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MySQL 职能: 选择和返回栏
原标题:MySQL Function: Selecting and returning column

我有一个表格处理程序,其中包括<代码>ID(未签字的编码、自动编码)和<代码>Name(独有的)。

以任何投入(无论在表格中)来选择以下职能,总是会产生结果。

ERROR 1172 (42000): 结果包括一行以上。

CREATE FUNCTION LookupOrInsertProcess(nametwo VARCHAR(255))
  RETURNS INT UNSIGNED
BEGIN
  DECLARE myid INT UNSIGNED;
  SELECT ID INTO myid FROM Processes WHERE Name=nametwo;
  RETURN myid;
END$$

但是,选择以下职能总是返回全国人民代表大会:

CREATE FUNCTION LookupOrInsertProcess(nametwo VARCHAR(255))
  RETURNS INT UNSIGNED
BEGIN
  DECLARE myid INT UNSIGNED;
  SELECT ID INTO myid FROM Processes WHERE Name=nametwo;
  RETURN myid;
END$$

此外,请指出以下事实确实反映了正确的结果(第30和50号是任意的):

CREATE FUNCTION LookupOrInsertProcess(nametwo VARCHAR(255))
  RETURNS INT UNSIGNED
BEGIN
  DECLARE myid INT UNSIGNED;
  SELECT ID INTO myid FROM Processes WHERE Name=nametwo;
  IF myid IS NULL THEN
    RETURN 30;
  ELSE
    RETURN 50;
  END IF;
END$$

感谢任何帮助。

UPDATE:Edited,以消除表格栏和功能段之间的冲突。 我不相信这个问题。

UPDATE2:请指出,在投入部分已经或没有列入表格的情况下,以下两个部分似乎都在工作。 为什么没有表层的表层输入灯片,联合国扫盲十年的功能甚至回报?

CREATE FUNCTION LookupOrInsertProcess(nametwo VARCHAR(255))
  RETURNS INT UNSIGNED
BEGIN
  DECLARE myid INT UNSIGNED;
  SELECT ID INTO myid FROM Processes WHERE Name=nametwo;
  RETURN COALESCE(myid, 0);
END$$
问题回答

Column names are not case-sensitive in MySql so you may find that where Name = name means give me every row .

将输入参数更改为您的程序(以及查询的相应条件)如下:srchName 页: 1

Try to rename input name to different name like inputName.

Update: Another suggestion.

DECLARE myid INT UNSIGNED;
SET myid = (SELECT ID FROM Processes WHERE Name=nametwo);
RETURN myid;




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