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MySQL选择,存储在变量中_Mysql_Stored Procedures_Select_Local Variables - Fatal编程技术网

MySQL选择,存储在变量中

MySQL选择,存储在变量中,mysql,stored-procedures,select,local-variables,Mysql,Stored Procedures,Select,Local Variables,对于一个存储过程,我想做一个选择,并将列的值存储到一个变量中 我该怎么做 我想这样做: DECLARE countTemp INT; SET countTemp=(SELECT COUNT(Name) FROM mytable WHERE Name= var_name LIMIT 0,1); 或者,像这样: DECLARE countTemp INT; SELECT countTemp=ColumnXYZ FROM mytable WHERE Name= var

对于一个存储过程,我想做一个选择,并将列的值存储到一个变量中

我该怎么做

我想这样做:

    DECLARE countTemp INT;
    SET countTemp=(SELECT COUNT(Name) FROM mytable WHERE Name= var_name LIMIT 0,1);
或者,像这样:

    DECLARE countTemp INT;
    SELECT countTemp=ColumnXYZ FROM mytable WHERE Name= var_name LIMIT 0,1;
但是,我尝试了这些,MySQL说我的语法不正确;我如何做这样的事情?

像这样:

DECLARE myvar nvarchar(50);

SELECT ATextColumn INTO myvar FROM myTable LIMIT 1,1;

SELECT CONCAT('myvar is ',myvar ,' .');

您可以在select查询中轻松设置变量

SELECT @countTemp := ColumnXYZ FROM mytable WHERE Name= var_name LIMIT 0,1;

@countTemp是您的变量

您是否收到特定的错误消息?
脚本行:您的SQL语法有错误;检查与您的MySQL服务器版本对应的手册,以了解可在“DECLARE countTemp INT”附近使用的正确语法;SET countTemp=从第行的myTa'中选择COUNT(Name)
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