我的$stmtInsert执行两次,mysqli只有一次相同的数据

我有这两个功能.$ stmt查询在我的dbase中插入数据两次,即使使用检查例程也是如此.

一直试图抓住错误,但我找不到它.

常规的MySQLi例程以完全相同的方式调用,确实只插入一次数据.

我无法绕过原因.做了一些回声检查,echo'ed SQL insert etcetera.我只是不知道发生了什么.我还没有使用$ stmt,但是想要更好地利用程序员的生活.这没有帮助:-)

在XAMPP上运行测试:Apache/2.4.29(Win32)OpenSSL/1.1.0g PHP/7.2.1 Clientversie van数据库:libmysql - mysqlnd 5.0.12-dev - 20150407 PHP-versie:7.2.1

此代码插入两次:

$result = mysql_insert_array_serialised($conn, $dbt_serialsets, $_POST, $addon);


function mysql_insert_array_serialised($conn, $table, $data, $addon = "mc_api") {

        $random_code = substr(str_shuffle(MD5(microtime())), 0, 12);
        $settings    = serialize($data);
        $uid         = $_SESSION['uid'];

        $sql_chk = "SELECT sn FROM $table WHERE sn = ?";

        if ($stmt = $conn->prepare($sql_chk)) {
            $stmt->bind_param('s', $random_code);
            $stmt->execute();
            $stmt->store_result();

            if ($stmt->num_rows == 0) {
                // SN doesnt already exist
                $sql_insert = 'INSERT INTO ' . $table . ' (uid, addon, sn, settings) VALUES (?, ?, ?, ?)';
                if ($stmt       = $conn->prepare($sql_insert)) {
                    $stmt->bind_param('ssss', $uid, $addon, $random_code, $settings);
                    $stmt->execute();
                    if (!$stmt->execute()) {
                        echo $stmt->error;
                    }else{
                        return array(
                        "mysqli_error"         => false,
                        "sn"                   => $random_code
                        );  
                    }

                } else {
                    echo 'Could not prepare statement!';
                }
            }
        }
    }

虽然这个具有完全相同的数据,但应该完成工作.一个插页.

$result = mysql_insert_array($conn, $dbt_serialsets, $_POST, $addon);

function mysql_insert_array($conn, $table, $data, $addon) {

    $random_code = substr(str_shuffle(MD5(microtime())), 0, 12);
    $settings    = serialize($data);
    $uid         = $_SESSION['uid'];

    if ($conn->query("INSERT INTO $table (uid, addon, sn, settings) VALUES ('" . $uid . "', '" . $addon . "' , '" . $random_code . "', '" . $settings . "')")) {
        return array(
            "mysqli_error"         => false,
            "mysqli_insert_id"     => $conn->insert_id,
            "mysqli_affected_rows" => $conn->affected_rows,
            "mysqli_info"          => $conn->info,
            "sn"                   => $random_code
        );

    } else {
        return array("mysqli_error" => $conn->error);
    }
}   

我很困惑.关于类似问题的回复很少,有人说一旦上传到http服务器就解决了,我只在我的本地测试XAMPP上运行它

由于你们中的一些人认为双$ stmt->执行是问题,但事实并非如此.一个用于选择(检查exisitng),一个用于插入新数据.

原来的功能是这样的,改回来并仍然插入两次:

function mysql_insert_array_serialised_test($conn, $table, $data, $addon = "mc_api") {

    $random_code = substr(str_shuffle(MD5(microtime())), 0, 12);
    $settings    = serialize($data);
    $uid         = $_SESSION['uid'];

    $sql_insert = 'INSERT INTO ' . $table . ' (uid, addon, sn, settings) VALUES (?, ?, ?, ?)';
    if ($stmt       = $conn->prepare($sql_insert)) {
        $stmt->bind_param('ssss', $uid, $addon, $random_code, $settings);
        $stmt->execute();
        if (!$stmt->execute()) {
            echo $stmt->error;
        } else {
            return array(
                "mysqli_error" => false,
                "sn"           => $random_code
            );
        }
    } else {
        echo 'Could not prepare statement!';
    }        
}

显然,我读得不好.我很抱歉.帖子中提到的双重执行是存在的,删除1确实有帮助.我以为一个用于检查是否有有效结果,而不是实际再次执行查询.

去耻辱的角落*

Nick.. 5

你有两个调用$stmt->execute()这两个调用将执行INSERT:

                $stmt->execute();
                if (!$stmt->execute()) {

你可以删除第一个.

你有两个调用$stmt->execute()这两个调用将执行INSERT:

                $stmt->execute();
                if (!$stmt->execute()) {

你可以删除第一个.