返回一个结果的javascript innerhtml

单击复选框时，只有第一行显示文本框

当我选中复选框时，文本框将显示在该行中

当我单击submit按钮时，它只插入第一行的数据

例如，当我选中3个框时，它将插入第一行的数据3次

如何修复我的代码

<script type="text/javascript">
    function changeText(codeasset){
          if(codeasset.checked){

        document.getElementById('hotStuff').innerHTML = '<input type="text" id="hotStuff" style="width: 150px;" />';

    }else {
        document.getElementById('hotStuff').innerHTML='';
    }}
</script>


<?php

     $assets_supplier_query = mysql_query("SELECT * FROM supplier_assets WHERE status = '0' AND supplier_id = '$supplier_id'");

      while ($get_supplier_asset = mysql_fetch_array($assets_supplier_query))
                                {

        $asset_code = $get_supplier_asset['asset_code'];
        $asset_qty = $get_supplier_asset['supply_quantity'];

        $asset_query = mysql_query("SELECT * FROM ref_assetmodel WHERE modelID = '$asset_code' ");
        $get_asset = mysql_fetch_array($asset_query);
        $asset_model = $get_asset['assetModel'];
        $specifications = $get_asset['specifications'];
        $category_id = $get_asset['categoryID'];
        $type_id = $get_asset['typeID'];

        $category_query = mysql_query("SELECT * FROM ref_assetcategory WHERE categoryID = '$category_id' ORDER BY categoryID");
        $get_category = mysql_fetch_array($category_query);
        $category = $get_category['assetCategory'];

        $type_query = mysql_query("SELECT * FROM ref_assettype WHERE typeID = '$type_id' ");
        $get_type = mysql_fetch_array($type_query);
        $type = $get_type['assetType'];
?>

<form action="../process/approve_supplies_process.php" method="post" name="supplier">

       <td><input type="checkbox" id="codeasset" name="supply[]" onclick="changeText(this)" value="<?php echo $asset_code ?>">&nbsp;</td>

        <td name="supply[]"><?php echo $type; ?></td>
        <td name="supply[]"><?php echo $asset_model; ?></td>
        <td name="supply[]"><?php echo $asset_qty; ?></td>
        <td name="supply[]" id="hotStuff" value="<?php echo $asset_code ?>"></td>

        <?php

                    }
        ?>

      <tr>

           <td align="center"><input type="submit" class="edit_butt" name="action" value="Submit" onclick=""></td>

            <input type="hidden" name="supplier_id" value="<?php echo $supplier_id; ?>">          

        </tr> 
     </form>




<?php
        $asset_code = $_POST['asset_code'];
        $categoryID = $_POST['category_id'];
        $type_id = $_POST['type_id'];
        $supplier_id = $_POST['supplier_id'];

        date_default_timezone_set('Asia/Manila');
        $inventoryDate= date("Y-m-d");

        if(isset($_POST['supply'])){


            foreach ($_POST['supply'] as $_value)
            {
                echo "Box #{$_value} was selected!\n";
                mysql_query("INSERT INTO ref_inventory(inventoryDate, categoryID, typeID, modelID, serialNumber, qrCode, is_owned, supplier_id)
                VALUES('$inventoryDate', '$categoryID', '$type_id', '$asset_code', '954', '', 1, '$supplier_id') ");
            }
        }
?>

---

id name id=hotStuff不应该有多个同名元素。您应该使用css类或唯一id名称来循环表列。改进的问题文本和代码段