[英]passing multiple option values in php
我正在嘗試為php中的選項分配多個值,以便以后可以分別提取它們並插入到mysql中。
我要如何分配。
<select name="val[parent]">
<?php
if($cnt > 0){
while($parent = mysqli_fetch_assoc($sql))
{
?>
<option value="<?php echo $parent['category_id']. "-" . $parent['trade_id'];?> "> <?php echo $parent['trade_name']; ?> </option>
<?php }} ?>
</select>
這就是我試圖提取它們的方式
if(isset($_POST['val']))
{
list($category_id, $trade_id) = explode("-", $_POST['val'], 2);
$aVals = $_POST['val'];
$category_id = 0;
$category_name = $aVals['category_name'];
$parent = $aVals['parent'];
$trade_id =$aVals['trade_id'];
if(isset($aVals['category_id']))
{
$category_id = $aVals['category_id'];
}
$sql = "INSERT INTO category SET category_name = '$category_name', parent = $parent, vcat_id = '$trade_id', status = 1";
$sql = mysqli_query($databaseLink,$sql);
header('location:category.php');
}
我不知道我如何提取他們分配給他們。 請幫忙。
這是完整版
<?php
require_once "connection.php";
$bEdit = false;
if(isset($_POST['val']))
{
$Value = explode("-",$_POST['val']['parent']);
$aVals = $_POST['val'];
$category_id = 0;
$category_name = $aVals['category_name'];
$parent = $aVals['parent'];
$trade_id = $Value[1];
if(isset($aVals['category_id']))
{
$category_id = $aVals['category_id'];
}
$sql = "INSERT INTO category SET category_name = '$category_name', parent = $parent, vcat_id = '$trade_id', status = 1";
$sql = mysqli_query($databaseLink,$sql);
header('location:category.php');
}
?>
<form role="form" method="post">
<table cellpadding="5" cellspacing="5">
<tr><td>
<h2><?php echo $bEdit ? 'Edit' : 'Add '; ?> Category</h2>
<?php if($bEdit){ ?>
<input type="hidden" name="val[category_id]" value="<?php echo $aRow['category_id']; ?>">
<?php } ?>
</td></tr>
<tr><td>
Select Parent : <br />
<?php
$sql = mysqli_query($databaseLink,"SELECT * FROM `category` WHERE 1 ");
$cnt = mysqli_num_rows($sql);
?>
<select name="val[parent]">
<?php
if($cnt > 0){
while($parent = mysqli_fetch_assoc($sql))
{
?>
<option value="<?php echo $parent['category_id']. "-" . $parent['trade_id'];?> "> <?php echo $parent['trade_name']; ?> </option>
<?php }} ?>
</select>
</td></tr>
<tr><td>
Name : <br />
<input class="form-control" placeholder="Name" name="val[category_name]" type="text" autofocus required value="<?php echo $bEdit ? $aRow['name'] : ''; ?>">
</td></tr>
<tr><td><br />
<button type="submit" class="btn btn-lg btn-success btn-block">Save</button>
</td></tr>
</table>
</form>
您可以從表單中分配PHP變量,如下所示:
if(isset($_POST['val']['parent'])){
$parent = $_POST['val']['parent'];
# split the value by "-", and assign to category_id & trade_id
list($category_id, $trade_id) = explode("-", $_POST['val']['parent'], 2);
}
if(isset($_POST['val']['category_name'])){
# set the category_name variable
$category_name = $_POST['val']['category_name'];
}
if(isset($_POST['val']['category_id'])){
# override category_id variable
$category_id = $_POST['val']['category_id'];
}
您可能想收緊MySQL查詢,但似乎存在安全漏洞。
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