Question

我需要使用PHP代码从数据库中显示一个表。我在PHP中的while循环之后放置的提交按钮显示在结果表显示之前这是我的代码......

     <?php
     echo '<form action="" id="todaymenu" method="post"                             onSubmit="setData()">';
         include("config.php");
         session_start();
         $sql="select * from menu";
        $result=mysqli_query($conn,$sql);
        if (!$result) {
                die("Query to show fields from table failed");
        } 
        $fields_num = mysqli_num_fields($result);
        echo "<table><tr>";
        // printing table rows
        while($row = mysqli_fetch_array($result))
        {
            //echo $row['image'];
                echo "<td>";
                echo '<img src="'.$row['image'].'" alt="" style="width:          300px; height: 300px;"" />';
                echo "</td>";
                echo "<td>";
                echo $row['name'];
                echo "</td>";
                echo "<td>";
                echo $row['description'];
                echo "</td>";
                echo "<td>";
                echo $row['price'];
                echo "</td>";
                echo "<td>";
                  echo "<select name=\"quantity\" required> <option value=\"0\"> 0 </option> <option value=\"1\"> 1 </option> <option value=\"2\"> 2 </option> <option value=\"3\"> 3 </option> <option value=\"4\"> 4 </option> <option value=\"5\"> 5 </option> <option value=\"6\"> 6 </option> /select>"; 
                 echo "</td>";
                echo "</tr>";

     }


                echo '<button type="submit"          onclick="window.location.href=\'placeorder.php\'" class="placeorder"><i          class="fa fa-shopping-cart fa-lg"></i> &nbsp; &nbsp;Place Order</button>';

     mysqli_close($conn);

     echo '<a href="#menu-toggle" class="btn btn-default" id="menu-         toggle">Toggle Menu</a> </form>';
     ?>

Answer 1

确保您的PHP输出的HTML有效。看起来您的结构已关闭，您缺少一些开始和结束标记。

确保使用启动表格并以结束，否则浏览器可能会错误地解释显示。

Answer 2

对代码进行这些更改，它应该可以正常工作！

    echo '<table>';
    // printing table rows
    while($row = mysqli_fetch_array($result))
    {
        //echo $row['image'];
            echo '<tr>';
            echo "<td>";
            echo '<img src="'.$row['image'].'" alt="" style="width:300px; height: 300px" />';
            echo "</td>";
            echo "<td>";
            echo $row['name'];
            echo "</td>";
            echo "<td>";
            echo $row['description'];
            echo "</td>";
            echo "<td>";
            echo $row['price'];
            echo "</td>";
            echo "<td>";
            echo '<select name="quantity" required> <option value="0"> 0 </option> <option value="1"> 1 </option> <option value="2"> 2 </option> <option value="3"> 3 </option> <option value="4"> 4 </option> <option value="5"> 5 </option> <option value="6"> 6 </option> </select>'; 
             echo "</td>";
            echo "</tr>";

 }
 echo '</table>';
 echo '<button type="submit" onclick="window.location.href=\"placeorder.php\"" class="placeorder"><i class="fa fa-shopping-cart fa-lg"></i> &nbsp; &nbsp;Place Order</button>';
 echo '<a href="#menu-toggle" class="btn btn-default" id="menu-toggle">Toggle Menu</a> </form>';
 mysqli_close($conn);

Answer 3

您可以使用替代语法使输出更容易。您还错过了标记中的一些标记。尝试这样的事情：

<?php
    include("config.php");
    session_start();
    $sql="select * from menu";
    $result=mysqli_query($conn,$sql);
    if (!$result) {
        die("Query to show fields from table failed");
    } 
    $fields_num = mysqli_num_fields($result);
?>
<form action="" id="todaymenu" method="post" onSubmit="setData()">
    <table>
    <?php while($row = mysqli_fetch_array($result)): ?>
        <tr>
            <td><img src="'.$row['image'].'" alt="" style="width: 300px; height: 300px;" /></td>
            <td><?= $row['name']; ?></td>
            <td><?= $row['description']; ?></td>
            <td><?= $row['price']; ?></td>
            <td>
                <select name="quantity" required>
                    <option value="0"> 0 </option>
                    <option value="1"> 1 </option>
                    <option value="2"> 2 </option>
                    <option value="3"> 3 </option>
                    <option value="4"> 4 </option>
                    <option value="5"> 5 </option>
                    <option value="6"> 6 </option>
                 </select>
             </td>
        </tr>
    <?php endwhile; ?>
    </table>
    <button type="submit" onclick="window.location.href='placeorder.php'" class="placeorder">
        <i class="fa fa-shopping-cart fa-lg"></i> &nbsp; &nbsp;Place Order
    </button>

    <?php mysqli_close($conn); ?>

    <a href="#menu-toggle" class="btn btn-default" id="menu-toggle">Toggle Menu</a>
</form>

php代码执行后无法显示提交按钮

3 个答案: