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