如何通过php脚本将输入参数传递给mysql存储过程 [英] how to pass input parametr to a mysql stored procedure through php script

问题描述

我有一个商店程序...



DELIMITER $$

创建程序getdatabyinput1

（



IN Uname int



）

BEGIN

SELECT UserNameID，userName，pass FROM username

where userName = Uname;

END $$



DELIMITER;



和我的php脚本是......





$ Uname = $ _ POST ['user'];

$ Upass = $ _ POST ['pass'] ;



$ result = mysqli_query（$ con，'CALL getdatabyinput1（$ Uname）'）或死（wrong1）;





while（$ row = mysqli_fetch_array（$ result））

{

echo $ row ['userName ']。 。 $ row ['pass'];

echo
;

}




i也搜索google ..但无法理解，实际上我是php的新手..所以请尽可能描述

解决方案

创建程序getdatabyinput1

（



IN Uname int



）

BEGIN

SELECT UserNameID，userName，pass FROM username

其中userName = Uname;

END

DELIMITER;



和我的PHP脚本是......

UNAME =

i have an store procedure ...

DELIMITER $$
create procedure getdatabyinput1
(

IN Uname int

)
BEGIN
SELECT UserNameID,userName,pass FROM username
where userName=Uname;
END $$

DELIMITER ;

and my php script is ...


$Uname=$_POST['user'];
$Upass=$_POST['pass'];

$result= mysqli_query($con,'CALL getdatabyinput1($Uname)') or die("wrong1");


While($row = mysqli_fetch_array($result))
{
echo $row['userName'] . " " . $row['pass'];
echo "
";
}


i also search on google .. but unable to understand , actually i am new to php .. so pls describe if possible

解决方案

create procedure getdatabyinput1
(

IN Uname int

)
BEGIN
SELECT UserNameID,userName,pass FROM username
where userName=Uname;
END

DELIMITER ;

and my php script is ...

Uname=

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