Question

我有一个下拉列表，该列表使用数据库中的表中的数据导入。下拉值由值MR_ID组成，该值与位于不同表中的MR_Name连接。根据所选内容，我想显示一个表格，该表格显示与下拉列表中所选内容相关的必要信息。因此，例如，如果我在下拉列表中选择“1 - Company A”，我希望Supp_ID值（可以超过1个值）与要显示在表中的“1 - Company A”值相关联。我怎么能这样做？

该表只有2列，MR_ID（显示在下拉列表中）和Supp_ID。

之前我有这个工作，但我不得不改变我的查询而不是使用CTE而且它让我失望。

HTML下拉列表：

<!-- Form -->    
<form name="myForm" action="">

<!-- Dropdown List -->
    <select name="master_dropdown" id="mr_id">

    <option selected value="select">Choose a MR_ID</option>
        <?php foreach($users->fetchAll() as $user) { ?>
            <option data-name="<?php echo $user['MR_ID'];?>">
                <?php echo $user ['MR_ID'];?>
            </option>
        <?php } ?>
    </select>
</form>

以下是我的主页上的新旧查询（index.php）......

// Old query on main page (index.php)
$sql = "SELECT DISTINCT CAST(MR_ID AS INT) AS MR_ID FROM Stage_Rebate_Index";

// New query on main page (index.php)
$sql = "WITH cte AS (
SELECT DISTINCT CONCAT(CAST(Stage_Rebate_Index.MR_ID AS INT),' - ', Stage_Rebate_Master.MR_Name) AS MR_ID
    , Stage_Rebate_Index.MR_ID AS sort_column
FROM Stage_Rebate_Index
LEFT JOIN Stage_Rebate_Master
ON Stage_Rebate_Master.MR_ID=Stage_Rebate_Index.MR_ID
)
SELECT MR_ID
FROM
    cte
ORDER BY
    sort_column;";

以下是我在test-table.php页面上的旧查询和新查询

// Query that is used in ajax call (test-table.php)
    $sql_one = "SELECT CAST(Supp_ID AS INT) AS Supp_ID, CAST(MR_ID AS INT) AS MR_ID FROM Stage_Rebate_Index WHERE MR_ID = '$mr_id'";

// Query that is used in ajax call (test-table.php)
$sql_one = "WITH cte AS (
SELECT DISTINCT CONCAT(CAST(Stage_Rebate_Index.MR_ID AS INT),' - ', Stage_Rebate_Master.MR_Name) AS MR_ID
    , Stage_Rebate_Index.MR_ID AS sort_column, CAST(Stage_Rebate_Index.Supp_ID as INT) AS Supp_ID
FROM Stage_Rebate_Index
LEFT JOIN Stage_Rebate_Master
ON Stage_Rebate_Master.MR_ID=Stage_Rebate_Index.MR_ID
)
SELECT MR_ID, Supp_ID
FROM
    cte
WHERE
    MR_ID = '$mr_id'
ORDER BY
    sort_column;";

这是我的ajax（index.js）以及一行带来$ _POST值的代码......

// Reads what the user selects from the drop down list and displays table when a selection is made
function updatetable(myForm) {

    function show() { document.getElementById('index-table').style.display = 'block'; }


    var selIndex = myForm.selectedIndex;
    console.log();
    var selName = $( "#mr_id option:selected" ).text();

// Ajax sends POST method to Stage_Rebate_Index table and pulls information based on drop down selection

$.ajax ({
    url: "test-table.php",
    method: "POST", //can be post or get, up to you
    data: {
        mr_id : selName
    },
    beforeSend: function () {
        //Might want to delete table and put a loading screen, otherwise ignore this
    },
    success: function(data){
        $("#table_div").html(data); // table_div is the div you're going to put the table into, and 'data' is the table itself.
    }
 });

}

$ _ POST方法引入test-table.php ...

$mr_id = $_POST['mr_id'];

test-table.php脚本......

<?php
$host="xxxxxxxxxxxx"; 
$dbName="xxxxx"; 
$dbUser="xxxxxxxx"; 
$dbPass="xxxxxxxxxxxxxx";

$mr_id = $_POST['mr_id'];

$dbh = new PDO( "sqlsrv:server=".$host."; Database=".$dbName, $dbUser, $dbPass);
$dbh->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );

$sql_one = "WITH cte AS (
SELECT DISTINCT CONCAT(CAST(Stage_Rebate_Index.MR_ID AS INT),' - ', Stage_Rebate_Master.MR_Name) AS MR_ID
    , Stage_Rebate_Index.MR_ID AS sort_column, CAST(Supp_ID as INT) AS Supp_ID
FROM Stage_Rebate_Index
LEFT JOIN Stage_Rebate_Master
ON Stage_Rebate_Master.MR_ID=Stage_Rebate_Index.MR_ID
)
SELECT MR_ID, Supp_ID
FROM
    cte
WHERE
    MR_ID = '$mr_id'
ORDER BY
    sort_column;";


//$users = $dbh->query($sql);
$users_one = $dbh->query($sql_one);
?>

<html>
    <body>

        <!-- Table -->
<p> 
    <div id="table_div">
        <table border="1" id="index_table" class="ui-widget ui-widget-content">
            <thead>
                <tr class="ui-widget-header">
                <td>MR ID</td>
                <td>Supplier ID</td>
                </tr>
            </thead>
            <?php foreach($users_one->fetchAll() as $supp) { ?>
            <tr>
                <td class="mr_id"><?php echo $supp['MR_ID'];?></td>
                <td class="supp_id"><?php echo $supp['Supp_ID'];?></td>
            </tr>
            <?php } ?>
        </table>
    </div>

    </body
    </html

Answer 1

有几种方法可以对此进行调试，并提供有关正在发生的事情的更多信息（通过查看从先前查询和当前查询获得的结果集的差异，只是孤立的查询）。如果您更新的查询只返回一组空结果或实际错误等，那将是一件好事。

然而，在顶部，我认为您的SQL中可能存在一个问题，即您尝试将int与字符串连接起来，这里：

CONCAT(CAST(Stage_Rebate_Index.MR_ID AS INT),' - ', Stage_Rebate_Master.MR_Name)

我假设MR_Name是一个varchar，所以我确定你需要做的一件事就是将MR_ID作为varchar而不是int转换。

查询可能还有其他问题，但最好还是了解实际结果是什么。

根据下拉列表选择显示表

1 个答案: