Question

我有此代码，我想开始与该数据库的新连接，但在以前的数据库连接中具有不同的值，但是无论何时我启动该连接，我在网站上收到的错误说“无法获取mysqli”，请帮助我不知道该怎么办

我尝试使用“ mysqli_close（$ conn）;”关闭与数据库的连接。但没有用

$result = mysqli_query($conn, "SELECT * FROM userissues, response WHERE response.idmatch = userissues.emailcount");
$count = mysqli_num_rows($result);

if($count == 0){
    mysqli_close($conn);

    $result = mysqli_query($conn,"SELECT * FROM userissues WHERE departmentSent = '$_SESSION[user_email]' ORDER BY dateSent DESC");
    $count = mysqli_num_rows($result);

    if($count == 0){
        echo '<p style="margin-left: 162px; width: 300px; float: left; height: 50px;">you have no emails</p>';
    }else{
        while($row = mysqli_fetch_array($result))
        {
            ?>
            <a href="email_preview.php?id=<?php echo $row['emailcount'];?>&site=inbox">
                <?php

            echo '
            <table border="1" id="'.$row["emailcount"].'" style="margin-left: 162px; width: 300px; float: left; height: 50px;">
            <tr><td>'.$row["user_email"].'</td></tr><br>
            <tr><td>'.$row["dateSent"].'</td></tr><br>
            <tr><td>'.$row["subject"].'</td></tr><br>
            <tr><td> '.$row["report"].'</td></tr><br><br>
            </table>
            </a>
            ';

                }
    }
    mysqli_close($conn);
}elseif($count > 0){
        mysqli_close($conn);

        $result = mysqli_query($conn,"SELECT * FROM userissues, response WHERE userissues.departmentSent = '$_SESSION[user_email]' AND response.user_receiver = '$_SESSION[user_email]' ORDER BY userissues.dateSent AND response.dateResponded DESC");
        $count = mysqli_num_rows($result);

        if($count == 0){
            echo '<p style="margin-left: 162px; width: 300px; float: left; height: 50px;">you have no emails</p>';
        }else{
            while($row = mysqli_fetch_array($result)){
                    echo'';
            }
        }
}

我只想建立到相同数据库但值不同的新连接

Answer 1

如果关闭连接mysqli_close($conn);，则无法再使用该连接发出查询。基本上，这将破坏您可以用来发出查询和检索结果的管道。

我认为您真正想做的是mysqli_free_result($result)

documentation for mysqli_free_result()

您的脚本向SQL Injection Attack开放   甚至if you are escaping inputs, its not safe!   在MYSQLI_或PDO API中使用prepared parameterized statements

如果确实需要关闭与数据库mysqli_close($conn);的连接，则必须发出另一个$con = mysqli_connect()来创建另一个连接。但是，这是一个相对较慢的过程，一个脚本实际上应该只打开一个连接。

想要连接到相同的数据库表但具有不同的值

1 个答案: