PHP 命令不同步错误 [英] PHP Commands Out of Sync error

问题描述

我在 PHP/MySQLi 中使用两个准备好的语句从 mysql 数据库中检索数据.但是，当我运行这些语句时，出现命令不同步，您现在无法运行该命令"错误.

I am using two prepared statements in PHP/MySQLi to retrieve data from a mysql database. However, when I run the statements, I get the "Commands out of sync, you can't run the command now" error.

这是我的代码:

    $stmt = $mysqli->prepare("SELECT id, username, password, firstname, lastname, salt FROM members WHERE email = ? LIMIT 1";
    $stmt->bind_param('s', $loweredEmail);
    $stmt->execute();
    $stmt->store_result();
    $stmt->bind_result($user_id, $username, $db_password, $firstname, $lastname, $salt);
    $stmt->fetch();

    $stmt->free_result();
    $stmt->close();

    while($mysqli->more_results()){
        $mysqli->next_result();
    }

    $stmt1 = $mysqli->prepare("SELECT privileges FROM delegations WHERE id = ? LIMIT 1");
    //This is where the error is generated
    $stmt1->bind_param('s', $user_id);
    $stmt1->execute();
    $stmt1->store_result();
    $stmt1->bind_result($privileges);
    $stmt1->fetch();

我尝试过的:

• 将准备好的语句移到两个单独的对象中.

• 使用代码:

• Moving the prepared statements to two separate objects.

• Using the code:

while($mysqli->more_results()){
    $mysqli->next_result();
}
//To make sure that no stray result data is left in buffer between the first
//and second statements

使用 free_result() 和 mysqli_stmt->close()

PS: 'Out of Sync' 错误来自第二条语句的 '$stmt1->error'

PS: The 'Out of Sync' error comes from the second statement's '$stmt1->error'

推荐答案

在 mysqli::query 如果您使用 MYSQLI_USE_RESULT 所有后续调用将返回错误 Commands out of sync 除非您调用 mysqli_free_result()

In mysqli::query If you use MYSQLI_USE_RESULT all subsequent calls will return error Commands out of sync unless you call mysqli_free_result()

在调用多个存储过程时，您可能会遇到以下错误:命令不同步；您现在无法运行此命令".即使在调用之间对结果对象使用 close() 函数时，也会发生这种情况.要解决此问题，请记住在每次调用存储过程后调用 mysqli 对象上的 next_result() 函数.请参见下面的示例:

When calling multiple stored procedures, you can run into the following error: "Commands out of sync; you can't run this command now". This can happen even when using the close() function on the result object between calls. To fix the problem, remember to call the next_result() function on the mysqli object after each stored procedure call. See example below:

<?php
// New Connection
$db = new mysqli('localhost','user','pass','database');

// Check for errors
if(mysqli_connect_errno()){
 echo mysqli_connect_error();
}

// 1st Query
$result = $db->query("call getUsers()");
if($result){
     // Cycle through results
    while ($row = $result->fetch_object()){
        $user_arr[] = $row;
    }
    // Free result set
    $result->close();
    $db->next_result();
}

// 2nd Query
$result = $db->query("call getGroups()");
if($result){
     // Cycle through results
    while ($row = $result->fetch_object()){
        $group_arr[] = $row;
    }
     // Free result set
     $result->close();
     $db->next_result();
}
else echo($db->error);

// Close connection
$db->close();
?>

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