php 如何使用php将上传的图像保存在数据库中
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how to save uploaded image in database with php
提问by Darya
hello write this code to upload and save uploaded image in data base but don't work , please help me !!
您好，编写此代码将上传的图像上传并保存在数据库中，但不起作用，请帮帮我！！
File 1:
文件 1：
<form action="upload.php" method="post" enctype="multipart/form-data" target="upload_target" onsubmit="startUpload();" >
<label>File:
<input name="myfile" type="file" size="30" />
</label>
<input type="submit" name="submitBtn" class="sbtn" value="Upload" />
<iframeid="upload_target"name="upload_target"src="#"style="width:0;height:0;border:0px solid #fff;"></iframe>
</form>
</div>
File 2:
文件2：
<?php
// Edit upload location here
$destination_path = getcwd().DIRECTORY_SEPARATOR;
$target_path="my/";
$result = 0;
$name=$_FILES['myfile']['name'];
$target_path = $target_path . basename( $_FILES['myfile']['name']);
if(@move_uploaded_file($_FILES['myfile']['tmp_name'], $target_path)) {
list($width, $height, $type, $attr) = getimagesize($target_path);
echo "Image width " .$width;
echo "<BR>";
echo "Image height " .$height;
echo "<BR>";
echo "Image type " .$type;
echo "<BR>";
echo "Attribute " .$attr;
$result = 1;
}
// sleep(1);
$link=mysql_connect('localhost','root','');
if(!$link)
{die('you cannot connect to database...');}
$db=mysql_select_db('final');
if(!$db)die('smile');
if (isset($_FILES['myfile']) && $_FILES['myfile']['size'] > 0) {
// define the posted file into variables
$name = $_FILES['myfile']['name'];
$tmp_name = $_FILES['myfile']['tmp_name'];
$type = $_FILES['myfile']['type'];
$size = $_FILES['myfile']['size'];
// if your server has magic quotes turned off, add slashes manually
//if(!get_magic_quotes_gpc()){
//$name = addslashes($name);
//}
// open up the file and extract the data/content from it
$extract = fopen($tmp_name, 'r');
$content = fread($extract, filesize($tmp_name));
$content = addslashes($content);
fclose($extract);
// connect to the database
// the query that will add this to the database
$s=mysql_query("SET NAMES 'utf8'");
$sql = "INSERT INTO `final`.`products` (`Productcode`, `Productname`, `Price`,`Descriptionofgood`, `image`) VALUES ('','','','','".$target_path."') WHERE `products`.`Productcode`='1371' ";
$results = mysql_query($sql);
if(!$result)die('not');
}
?>
回答by Dhruv Balhara
Technically, if it is a small project. You should not store images files in "Database" ; rather only their link (you may not even need that). Image or any media files are stored on server along with your other files (html,css,php). Of course, you need to put them on a dedicated folder for it. The reason for not storing on database : because they are meant for data retrieval only and more importantly they are of smaller size (bigger sizes do exist, i am speaking in case of a small project which requires least possible resources. Storing media files on database is just not efficient.
从技术上讲，如果它是一个小项目。您不应将图像文件存储在“数据库”中；而只是他们的链接（您甚至可能不需要那个）。图像或任何媒体文件与您的其他文件（html、css、php）一起存储在服务器上。当然，您需要将它们放在专门的文件夹中。不存储在数据库上的原因：因为它们仅用于数据检索，更重要的是它们的尺寸较小（确实存在较大的尺寸，我说的是一个需要最少资源的小项目。将媒体文件存储在数据库上只是效率不高。
Looking at your code, i can tell you are trying to store the files on your server.
查看您的代码，我可以告诉您正在尝试将文件存储在您的服务器上。
They have used a very simple script for uploading here. Try it on your localhost before trying on a server.
他们使用了一个非常简单的脚本来上传这里。在尝试服务器之前先在本地主机上尝试。
回答by Tirthraj Rao
if(@move_uploaded_file($_FILES['myfile']['tmp_name'], $target_path))is incorrect.
if(@move_uploaded_file($_FILES['myfile']['tmp_name'], $target_path))是不正确的。
It should be if(move_uploaded_file($_FILES['myfile']['tmp_name'], $target_path))
它应该是 if(move_uploaded_file($_FILES['myfile']['tmp_name'], $target_path))
Remove the @.
删除@.
回答by Mohsin Rasheed
I have created three page for it
我为它创建了三个页面
index.php (Form for uplad image)
upload.php (Save the image in upload folder and its path in database)
index.php（上传图片的表格）
upload.php（将图片保存在上传文件夹中，并保存在数据库中的路径）
3.showimage.php (Show the images from database)
3.showimage.php (显示数据库中的图片)
Database Structure
数据库结构
id int(4) auto increment - image varchar(100) - image_name varchar(50)
id int(4) 自动递增 - 图像 varchar(100) - image_name varchar(50)
here is the code:
这是代码：
index.php
索引.php
<form method="post" action="upload.php" enctype="multipart/form-data">
<label>Choose File to Upload:</label><br />
<input type="hidden" name="id" />
<input type="file" name="uploadimage" /><br />
<input type="submit" value="upload" />
</form>
uplad.php
上传文件
<?php
$target_Folder = "upload/";
$uid = $_POST['id'];
$target_Path = $target_Folder.basename( $_FILES['uploadimage']['name'] );
$savepath = $target_Path.basename( $_FILES['uploadimage']['name'] );
$file_name = $_FILES['uploadimage']['name'];
if(file_exists('upload/'.$file_name))
{
echo "That File Already Exisit";
}
else
{
// Database
con=mysqli_connect("localhost","user_name","password","database_name"); //Change it if required
//Check Connection
if(mysqli_connect_errno())
{
echo "Failed to connect to database" . mysqli_connect_errno();
}
$sql = "INSERT INTO image (id,image, image_name)
VALUES ('','$target_Folder$file_name','$file_name') ";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added successfully in the database";
echo '<br />';
mysqli_close($con);
// Move the file into UPLOAD folder
move_uploaded_file( $_FILES['uploadimage']['tmp_name'], $target_Path );
echo "File Uploaded <br />";
echo 'File Successfully Uploaded to: ' . $target_Path;
echo '<br />';
echo 'File Name: ' . $_FILES['uploadimage']['name'];
echo'<br />';
echo 'File Type: ' . $_FILES['uploadimage']['type'];
echo'<br />';
echo 'File Size: ' . $_FILES['uploadimage']['size'];
}
?>
<a href="showimage.php">Show Image</a>
showimage.php
显示图像.php
<?php
$con=mysqli_connect("localhost","user_name","password","test"); //Change it if required
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM image " );
while($row = mysqli_fetch_array($result))
{
echo '<img src="' . $row['image'] . '" width="200" />';
echo'<br /><br />';
}
mysqli_close($con);
?>
