我正在尝试提交一个简单的 PHP 表单并将一些数据插入 MySQL，我的表是：
- 类别：id，category_name
- 表 1：category_id(FK)、标题、描述
- table2：table1_id(FK)、文件类型、文件大小、文件日期、文件名
- 形式：日期、标题、描述、类别（下拉）、上传文件（获取文件信息，如类型、分机、大小、文件名）
这是代码：
形式：
<strong>Fill up the form:</strong><br /><br>
<form enctype="multipart/form-data" action="upfileone.php" method="POST">
Date: <?php echo date("d-M-Y") ?>
<p>Title:
<input type="text" name="title" value="<?php echo $sel_filepage['file_title']; ?>" id="file_title" />
</p>
<p>Description:<br>
<textarea name="description" rows="4" cols="24">
<?php echo $sel_filepage['content']; ?></textarea>
</p>
Category:
<select name="select_cat">
<?php $cat_set = get_all_categs();
while($category = mysql_fetch_array($cat_set)){
$catname = $category['cat_name'];
echo '<option value="'.$catname.'">'.$catname.'</option>';
}
?>
</select>
<br><br>
<label for="file">Choose File to Upload:</label>
<input type="file" name="upfile" id="upfile" > <br /><br />
<input type="hidden" name="MAX_FILE_SIZE" value="1000000" />
<input type="hidden" name="filepage" value="<?php echo $_GET['filepage']?>">
<input type="submit" name="upload" value="Add" class="pure-button pure-button-success">
<a href="content.php?filepage=<?php echo $sel_filepage['id']; ?>" class="pure-button">Cancel</a>
</form> <!-- END FORM -->
动作文件：
<?php
require_once("includes/functions.php");
// directory to be saved
$target = "server/php/files/";
$target = $target . basename($_FILES['upfile']['name']);
// gets info from FORM
$currentDate = date("Y-m-d");
$file_title = $_POST['title'];
$content = $_POST['description'];
$category = $_POST['select_cat'];
$upfile = ($_FILES['upfile']['name']);
// connects to db
$con = mysqli_connect("localhost", "root", "password", "database");
if(mysqli_connect_errno())
{
echo "error connection" . mysqli_connect_error();
}
// insert to database
$sql = "INSERT INTO filepages (category_id,file_title,content)
VALUES ('$category','$file_title','$content')";
/*$sql2 = "BEGIN
INSERT INTO filepages (category_id, file_title, content)
VALUES ('$category','$file_title','$content')
INSERT INTO fileserv (file_date)
VALUES ($currentDate)
COMMIT";*/
if(!mysqli_query($con, $sql))
{
die('Error ' . mysqli_error());
}
echo "1 File Added <br>";
if (file_exists("server/php/files/" . $_FILES["upfile"]["name"]))
{
echo $_FILES["upfile"]["name"] . " already exists. ";
}
else
{
insertFile( $_POST['filepage'],
$_FILES["upfile"]["type"],
($_FILES["upfile"]["size"] / 1024),
$_FILES["upfile"]["name"]);
move_uploaded_file($_FILES["upfile"]["tmp_name"],"server/php/files/" . $_FILES["upfile"]["name"]);
echo "The FILE " . basename($_FILES['upfile']['name']) . " has been uploaded.<br>";
echo "Stored in: " . "server/php/files/" . $_FILES["upfile"]["name"] . "<br>";
echo "Upload: " . $_FILES["upfile"]["name"] . "<br>";
echo "Type: " . $_FILES["upfile"]["type"] . "<br>";
echo "Size: " . ($_FILES["upfile"]["size"] / 1024) . " kB<br>";
echo "Temp file: " . $_FILES["upfile"]["tmp_name"] . "<br>";
}
?>
它确实插入到两个表中，现在我的主要问题是如何从类别的下拉菜单中获取 ID 并插入到Filepages.category_id.
如何获取Filepages.ID数据并将其插入Fileserve？