Question

正如标题所述，我试图将数据插入2个表中，如果表单通过了几个条件（用户登录后，用户之前没有对该产品投票，所有字段都已填写）。

当条件只是：

时，它正在将数据完美地插入到两个表中

$sql = "SELECT productid FROM votes WHERE username='$username' LIMIT 1";

但是我意识到，如果用户对任何产品进行了投票，那么他们的用户名就会显示在表格中，如果没有对产品进行投票，他们就会失败。所以我刚补充说：

$sql = "SELECT productid FROM votes WHERE username='$username' AND productid='$id' LIMIT 1";

现在，如果我尝试向数据库提交数据，它总是返回'插入投票表中的错误'消息，但不返回mysql_error（），显然不会在投票表中插入新行，但奇怪的是它会更新产品表。

我无法弄清楚发生了什么，所以如果有人能帮我诊断问题，我真的很感激！这是代码：

<?php
    if($_SERVER['REQUEST_METHOD'] == 'POST'){
        if($_POST['slider_surface'] !== "0" && $_POST['slider_edgewear'] !== "0" && $_POST['slider_centering'] !== "0" && $_POST['slider_corners'] !== "0"){
            $dbhost = 'localhost';
            $dbuser = 'root';
            $dbpass = 'root';
            $conn = mysql_connect($dbhost, $dbuser, $dbpass);
            if(! $conn )
                {
                    die('Could not connect: ' . mysql_error());
                }

            $slider_surface = $_POST['slider_surface'];
            $slider_edgewear = $_POST['slider_edgewear'];
            $slider_centering = $_POST['slider_centering'];
            $slider_corners = $_POST['slider_corners'];
            $id = preg_replace('#[^a-z0-9]#i', '', $_GET['id']);
            session_start();
            $username = $_SESSION['username'];
            //check if user has already voted
            mysql_select_db('products');
            $sql = "SELECT productid FROM votes WHERE username='$username' AND productid='$id' LIMIT 1";
            $query = mysql_query( $sql, $conn );
            $uname_check = mysql_num_rows($query);
            if ($username){
                if ($uname_check < 1) {


                    $sql =  "INSERT INTO votes ".
                            "(username,productid,votesurface,voteedgewear,votecentering,votecorners,datetime) ".
                            "VALUES('$username','$id','$slider_surface','$slider_edgewear','$slider_centering','$slider_corners', now())";

                    $retval = mysql_query( $sql, $conn );

                    $id='';

                    // Make sure the _GET product ID is set, and sanitize it
                    $id = preg_replace('#[^a-z0-9]#i', '', $_GET['id']);

                    //Retrieves data from MySQL 
                    $data = mysql_query("SELECT * FROM products WHERE id='$id'") or die(mysql_error()); 
                    $product = mysql_fetch_array( $data );

                    $newvotecount = $product['votecount'] + 1;
                    $newsum_surface = $product['sumsurface'] + $slider_surface;
                    $newsum_edgewear = $product['sumedgewear'] + $slider_edgewear;
                    $newsum_centering = $product['sumcentering'] + $slider_centering;
                    $newsum_corners = $product['sumcorners'] + $slider_corners;

                    $sql =  "UPDATE products SET votecount='{$newvotecount}', sumsurface='{$newsum_surface}', sumedgewear='{$newsum_edgewear}', sumcentering='{$newsum_centering}', sumcorners='{$newsum_corners}' WHERE id='$id'";

                    $retval2 = mysql_query( $sql, $conn );


                    if(! $retval){
                        die('Error inserting into votes table: ' . mysql_error());
                    }
                    else if(! $retval2){
                        die('Error inserting into products table: ' . mysql_error());
                    }
                    $grading_error = 'success';
                    mysql_close($conn);
                } else
                    $grading_error = 'duplicateuser';
        } else
            $grading_error = 'nouser';
        }
    else
        $grading_error = 'emptyfields';}
?>

Answer 1

将数据插入表时出现问题，因此INSERT语句必定存在错误。正如@ user2910809在之前的评论中所说，sql评估为：

INSERT INTO votes (username,
                   productid,
                   votesurface,
                   voteedgewear,
                   votecentering,
                   votecorners,
                   datetime) 
           VALUES ('magmar',
                   '52',
                   '7',
                   '6',
                   '5',
                   '5',
                   now())

这句话在句法上是正确的。因此，如果出现错误，则必须使用插入的值。

正如@ user2910809对其评论所述，UNIQUE列上有一个username键，这就是抛出错误的原因。

因此解决方案是修改表的索引以允许多行具有相同的username。

修改：根据评论中的建议，解决此问题的SQL是：ALTER TABLE votes DROP INDEX username;

MySQL PHP - 将表单数据插入表中。未知错误

1 个答案: