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Mysql查询if语句，更新错误_Mysql_Sql - Fatal编程技术网
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Mysql查询if语句，更新错误

mysql sql

Mysql查询if语句，更新错误,mysql,sql,Mysql,Sql,早上好，我正在尝试在遍历树中上下移动节点，但问题是，如果我有向上移动节点的查询，它会在sql语法中抛出一个错误，请问，您知道问题在哪里吗 UPDATE categories SET lft = IF( lft >= 17 AND rgt <= 18, lft + 1, IF(lft >= 15 AND rgt <= 16, lft - 1)), rgt = IF( lft >= 17 AND rgt <= 18, rgt + 1, IF(lft &

早上好，我正在尝试在遍历树中上下移动节点，但问题是，如果我有向上移动节点的查询，它会在sql语法中抛出一个错误，请问，您知道问题在哪里吗

UPDATE categories 
SET lft = IF( lft >= 17 AND rgt <= 18, lft + 1, IF(lft >= 15 AND rgt <= 16, lft - 1)), 
    rgt = IF( lft >= 17 AND rgt <= 18, rgt + 1, IF(lft >= 15 AND rgt <= 16, rgt - 1))
WHERE (rgt <= 18 AND lft >= 15)

更新类别

设置lft=IF（lft>=17和rgt=15和rgt=17和rgt=15和rgt在内部IF（）函数调用中缺少第三个参数，例如


IF(lft >= 15 AND rgt <= 16, lft - 1)
                          ^        ^^^^



IF（lft>=15且rgt=15且rgt您错过了内部IF（）中的第三个参数

试试这个


   UPDATE categories 
    SET lft = IF( lft >= 17 AND rgt <= 18, lft + 1, IF(lft >= 15 AND rgt <= 16, lft - 1,0)), 
        rgt = IF( lft >= 17 AND rgt <= 18, rgt + 1, IF(lft >= 15 AND rgt <= 16, rgt - 1,0))
    WHERE (rgt <= 18 AND lft >= 15)



更新类别

设置lft=IF（lft>=17和rgt=15，rgt=17和rgt=15，rgt@JanKožušník替换
else_stmt_expre
）

   UPDATE categories 
    SET lft = IF( lft >= 17 AND rgt <= 18, lft + 1, IF(lft >= 15 AND rgt <= 16, lft - 1,0)), 
        rgt = IF( lft >= 17 AND rgt <= 18, rgt + 1, IF(lft >= 15 AND rgt <= 16, rgt - 1,0))
    WHERE (rgt <= 18 AND lft >= 15)