Question

我有3个选择。当用户选择第三个选择并将变量发布到另一个PHP文件时，将发生jQuery onchange，因此我可以根据选择的ID检查数据库。问题是，如何获取第一个和第二个选择变量并将其与第三个选择一起发布？

这些是我使用的表：

    fromlist table : 
    | id | fromname |
    |  1 |     A    |
    |  2 |     B    |
    |  3 |     C    |

    tolist table:
    | id | toname |
    |  1 |    Z   |

    productlist table:
    | id | name |
    |  1 | CPO  |
    |  2 | PK   |

    transport table:
    | id  | product_id | product | to_id | from_id | price  |
    |  1  |     100    |    1    |   1   |    1    | 100.00 |
    |  2  |     101    |    2    |   1   |    1    | 200.00 |
    |  3  |     102    |    1    |   1   |    2    | 300.00 |
    |  4  |     103    |    2    |   1   |    2    | 400.00 |
    |  5  |     104    |    2    |   1   |    3    | 500.00 |

代码如下：

<form name = "transport" method = "GET" action = "http://interlink2u.com/transport/transportfiles/verify.php">
        <select name = "phive_product_id" id = "phive_product_id" hidden></select>  

        <label>From:</label>
            <select name = "select_from" id = "select_from" required>
                <option id = "" value = "">None</option>
                <?php
                    $fromlist = mysqli_query($connection, "SELECT * FROM fromlist ORDER BY fromname ASC");

                    while ($row = mysqli_fetch_assoc($fromlist))
                    {
                        $getID = $row['id'];
                        $getName = $row['fromname'];
                        ?><option id = "<?php echo $getID; ?>" value = "<?php echo $getName; ?>"><?php echo $getName;?></option><?php
                    }
                ?>
            </select><br><br>

        <label>To:</label>
            <select name = "select_to" id = "select_to" required>
                <option id = "" value = "">None</option>
                <?php
                    $tolist = mysqli_query($connection, "SELECT * FROM tolist ORDER BY toname ASC");

                    while ($row = mysqli_fetch_assoc($tolist))
                    {
                        $getID = $row['id'];
                        $getName = $row['toname'];
                        ?><option id = "<?php echo $getID; ?>" value = "<?php echo $getName; ?>"><?php echo $getName;?></option><?php
                    }
                ?>
            </select>
            <br><br>

        <label>Product:</label>
            <select name = "select_product" id = "select_product" required>
                <option id = "" value = "">None</option>
                <?php
                    $products = mysqli_query($connection, "SELECT * FROM products ORDER BY name ASC");

                    while ($row = mysqli_fetch_assoc($products))
                    {
                        $getID = $row['id'];
                        $getName = $row['name'];
                        ?><option id = "<?php echo $getID; ?>" value = "<?php echo $getName; ?>"><?php echo $getName;?></option><?php
                    }
                ?>
            </select>

            <select name = "phive_booked_price" id = "phive_booked_price" hidden></select>
            <br><br>

        <input type = "submit" value = "Submit" id="disablebutton">
</form><br><br><br>

这是脚本：

jQuery(document).on("click", 'select#select_product', function(e) 
        {
            var productID = jQuery(this).children(":selected").attr("id");
            var fromID = jQuery('#select_from').children(":selected").attr("id");
            var toID = jQuery('#select_to').children(":selected").attr("id");

            $.ajax
            ({
                type: "POST",
                data: {productID: productID, fromID: fromID, toID: toID},
                url: 'http://interlink2u.com/transport/transportfiles/getprice.php',
                dataType: 'json',
                success: function(json) 
                {
                    var $el4 = jQuery("select#phive_product_id");
                    $el4.empty();
                    $.each(json, function(k, v) 
                    {                     
                        $el4.append("<option id = '" + v.product_id + "'value = '" + v.product_id + "'>" + v.product_id + "</option>");
                    });

                    var $el5 = jQuery("select#phive_booked_price");
                    $el5.empty();                        
                    $.each(json, function(k, v) 
                    {    
                        $el5.append("<option id = '" + v.product_id + "'value = '" + v.price + "'>" + v.price + "</option>");
                    });
                }
            });                
        })

getprice.php代码：

<?php 
    include_once 'config/config.php';

    $productID= $_POST['productID'];

    $result = mysqli_query($connection, "SELECT * FROM transport 
                                            INNER JOIN tolist ON transport.to_name = tolist.id
                                            INNER JOIN products ON transport.product = products.id 
                                            WHERE product = '" .$productID. "'");
    $results = [];

    while ($row = mysqli_fetch_assoc($result)) 
    {
        $results[] = $row;
    }

    header('Content-Type: application/json');
    echo json_encode($results);
?>

我想获取第一个和第二个选定的值，然后将其发布到getprice.php，以便我可以像这样查询结果：

<?php 
    include_once 'config/config.php';

    $fromID = $_POST['fromID'];
    $toID = $_POST['toID'];
    $productID = $_POST['productID'];

    $result = mysqli_query($connection, "SELECT * FROM transport 
                                            INNER JOIN tolist ON transport.to_name = tolist.id
                                            INNER JOIN products ON transport.product = products.id
                                            WHERE from_id = '" .$fromID. "'
                                            AND to_name = '".$toID."'
                                            AND product = '".$productID."'");
    $results = [];

    while ($row = mysqli_fetch_assoc($result)) 
    {
        $results[] = $row;
    }

    header('Content-Type: application/json');
    echo json_encode($results);
?>

Answer 1

您仅在ajax中发送productID：

data: {productID: productID},

如果将行更改为：

data: $('form[name="transport"]').serialize(),

Read more on serialize.您的jquery调用现在看起来像：

jQuery(document).on("click", 'select#select_product', function(e) 
    {
        $.ajax
        ({
            type: "POST",
            data: $('form[name="transport"]').serialize(),
            url: 'http://interlink2u.com/transport/transportfiles/getprice.php',
            dataType: 'json',
            success: function(json) 
            {
                var $el4 = jQuery("select#phive_product_id");
                $el4.empty();
                $.each(json, function(k, v) 
                {                     
                    $el4.append("<option id = '" + v.product_id + "'value = '" + v.product_id + "'>" + v.product_id + "</option>");
                });

                var $el5 = jQuery("select#phive_booked_price");
                $el5.empty();                        
                $.each(json, function(k, v) 
                {    
                    $el5.append("<option id = '" + v.product_id + "'value = '" + v.price + "'>" + v.price + "</option>");
                });
            }
        });                
    })

然后在您的getprice.php代码中需要更新：

$productID = $_POST['select_product']; // now select_product as per your selectbox name

但是您现在也可以使用：

$_POST['select_from']
$_POST['select_to']
$_POST['phive_booked_price']
$_POST['phive_product_id']

编辑：

要按照您在评论中的要求将其包括在内，您需要进行更新：

var productID = $('#select_product').val();
var fromID = $('#select_from').val();
var toID= $('#select_to').val();

Answer 2

在脚本中，您正在发送如下数据：data: {productID: productID, fromID: fromID, toID: toID},

因此，在getprice.php中，您应该收到类似以下的内容：

 $fromID = $_POST['fromID'];
 $toID = $_POST['toID'];
 $productID = $_POST['productID'];

如何从另一个选择中将先前选择的值转换为当前选择onclick？

2 个答案: