm trying to run a query that has a few columns that are a CLOB datatype. If i run the query like normal, all of those fields just have (CLOB) as the value.

I tried using DBMS_LOB.substr(column) and i get the error

ORA-06502: PL/SQL: numeric or value error: character string buffer too small

How can i query the CLOB column?

解决方案

When getting the substring of a CLOB column and using a query tool that has size/buffer restrictions sometimes you would need to set the BUFFER to a larger size. For example while using SQL Plus use the SET BUFFER 10000 to set it to 10000 as the default is 4000.

Running the DBMS_LOB.substr command you can also specify the amount of characters you want to return and the offset from which. So using DBMS_LOB.substr(column, 3000) might restrict it to a small enough amount for the buffer.

See oracle documentation for more info on the substr command

    DBMS_LOB.SUBSTR (
       lob_loc     IN    CLOB   CHARACTER SET ANY_CS,
       amount      IN    INTEGER := 32767,
       offset      IN    INTEGER := 1)
      RETURN VARCHAR2 CHARACTER SET lob_loc%CHARSET;