基於下拉選擇的顯示錶格

我有一個下拉列表，它是從數據庫中的一個表格導入的。下拉值由與位於不同表中的MR_Name連接的值MR_ID組成。根據所選內容，我想顯示一個表格，其中顯示了與下拉列表中所選內容相關的必要信息。因此，例如，如果我在下拉列表中選擇「1 - 公司A」，我想要在表格中顯示與「1 - 公司A」值相關的Supp_ID值（可以多於1個值）。我怎樣才能做到這一點？基於下拉選擇的顯示錶格

該表只有2列，MR_ID（這是下拉列表中顯示的內容）和Supp_ID。

我有這個工作之前，但我不得不改變我的查詢使用CTE來代替，它把我扔了。

HTML下拉：

<!-- 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>

這裏是我的新老查詢我的主網頁（的index.php）上...

// 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;";

這裏是我的新老查詢我的測試-table.php頁

// 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;";

這是我的AJAX（index.js）用一行代碼，在$ _POST值帶來沿...

// 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方法帶來的測試table.php ...

$mr_id = $_POST['mr_id'];

測試table.php腳本...

<?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

來源

2016-12-15 Rataiczak24

有幾種方法你可以調試它，並提供更多關於正在發生的事情的信息（通過查看你從前面的查詢和當前查詢得到的結果集中的差異，僅僅是查詢）。這將是很好的知道，如果您更新的查詢是簡單地返回一組空的結果，或者一個實際的錯誤等

不過，我認爲可能是在您的SQL中可能存在的問題，因爲你是試圖連接int一個字符串，在這裏：

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

我認爲MR_Name是一個varchar，所以我敢肯定，你必須要做的事情之一鑄MR_ID爲varchar，而不是INT。

查詢也可能存在其他問題，但最好更好地瞭解實際結果。

來源

2016-12-15 18:56:29 Sabir

MR_Name是一個'NVARCHAR' ...如果我在我正在使用的SQLPro Studio中運行這兩個查詢，兩次都會返回我需要的結果 – Rataiczak24

並且當您在您的Ajax中console.log數據變量'onSuccess'函數 - 是否記錄您期待的結果？ – Sabir

在控制檯中，它會根據我的下拉列表選項顯示值的表格的HTML代碼 – Rataiczak24

基於下拉選擇的顯示錶格

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