Question

我在这里看不到问题。我在输入单元格中输入数据，提交后只刷新页面，不在MySQL中发布任何内容。我是通过观看旧的在线教程来做到的，所以也许有一些旧的方法，这可能是一个问题。

<?php
        include "../db/connect.php";
        if (isset($_POST['pavadinimas'])) {

            $pavadinimas = mysqli_real_escape_string($con, $_POST['pavadinimas']);
            $kaina = mysqli_real_escape_string($con, $_POST['kaina']);
            $info = mysqli_real_escape_string($con, $_POST['info']);
            $gamintojas = mysqli_real_escape_string($con, $_POST['gamintojas']);
            $gamintojas = mysqli_real_escape_string($con, $_POST['atmintis']);
            $tipas = mysqli_real_escape_string($con, $_POST['tipas']);
            $kiekis = mysqli_real_escape_string($con, $_POST['kiekis']);
            // See if that product name is an identical match to another product in the system
            $sql = mysqli_query($con, "SELECT id FROM prekes WHERE pavadinimas='$pavadinimas' LIMIT 1");
            $productMatch = mysqli_num_rows($sql); // count the output amount
            if ($productMatch > 0) {
                echo '<script type="text/javascript">alert("KLAIDA! Bandėte įkelti prekę, kurios pavadinimas jau yra įrašytas duomenų bazėje.");</script>';
                exit();
            }
            // Add this product into the database now
            $sql = mysqli_query($con, "INSERT INTO prekes (pavadinimas, kaina, info, gamintojas, atmintis, tipas, kiekis, laikas) 
                VALUES('$pavadinimas','$kaina','$info','$gamintojas','$atmintis','$tipas','$kiekis',now())") or die (mysqli_error($con));
             $pid = mysqli_insert_id();
            // Place image in the folder 
            $newname = "$pid.jpg";
            move_uploaded_file( $_FILES['fileField']['tmp_name'], "../inventory_images/$newname");
            header("location: itemList.php"); 
            exit();
        }
        ?>

Answer 1

只有一个愚蠢的错误。在html中，SUBMIT按钮名称设置为“按钮”，不提交，因此它没有将其发布到数据库中。

和这个

if (isset($_POST['pavadinimas']))

应该是这个

if (isset($_POST['submit']))

PHP不会从表单中输入的数据发布到DB？

1 个答案: