我已经看到这个问题被问了很多次,但是他们都很长,我只是无法理解他们正在做的事情......所以,有人能告诉我如何从中得到LAST_INSERT_ID()这个使用PDO进入php的过程:
CREATE TABLE names (
ID INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
name varchar(50) NOT NULL
)
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CREATE DEFINER=`root`@`localhost` PROCEDURE `simpleProcedure`(newname varchar(50), OUT returnid INT(11))
BEGIN
INSERT INTO names (name) VALUES (newname);
SET returnid = LAST_INSERT_ID();
END
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$stmt=$db->prepare("CALL simpleProcedure(:name,:returnid)");
$stmt->bindValue(':name',$name,PDO::PARAM_STR);
$stmt->bindParam(':returnid',$returnid,PDO::PARAM_INT,11);
$stmt->execute();
echo $returnid;
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但是,对于脑细胞比我更多的人来说,这可能是显而易见的.任何帮助赞赏.
事实证明,这是一个已经持续了很长时间的错误...自2005年以来!
这是最初的错误报告:2005年至2013年.这是新的错误报告:从2013年到现在.
有多种方法可以获得答案,我找到了其中一个并证明了这一点......
'技巧'是从'mysql'程序获取输出.这是一个"两个阶段"的过程.
第一部分是使用您的输入运行过程,并告诉它存储结果的MYSQL变量.
然后,运行单独的查询以"选择"那些"mysql"变量.
这里描述得非常清楚:php-calling-mysql-stored-procedures
更新(2017年1月):
这是一个示例,显示了'IN','INOUT'和'OUT'Mysql过程参数的变量的使用.
在我们开始之前,有一些提示:
当您尝试将变量绑定到INOUT和OUT参数时,您将获得一些非常奇怪的运行时错误.
像往常一样,我倾向于提供比所需更多的评论; - /
运行时环境(XAMPP):
源代码:
CREATE PROCEDURE `demoSpInOutSqlVars`(IN pInput_Param INT, /* PHP Variable will bind to this*/
/* --- */
INOUT pInOut_Param INT, /* contains name of the SQL User variable that will be read and set by mysql */
OUT pOut_Param INT) /* contains name of the SQL User variable that will be set by mysql */
BEGIN
/*
* Pass the full names of SQL User Variable for these parameters. e.g. '@varInOutParam'
* These 'SQL user variables names' are the variables that Mysql will use for:
* 1) finding values
* 2) storing results
*
* It is similar to 'variable variables' in PHP.
*/
SET pInOut_Param := ABS(pInput_Param) + ABS(pInOut_Param); /* always positive sum */
SET pOut_Param := ABS(pInput_Param) * -3; /* always negative * 3 */
END$$
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数据库连接:
$db = appDIC('getDbConnection', 'default'); // get the default db connection
$db->setAttribute(PDO::ATTR_EMULATE_PREPARES, true);
$db->setAttribute(PDO::ATTR_DEFAULT_FETCH_MODE, PDO::FETCH_ASSOC);
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注意:输出与EMULATE_PREPARES= false 相同.
设置将使用的所有PHP变量:
$phpInParam = 5;
$phpInOutParam = 404; /* PHP InOut variable ==> read and should be changed */
$phpOutParam = null; /* PHP Out variable ==> should be changed */
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定义并准备SQL过程调用:
$sql = "call demoSpInOut(:phpInParam,
@varInOutParam, /* mysql variable name will be read and updated */
@varOutParam)"; /* mysql variable name that will be written to */
$stmt = $db->prepare($sql);
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绑定PHP变量并设置SQL变量:
1)绑定PHP变量
$ stmt-> bindParam(':phpInParam',$ phpInParam,PDO :: PARAM_INT);
2)设置SQL用户INOUT变量
$ db-> exec("SET @varInOutParam = $ phpInOutParam"); //这是安全的,因为它只是将值设置为MySql变量.
执行程序:
$allOk = $stmt->execute();
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将SQL变量放入PHP变量中:
$sql = "SELECT @varInOutParam AS phpInOutParam,
@varOutParam AS phpOutParam
FROM dual";
$results = current($db->query($sql)->fetchAll());
$phpInOutParam = $results['phpInOutParam'];
$phpOutParam = $results['phpOutParam'];
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注意:也许不是最好的方法; - /
显示PHP变量
"$phpInParam:" => "5"
"$phpInOutParam:" => "409"
"$phpOutParam:" => "-15"
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