我想插入一個表中，從同一數據庫中的另一個表中獲取數據的PHP

Question

我為兩個表創建一個關系，一個包含項目，另一個包含項目的類別。 我可以用categorys表中的值填充option標記。 現在我想將從選項列表中選擇的值發布到items表中的category_id（我將其設置為int，與categorys表中的id一樣），但是我在添加其他信息的同時得到了category_id空白。 我怎樣才能解決這個問題？ 下面的代碼。

<?
    session_start();
    $_SESSION['message'] = "";

    $mysqli = new mysqli('localhost', 'root', '', 'auction');
    if($_SERVER['REQUEST_METHOD'] == 'POST') {
        $item_name = $mysqli->real_escape_string($_POST['item_name']);
        $category_id = $mysqli->real_escape_string($_POST['cat']);
        $item_description = $mysqli->real_escape_string($_POST['item_description']);
        $item_image_path = $mysqli->real_escape_string('Images/item_img/' . $_FILES['item_image']['name']);

        //make sure file is of image type
        if(preg_match("!image!", $_FILES['item_image']['type'])) {
            if(copy($_FILES['item_image']['tmp_name'], $item_image_path)) {
                $_SESSION['item_name'] = $item_name;
                $_SESSION['cat'] = $category_id;
                $_SESSION['item_description'] = $item_description;
                $_SESSION['item_image'] = $item_image_path;
                //inserting into the database
                $sql = "INSERT INTO items (item_name,category_id, item_image, item_description)VALUES('$item_name', '$category_id', '$item_image_path','$item_description')";
                if($mysqli->query($sql) === true) {
                    $_SESSION['message'] = "Item Upload Successful!";
                } else {
                    $_SESSION['message'] = "file upload failed";
                }
            } else {
                $_SESSION['message'] = "file copying failed";
            }
        } else {
            $_SESSION['message'] = "please upload gif, jpg, png";
        }

    }

    $result = $mysqli->query("SELECT * FROM items ORDER BY rand() LIMIT 10") or die($mysqli->error);

?>

<html>

    <head>
        <title>Upload item</title>
        <link rel="StyleSheet" href="Bootstrap/css/bootstrap.main.css">
        <link rel="StyleSheet" href="Bootstrap/css/bootstrap.min.css">
        <link rel="StyleSheet" href="style.css">
        <!--for countdown timer-->
        <script type="text/javascript">
            setInterval(function() {
                var xmlhttp = new XMLHttpRequest();
                xmlhttp.open("GET", "response.php", false);
                xmlhttp.send(null);
                document.getElementById("timer").innerHTML = xmlhttp.responseText;

            }, 1000);
        </script>
    </head>

    <body>
        <div>
            <!--to display records from database-->
            <div class="row col-sm-12 c-head mar-pad">
                <?php
                $A = 0;
                while($auction = $result->fetch_assoc()):
                    ?>
                    <div class="grid ">
                        <h4><?= $auction['item_name'] ?></h4>
                        <img src='<?= $auction[' item_image '] ?>' class='img-responsive'>
                        <span id="timer" class="timer"></span>
                        <button class="c-button" name='bid'>Bid Now!</button>
                    </div>
                    <?php
                    if($A % 4 == 0)
                        echo "<br/>";
                    $A++;
                endwhile;
                ?>
            </div>

            <!--for file upload form-->
            <form class="form-horizontal" role="form" action="auction_upload.php" method="POST" enctype="multipart/form-data">
                <h1><?=$_SESSION['message']?></h1>
                <div class=" form-group">
                    <label class="control-label col-sm-2">Item Name:</label>
                    <div class="col-sm-8">
                        <INPUT type="text" class="form-control" name="item_name" required/>
                    </div>
                </div>
                <div class="form-group">
                    <label class="control-label col-sm-2">Category:</label>
                    <div class="col-sm-8 ">

                        <select class='form-control'>
                            <?php
                                $mysqli = new mysqli('localhost','root','','auction');
                                $result1 = $mysqli->query("SELECT * FROM `categories`");

                                while ($row = mysqli_fetch_array($result1)):;?>
                                    <option name="cat" value="<?=$row[0];?>">
                                        <?=$row[1];?>
                                    </option>

                                <?php endwhile;?>
                        </select>
                    </div>
                </div>
                <div class="form-group">
                    <label class="control-label col-sm-2">Item Image:</label>
                    <div class="col-sm-8">
                        <INPUT type="file" class="form-control" name="item_image" accept="image/*" required/>
                    </div>
                </div>
                <div class="form-group">
                    <label class="control-label col-sm-2">Item Description:</label>
                    <div class="col-sm-8">
                        <textarea class="form-control" name="item_description" required>
                        </textarea>
                    </div>
                </div>

                <div class="form-group">
                    <div class="col-sm-offset-2 col-sm-8">
                        <button type="submit" class="btn btn-default" name="upload">Upload</button>
                    </div>
                </div>
            </form>
        </div>
    </body>
</html>

Answer 1

您的select元素沒有name ：

<select class ='form-control' >

因此，該值根本不會發送到服務器。 為了在表單發布中包含該元素的值，表單元素需要一個name （這是鍵/值對中的鍵）。

由於您要在$_POST['cat']查找，因此名稱為“ cat”：

<select class ='form-control' name="cat" >

我想插入一個表中，從同一數據庫中的另一個表中獲取數據的PHP

問題描述

1 個解決方案

解決方案1
0 2017-08-07 15:27:41

我想插入一個表中，從同一數據庫中的另一個表中獲取數據的PHP

問題描述

1 個解決方案

解決方案1 0 2017-08-07 15:27:41

解決方案1
0 2017-08-07 15:27:41