我正在研究php。我的问题是如何限制一个用户一个月只能上传三张图片。
我的mysql数据库表--
CREATE TABLE `images` (
`id` int(40) NOT NULL,
`user_name` varchar(40) NOT NULL,
`mobile` varchar(30) NOT NULL,
`email` varchar(50) NOT NULL,
`name` longblob NOT NULL,
`position` int(40) NOT NULL,
`date` date NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
我想一个用户只能上传3个月的图像。
请帮助我的php脚本。我是php的初学者。共享最佳解决方案。。
我用这个--
<?php
include("admin/config.php");
if(isset($_POST["insert"]))
{
$mobile = $_POST['mobile'];
$email = $_POST['email'];
$user_name = $_POST['user_name'];
$fileinfo = @getimagesize($_FILES["image"]["tmp_name"]);
$width = $fileinfo[0];
$height = $fileinfo[1];
$allowed_image_extension = array(
"png",
"jpg",
"jpeg"
);
$file_extension = pathinfo($_FILES["image"]["name"], PATHINFO_EXTENSION);
$sql="select * from images where (name='$name');";
$count=mysqli_query($mysqli,$sql);
$count=count($_FILES['name']);
if($count>3)
{
echo "<font color='red'>3 image upload </font>";
} else{
$file_extension = pathinfo($_FILES["image"]["name"], PATHINFO_EXTENSION);
if (! file_exists($_FILES["image"]["tmp_name"])) {
$response = array(
"type" => "error",
"message" => "Choose image file to upload."
);
}
else if (! in_array($file_extension, $allowed_image_extension)) {
$response = array(
"type" => "error",
"message" => "<font color='red'>Upload valiid images. Only PNG and JPEG are allowed.</font>"
);
echo $result;
} // Validate image file size
else if (($_FILES["image"]["size"] > 2000000)) {
$response = array(
"type" => "error",
"message" => "Image size exceeds 2MB"
);
} // Validate image file dimension
else if ($width > "1250" || $height > "720") {
$response = array(
"type" => "error",
"message" => "<font color='red'>Image dimension should be within 1250X720</font>"
);
} else {
$target = '/image';
$target = "image/" . basename($_FILES["image"]["name"]);
$file = addslashes(file_get_contents($_FILES["image"]["tmp_name"]));
$query = mysqli_query($mysqli,"INSERT INTO images VALUES ('','$user_name','$mobile','$email','$file','',NOW())");
if (move_uploaded_file($_FILES["image"]["tmp_name"], $target)) {
$response = array(
"type" => "success",
"message" => '<font color="green">Image uploaded successfully </font>'
);
} else {
$response = array(
"type" => "error",
"message" => "<font color='red'>Problem in uploading image files.</font>"
);
}
}
// if(mysqli_query($connect, $query))
// {
// echo '<script>alert("Image Inserted into Database")</script>';
// }
}
}
?>
还有我的html表单-
<form method="post" id="frm-image-upload" action="my-account.php#parentHorizontalTab3" name='img'
method="post" enctype="multipart/form-data">
<div class="agileits_w3layouts_contact_left"style="margin-left:20%;">
<input type="hidden" name="user_name" value="<?php $space = " ";
echo $row["fname"].$space.$row["lname"]; ?>" id="user_name" Placeholder="Your Name" required />
<input type="hidden" name="mobile" value="<?php echo $row["mobile"]; ?>" id="mobile"Placeholder="Mobile" required />
<input type="hidden" name="email" value="<?php echo $row["email"]; ?>" id="email" Placeholder="Email" required/>
</div>
<center>
<input type="file" name="image" id="image" />
</center>
<br />
<center><input type="submit" name="insert" id="insert" value="Upload" class="btn btn-info" /> </center>
</form>
请帮我分享最好的解决方案——一个用户一个月可以上传三张图片。
3条答案
按热度按时间5cg8jx4n1#
基本思想是统计给定用户在给定月份上传的所有图像,如:
如果上面的查询返回
3
,您可以阻止上载。hfsqlsce2#
考虑以下几点。。。
x4shl7ld3#
@fabrik解决方案是最简单的方法,但在处理大量图像/用户时,它的性能会急剧下降。
如果你不介意对解决方案再深入一点,我建议加上
image_upload_credit
字段到用户表。每次用户上传的图像,你会减少1学分。因为你已经可以访问
User
对象时,插入的复杂性是O(1)
.要完成解决方案,您需要编写一个
cron
(定期运行)重置image_upload_credit
每月的第一天UPDATE user SET image_upload_credit = 3 WHERE image_upload_credit <> 3;