我正在尝试将上传的文件详细信息存储在我的数据库中。我编写了以下代码，但我无法理解为什么它不读取查询 block 。它不会生成任何 MySQL 错误消息或任何其他语法错误。请检查一下。
<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";
}
?>
最佳答案
尝试
$sql = "INSERT INTO `category_images` (`image_name`,`image_location`) VALUES ('".$image_name."', '".$path."')";
$result = mysql_query($sql);
if ($result)
{
// Successful query execution
echo "successfull";
}
else {
// Some error occured while executing query.
// Show some useful information using echo/print.
// Then stop execution after taking other necessary steps
die(mysql_error());
}
此外，您的数据库很容易受到 SQL 注入(inject)攻击，因为您没有清理输入。您应该至少使用 mysql_real_escape_string()方法来确保这种情况不会发生。
如果上述方法不起作用，请尝试检查您的连接参数是否正常以及 MySQL 是否正在运行。
关于php - 无法在数据库中存储文件详细信息，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/17826797/