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PHP从存储过程返回输出

[英]PHP returning output from a Stored Procedure

我已经看到并阅读了有关此问题的一些问题,但我无法弄清楚发生了什么。 我有一个插入到表中的SP,然后仅返回带有最后一个插入ID的单列(因为我正在执行多个语句,所以lastInsertID()不起作用。

CREATE DEFINER=`graffixnyc`@`%` PROCEDURE `createUser`(
IN userName VARCHAR(50),
IN firstName VARCHAR(25),
IN lastName VARCHAR(25),
IN pass text)
BEGIN
SELECT FLOOR(RAND() * 0xFFFFFFFF) into @salt;
INSERT INTO `users` (`username`, `first_name`,`last_name`,`salt`,`password`)
VALUES (userName, firstName,lastName, (select
@salt),sha2(concat(md5(pass(select @salt)),256));
SELECT LAST_INSERT_ID()  as lastinsert;

END

现在,当我执行此操作时,MySql将返回最后插入的记录的值。 当我尝试使用PHP访问它时,得到Null。

$paramUsername = $req->params('username'); // Getting parameter with names
$paramFirstName = $req->params('firstname');
$paramLastName = $req->params('lastname');
$paramPassword = $req->params('password');
$sql = "CALL createUser(:username, :firstname,:lastname,:password)";
try {
    $dbCon = getConnection();
    $stmt = $dbCon->prepare($sql);  
    $stmt->bindParam("username", $paramUsername);
    $stmt->bindParam("firstname", $paramFirstName);
    $stmt->bindParam("lastname", $paramLastName);
    $stmt->bindParam("password", $paramPassword);
    $stmt->execute();
    $row= $stmt->fetch();
    $last_id=$row["lastinsert"];
    $user->id =$last_id;

我也尝试过使用类似这样的输出参数:

CREATE DEFINER=`graffixnyc`@`%` PROCEDURE `createUser`(
IN userName VARCHAR(50),
IN firstName VARCHAR(25),
IN lastName VARCHAR(25),
IN pass text,
OUT lastinsert INT)
BEGIN
SELECT FLOOR(RAND() * 0xFFFFFFFF) into @salt;
INSERT INTO `users` (`username`, `first_name`,`last_name`,`salt`,`password`)
VALUES (userName, firstName,lastName, (select
@salt),sha2(concat(md5(pass(select @salt)),256));
SET lastinsert=(SELECT LAST_INSERT_ID());
SELECT lastinsert;

END

$paramUsername = $req->params('username'); 
$paramFirstName = $req->params('firstname');
$paramLastName = $req->params('lastname');
$paramPassword = $req->params('password');
$sql = "CALL createUser(:username, :firstname,:lastname,:password,
@lastinsert)";

try {
    $dbCon = getConnection();
    $stmt = $dbCon->prepare($sql);  
    $stmt->bindParam("username", $paramUsername);
    $stmt->bindParam("firstname", $paramFirstName);
    $stmt->bindParam("lastname", $paramLastName);
    $stmt->bindParam("password", $paramPassword);
    $stmt->execute();
    $row = $dbCon->query("select @lastinsert;")>fetch();
    $last_id=$row["@lastinsert"];
    $user->id =$last_id;

当我这样尝试时,出现此错误:

{"error":{"text":SQLSTATE[HY000]: General error: 2014 Cannot execute queries
while other unbuffered queries are active. Consider using
PDOStatement::fetchAll(). Alternatively, if your code is only ever going to
run against mysql, you may enable query buffering by setting the
PDO::MYSQL_ATTR_USE_BUFFERED_QUERY attribute.}}

因此,我尝试了FetchAll并得到了相同的错误。.然后,我尝试了此操作,并在这里找到了可行的方法,但是由于它仅返回1行和1个字段,因此似乎并不需要遍历任何内容:

$stmt->execute();
$stmt = $dbCon->prepare("select @lastinsert;");  
$outputArray = $dbCon->query("select @lastinsert;")->fetchAll();
foreach($outputArray as $row)
{
    $last_id=$row["@lastinsert"];
}

那么,是否有更好的方法可以做到这一点? 我确定有。 我对PHP和mySQL还是很陌生,但是在SQLServer方面有很多经验,所以任何见识都会受到赞赏。

我想通了..这很好用,我不需要输入参数或循环:

SP:

CREATE DEFINER=`graffixnyc`@`%` PROCEDURE `createUser`(
IN userName VARCHAR(50),
IN firstName VARCHAR(25),
IN lastName VARCHAR(25),
IN pass text)
BEGIN
SELECT FLOOR(RAND() * 0xFFFFFFFF) into @salt;
INSERT INTO `users` (`username`, `first_name`,
`last_name`,`salt`,`password`)
VALUES (userName, firstName,lastName, (select @salt),sha2(concat(md5(pass
),(select @salt)),256));
SELECT LAST_INSERT_ID()as lastinsert;

END

PHP:

$paramUsername = $req->params('username'); 
$paramFirstName = $req->params('firstname');
$paramLastName = $req->params('lastname');
$paramPassword = $req->params('password');
$sql = "CALL createUser(:username, :firstname,:lastname,:password)";
try {
    $dbCon = getConnection();
    $stmt = $dbCon->prepare($sql);  
    $stmt->bindParam("username", $paramUsername);
    $stmt->bindParam("firstname", $paramFirstName);
    $stmt->bindParam("lastname", $paramLastName);
    $stmt->bindParam("password", $paramPassword);
    $stmt->execute();
    $user->id =$stmt->fetchColumn(0);
    $dbCon = null;

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