Question

这是我的家庭作业，我目前停留在第5部分： http://yorktown.cbe.wwu.edu/sandvig/mis314/assignments/Assignment06.aspx

结果应该是这样（选择一个类别，并在右边的div中填充相关产品）： http://yorktown.cbe.wwu.edu/sandvig/MIS314/Assignments/A06/CategoryItems.php?catID=1

这是我的版本的当前状态： https://yorktown.cbe.wwu.edu/students/192/roversg/A06/categoryitems.php?category=T-shirts

这里的三个表格均为PDF格式： https://1drv.ms/f/s!AqSyu9NNp2DFtZMQgnJuOWgwhVELsQ

我不知道如何在第二个$ sql =“ SELECT [...]”语句中调用一个选定类别中的每个项目。我知道我必须加入表，但是当指定CatName时，如何使echo语句显示与特定CatID相关的ItemID等？

下面，我将发布我的.php文件。

第一部分工作正常

<!DOCTYPE html>
<html><head>
        <title>Product Categories</title>
        <link href="/sandvig/mis314/assignments/style.css" rel="stylesheet" type="text/css">
        <link rel="stylesheet" type="text/css" href="main.css">
    </head>
    <body>
        <?php
        include 'databaseconnection.php';

        //connect to database
        $link = fConnectToDatabase();

        $category = $_GET['category'];
        ?>

        <div class="pageContainer">
            <div class="centerText">
                <h3>Product Category Count</h3>
                <hr>
            </div>
            <div class='equalColumnWraper'>
                <div class='leftColumn'>     
                    <div class="centerText">
                        <h3>Categories</h3>
                    </div>


                    <?php
                    //List records
                    $sql = "SELECT ab.ItemID, ab.CatID, CatName 
                        ,COUNT(CatName) as count
                        from geekproducts a, geekcategories b, geekproductcategories ab
                        WHERE b.CatID = ab.CatID AND a.ItemID = ab.ItemID
                        GROUP BY CatID"
                    ;

                    //$result is an array containing query results
                    $result = mysqli_query($link, $sql)
                            or die('SQL syntax error: ' . mysqli_error($link));

// iterate through the retrieved records
                    while ($row = mysqli_fetch_array($result)) {
                        //Field names are case sensitive and must match
                        //the case used in sql statement
                        echo "
        <a class='menuLink' href='?category={$row['CatName']}'>{$row['CatName']} ({$row['count']})</a>";
                    }
                    ?> 

                </div>

以下是我的问题

                    //List records
                    if (isset($category)) {
                        $sql = "SELECT ab.ItemID, ab.CatID, CatName, Name, Image, price

                        from geekproducts a, geekproductcategories ab, geekcategories b
                        WHERE b.CatName = '$category'
                        ";
                    } else {
                        $sql = "SELECT Name, Image, price
                FROM geekproducts, geekcategories 
                ORDER by Name";
                    }

                    //$result is an array containing query results
                    $result = mysqli_query($link, $sql)
                            or die('SQL syntax error: ' . mysqli_error($link));

                    echo "<div class='itemCount'>We sell " . mysqli_num_rows($result) . " items in '$category'</div>";

                    while ($row = mysqli_fetch_array($result)) {
                        //Field names are case sensitive and must match
                        //the case used in sql statement\
                         echo "                            
                            <div class='productItem'>
                            <img src='/sandvig/mis314/assignments/a06/images/m_{$row['Image']}'  class='productImage'>
                            <div class='productName'>{$row['Name']}</div>
                            <div class='productPrice'>Price: ${$row['price']}</div>
                            <div class='productDesc'> {$row['ShortDesc']}</div>
                            </div>
                            ";
                    }
                    ?>                    



                    <div class="clearfix"></div>
                </div>

                </body></html>

谢谢！

Answer 1

您可以重新编写或将查询更改为此：

SELECT gp.ItemID, gc.CatID, gc.CatName, gp.Name, gp.Image, gp.Price 
FROM geekproducts gp 
JOIN geekproductcategories gpc ON gpc.ItemID = gp.ItemID
JOIN geekcategories gc ON gc.CatID = gpc.CatID
WHERE gc.CatName = "{$category}"

PHP和MySQL-仅显示选定类别的产品和相关信息-三个表

1 个答案: