如何从php中第一个动态生成的第二个下拉列表将数据存储到数据库中


how to store data into database from dynamically generated second drop down from first in php?

这里我试图存储第二个下拉列表中的数据,该列表是使用javascript从第一个下拉列表动态生成的。它会存储第一个下拉列表命名类别的值,但不会将第二个下拉列表Sb_category的值存储到我的数据库表中,有一个代码我已经尝试过了。。。请检查一下,给我正确的路!!!!!

我使用过的表格:

  <div class="form-group" id="style_container_div">
                                    <label>Choose Category: </label>
                                    <select size="1" id="Category" class="form-control" title="" type="select" name="Category" value="-Select Your Category-">
                                        <optgroup>
                                            <option value="">-Select Your Category-</option>
                                            <option value="Charging">Charging</option>
                                            <option value="Consent">Consent</option>
                                            <option value="Content">Content</option>
                                            <option value="Feature">Feature</option>
                                            <option value="Navigation_flow">Navigation flow</option>
                                            <option value="Service_Consuption">Service Consuption</option>
                                            <option value="Service_Provisioning">Service Provisioning</option>
                                        </optgroup>
                                    </select>
                                    <div class="clear"></div>
                                    <div id="error-message-style"></div>
                                   </div>
                                    <div id="Charging"  class="style-sub-1"  style="display: none;" name="stylesub1" onchange="ChangeDropdowns(this.value)">
                                        <label>Which Sub-Category? </label>
                                        <select id="Charging" name="Charging" class="form-control">
                                            <optgroup>
                                            <option value="">-Choose A Sub-Category-</option>
                                            <option value="Charging">Charging</option>
                                            </optgroup>
                                        </select>
                                    </div>
                                    <div id="Consent"  class="style-sub-1"  style="display: none;" name="stylesub1" onchange="ChangeDropdowns(this.value)">
                                        <label>Which Sub-Category? </label>
                                        <select id="Consent" name="Consent" class="form-control">
                                            <optgroup>
                                            <option value="">-Choose A Sub-Category-</option>
                                            <option value="Accuracy">Accuracy</option>
                                            <option value="Double_Confirmation">Double Confirmation</option>
                                            <option value="Single_Confirmation">Single Confirmation</option>
                                            </optgroup>
                                        </select>
                                    </div>

我在表单操作中使用的PHP文件:''

        if(isset($_POST['submit']))
        {
         }else{
            header('Location:AddTestCase.php');
        }

        $conn = mysqli_connect('localhost', 'root', '','tmtool');
            if($conn -> connect_errno )
            {
              die('coudn''t connect to the database' . mysqli_connect_error());
            }
            if(! get_magic_quotes_gpc() )
            {
               $Category = addslashes(filter_input(INPUT_POST, 'Category'));
               $Sub_category = addslashes(filter_input(INPUT_POST, 'stylesub1'));
            }
            else
            {

               $Category = (filter_input(INPUT_POST, 'Category'));
               $Sub_category = (filter_input(INPUT_POST, 'stylesub1'));
            }



        if(($sql = $conn->prepare("INSERT INTO tmtool.testcase_master (`Category`, `Sub_category`) VALUES (?, ?)"))== FALSE)
        {
            echo "false";
        }        
        $sql->bind_param('ss',$Category , $Sub_category);
            if($sql->execute())
            {
                echo "Entered data successfully'n";
                mysqli_close($conn);
            }
            else {
              die('Could not enter data: ' . mysqli_error($conn));
            }    

第二个下拉列表的名称是Charging而不是stylesub1。请检查一下。

<div id="Charging"  class="style-sub-1"  style="display: none;" name="stylesub1" onchange="ChangeDropdowns(this.value)">
    <label>Which Sub-Category? </label>
        <select id="Charging" name="Charging" class="form-control">
            <optgroup>
                <option value="">-Choose A Sub-Category-</option>
                <option value="Charging">Charging</option>
            </optgroup>
        </select>
</div>

因此,在PHP中接受时,使用Charging关键字代替stylesub1

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