我正嘗試將上傳的文件詳細信息存儲在我的數據庫中。我寫了下面的代碼,但我無法理解它爲什麼不讀取查詢塊。它不會生成任何MySQL錯誤消息或任何其他語法錯誤。請檢查它。無法在數據庫中存儲文件詳細信息
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<form action="file_upload_test2.php" method="post"
enctype="multipart/form-data">
<label for="file">Filename:</label>
<input type="file" name="file" id="file"><br>
<input type="file" name="file2" id="file"><br>
<input type="file" name="file3" id="file"><br>
<input type="submit" name="submit" value="Submit">
</form>
</body>
</html>
<?php
include 'connect.php';
$allowedExts = array("gif", "jpeg", "jpg", "png");
$extension = end(explode(".", $_FILES["file"]["name"]));
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/jpg")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/x-png")
|| ($_FILES["file"]["type"] == "image/png"))
//&& ($_FILES["file"]["size"] < 200000)
&& in_array($extension, $allowedExts))
{
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
}
else
{
echo "Upload: " . $_FILES["file"]["name"] . "<br>";
echo "Type: " . $_FILES["file"]["type"] . "<br>";
echo "Size: " . ($_FILES["file"]["size"]/200000) . " kB<br>";
$image_name= $_FILES["file"]["name"];
$path= move_uploaded_file($_FILES["file"]["tmp_name"],
"upload/" . rand().$_FILES["file"]["name"]);
echo "Stored in: " . "upload/" . $_FILES["file"]["name"];
if (mysql_query ("Insert into category_images (image_name,image_location) VALUES ('$image_name', '$path')"))
{
echo "successfull";
}
else {
mysql_error();
}
}
}
else
{
echo "Invalid file";
}
?>
首先,不要使用mysql_ *,因爲它已折舊。其次,是回聲「存儲在:」迴應什麼? –
好的下一次我會照顧它,是'回聲存儲在'顯示路徑;存儲在:upload/paint.png –
嘗試'echo mysql_error();'顯示mysql錯誤。 – jcubic