正如标题所示,我在尝试从 Java 代码调用 Oracle 函数时遇到异常。我尝试了一些事情但没有成功。此时我不知道该怎么办。任何帮助或建议将不胜感激。提前致谢。
Java
CallableStatement cstmt = p_con.prepareCall("{? = CALL FXRATE_ENTITY(?, ?, ?, ?)}");
cstmt.registerOutParameter(1, java.sql.Types.INTEGER);
cstmt.setString(2, "SCR");
cstmt.setString(3,l_srccur);
cstmt.setString(4, l_tfcurr);
cstmt.setString(5, "B001");
cstmt.execute();
res = cstmt.getString(1);
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Oracle Function
FUNCTION FXRATE_ENTITY(CODCURRBASE IN VARCHAR, LCODCURRFROM IN VARCHAR, LCODCURRTO IN VARCHAR, LIDENTITY IN VARCHAR)
RETURN NUMBER IS
CODCUR_L VARCHAR2(3);
BEGIN
IF (LCODCURRTO = '*') THEN
RETURN NVL(FXRATE(CODCURRBASE, LCODCURRFROM, CODCURRBASE, LIDENTITY),
0);
ELSE
RETURN NVL(FXRATE(CODCURRBASE, LCODCURRFROM, LCODCURRTO, LIDENTITY), 0);
END IF;
END;
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您有多种选择:
A) 您正在使用 SQL92 语法(带有大括号 {})。我认为对于这个选项,“call”需要小写。
CallableStatement cstmt = p_con.prepareCall("{? = call FXRATE_ENTITY(?, ?, ?, ?)}")
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B) Oracle SQL 语句CALL 可以用于函数,但需要INTO子句。由于 OUT 参数现在位于末尾,因此请重新排序参数语句。
CallableStatement cstmt = p_con.prepareCall("CALL FXRATE_ENTITY(?, ?, ?, ?) INTO ?");
cstmt.setString(1, "SCR");
cstmt.setString(2,l_srccur);
cstmt.setString(3, l_tfcurr);
cstmt.setString(4, "B001");
cstmt.registerOutParameter(5, java.sql.Types.INTEGER);
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B) 您可以使用 PL/SQL 匿名块代替 CALL,如下例所示:
CallableStatement cstmt = p_con.prepareCall("begin ? := FXRATE_ENTITY(?, ?, ?, ?); end;");
cstmt.registerOutParameter(1, java.sql.Types.INTEGER);
cstmt.setString(2, "SCR");
cstmt.setString(3,l_srccur);
cstmt.setString(4, l_tfcurr);
cstmt.setString(5, "B001");
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