這個 MySQL 5.6 過程定義有什麼問題？使用變數作為表名

只是不明白為什麼會拋出這個錯誤？

錯誤程式碼：1064。您的 SQL 語法有錯誤；檢查與您的 MySQL 伺服器版本相對應的手冊，以在第 1 行的“NULL”附近使用正確的語法

drop procedure if exists dividir_tabla_por_usuario;

delimiter //
create procedure dividir_tabla_por_usuario(tabla varchar(100), columna_usuario varchar(100), columna_fecha varchar(100))
begin
 set @i = 0;
 set @q1 = concat('select count(*) from (select distinct ', columna_usuario ,' from ', tabla , ' order by ', columna_usuario ,' asc) as t1 into @c');
 prepare p1 from @q;
 execute p1;
while @i <= @c do
 set @r = 0;
 set @q2 = concat('select * from ', tabla , ' where ', columna_usuario ,' = (select ', columna_usuario ,' from (select @r:=@r +1 as ord, ', columna_usuario ,' from (select distinct ', columna_usuario ,' from ', tabla , 'order by ', columna_usuario ,' asc) as t2) as t3 where ord = ', @i ,' ) order by ', columna_fecha ,' asc limit 30000'); 
 prepare p2 from @q2;
 execute p2;
 set @i := @i + 1;
end while;
end//
delimiter ;

call dividir_tabla_por_usuario('mytable', 'mycolumn_1', 'mycolumn_2');

錯別字，總是錯別字。

這是最終的工作程式碼：

drop procedure if exists dividir_tabla_por_usuario;

delimiter //
create procedure dividir_tabla_por_usuario(tabla varchar(100), columna_usuario varchar(100), columna_fecha varchar(100))
begin
 set @i = 0;
 set @q1 = concat('select count(*) from (select distinct ', columna_usuario ,' from ', tabla , ' order by ', columna_usuario ,' asc) as t1 into @c');
 prepare p1 from @q1;
 execute p1;
while @i <= @c do
 set @r = 0;
 set @q2 = concat('(select distinct ', columna_usuario ,' from ', tabla , ' order by ', columna_usuario ,' asc) as t2)');
 set @q3 = concat('(select @r:=@r +1 as ord, ', columna_usuario ,' from ',@q2, ' as t3 where ord = ', @i ,' )');
 set @q4 = concat('(select ', columna_usuario ,' from ', @q3);
 set @q5 = concat('select * from ', tabla , ' where ', columna_usuario ,' = ', @q4 ,' order by ', columna_fecha ,' asc limit 30000'); 
 prepare p2 from @q5;
 execute p2;
 set @i := @i + 1;
end while;
end//
delimiter ;

call dividir_tabla_por_usuario('mytable', 'mycolumn1', 'mycolumn2');

引用自：https://dba.stackexchange.com/questions/59115