如何使用php腳本將圖像上傳到數據庫？
[英]How to upload an image to a database using php script?
問題描述
這是我上載的代碼..但是它不起作用..我使用了file_get_contents函數。
</head>
<body>
<form action="upload1.php" method="POST" enctype="multipart/form-data">
File:
<input type="file" name="image"/>
<input type="submit" value="Upload image" />
</form>
<?php
//connect to the database
$con = mysql_connect("localhost","root", "");
if(!$con)
{
die('Could not connect to the database:' . mysql_error());
echo "ERROR IN CONNECTION";
}
mysql_select_db("imagedatabase", $con);
//file properties
echo $file = $_FILES['image']['tmp_name'];
echo '<br />';
if(!isset($file))
echo "Please select an image";
else
{
$image = file_get_contents($_FILES['image']['tmp_name']);
echo $image_name = addslashes($_FILES['image']['name']); echo '<br \>';
echo $image_size = getimagesize($_FILES['image']['tmp_name']);
if($image_size == FALSE)
echo "That's not an image";
else
{
$insert = mysql_query("INSERT INTO images (image) VALUES ($image)",$con);
if(!$insert)
echo "Problem uploding the image. Please check your database";
else
{
$last_id = mysql_insert_id();
echo "Image Uploaded. <p /> Your image: <p /><img src=display.php? id=$last_id>";
}
}
}
mysql_close($con);
?>
</body>
</html>
檢索/顯示的代碼就是這種方式。
<?php
//connect to the database
mysql_connect("localhost","root", "") or die(mysql_error());
mysql_select_db("mydb") or die(mysql_error());
//requesting image id
$id = addslashes($_REQUEST['id']);
$image = mysql_query("SELECT * FROM images WHERE id = $id");
$image = mysql_fetch_assoc($image);
$image = $image['image'];
header("Conten-type: image/jpeg");
echo $image;
mysql_close($connect);
?>
我創建了一個名為“ imagedatabase”的數據庫和一個表
4 個解決方案
解決方案1
8 2012-01-04 08:05:24
在數據庫中存儲圖像是一個壞主意。 將其路徑存儲在數據庫中，通過.htaccess關閉包含圖像的目錄，然后在硬盤上使用它。
為什么不應該在DB中存儲文件？
如果您將使用DB存儲圖像，則將具有：
- 慢查詢
- 災難性指標的大小
- 橋php <-> mysql上的高負載
- 編輯照片的問題（您需要獲取圖像，進行一些修改並再次插入所有數據。哦，不了）
- 文件從一個地方轉移到另一個地方的問題
- 關於StackOverflow的新問題«如果文件不是文件而是字符串，如何使用文件»
解決方案2
1 2014-03-08 12:45:09
您應該在上傳過程中將文件保存在某個文件夾中，並將文件名保存在數據庫中，以便稍后可以從數據庫中調用文件名並將其鏈接為要下載的超鏈接，我正在使用以下代碼上傳圖像在名為files的文件夾中，並將文件名保存在數據庫中。 最后，我在變量$ newname中有文件名
if ($_FILES['file']['name']) {
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = end($temp);
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"))
&& ($_FILES["file"]["size"] < 500000)
&& in_array($extension, $allowedExts)
) {
if ($_FILES["file"]["error"] > 0) {
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
} else {
$ext = end(explode(".", $_FILES["file"]["name"]));
$filename = current(explode(".", $_FILES["file"]["name"]));
$newname = $filename . '_' . time() . '.' . $ext;
move_uploaded_file($_FILES["file"]["tmp_name"],
"files/" . $newname);
}
} else {
echo "<div class='alert alert-success'>Image type or size is not valid.</div>";
}
}
解決方案3
0 2012-01-04 10:16:16
將圖像直接上傳到數據庫不是一個好主意。 而是將照片上傳到文件夾中，然后僅將照片名稱插入數據庫中，然后在以后需要時調用它。 您可以根據需要嘗試以下代碼。
為了使代碼適合您，您必須執行以下步驟：
在代碼內部將“ your_photo”替換為您的輸入名稱（這種情況我想是“ image”）
創建一個要在其中上傳圖像的文件夾，然后在->> $ newname =“ support / images / profile /”中進行更改，在此處寫下您的圖像文件夾名稱
編寫正確的數據庫查詢。 並記住圖像名稱將自動創建，並且該名稱保留在此變量-> $ image_name內。 將名稱插入數據庫時，只需使用$ image_name作為值。
上傳腳本：
<?
// If Everything is good- process the form - write the data into the database
$photo=$this->input->post('your_photo');
if($photo==NULL){$image_name='0';}// if no photo is selected the default value of the photo would be 0
//photo upload starts
$errors=0;
if($_FILES['your_photo']){
$image=$_FILES['your_photo']['name'];
if($image) {
define ("MAX_SIZE","100");
function getExtension($str) {
$i = strrpos($str,".");
if (!$i) { return ""; }
$l = strlen($str) - $i;
$ext = substr($str,$i+1,$l);
return $ext; }
//reads the name of the file the user submitted for uploading
$image=$_FILES['your_photo']['name'];
//if it is not empty
if ($image)
{
//get the original name of the file from the clients machine
$filename = stripslashes($_FILES['your_photo']['name']);
//get the extension of the file in a lower case format
$extension = getExtension($filename);
$extension = strtolower($extension);
//if it is not a known extension, we will suppose it is an error and will not upload the file,
//otherwise we will do more tests
if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif"))
{
//print error message
$msg="Sorry! Unknown extension. Please JPG,JPEG,PNG and GIF only ";
$errors=1;
}
else
{
//get the size of the image in bytes
//$_FILES['image']['tmp_name'] is the temporary filename of the file
//in which the uploaded file was stored on the server
$size=filesize($_FILES['your_photo']['tmp_name']);
//compare the size with the maxim size we defined and print error if bigger
if ($size < MAX_SIZE*1024)
{
//we will give an unique name, for example the time in unix time format
$image_name=time().'.'.$extension;
//the new name will be containing the full path where will be stored (images folder)
$newname="support/images/profile/".$image_name;
//we verify if the image has been uploaded, and print error instead
$copied = copy($_FILES['your_photo']['tmp_name'], $newname);
if (!$copied)
{
$msg="Sorry, The Photo Upload was unsuccessfull!";
$errors=1;
}
}
else
{
$msg="You Have Exceeded The Photo Size Limit";
$errors=1;
}
}}}
/*Image upload process ends here- If any problem occurs it will display error message via the $msg,
otherwise it will upload the image to the image folder. To insert the photo into database $image_name has been used*/
}
if(($_FILES['your_photo'])&& ($errors))/* If any photo is selected and any problem occurs while uploading it will
display an error message, otherwise transfer the data to Mod_addstudent model */
{
echo $msg;
}
else {
//Insert into database.Just use this particular variable "$image_name" when you are inserting into database
$sql="INSERT INTO your_table (field1, your_image_field) VALUES ('','$image_name')";
}
?>
然后查看圖片：
<?php
// Retrieve information from Database First and then ..
if (empty($your_photo))
{ $image_location="images/avatar.jpg";} //if there is no image in database
else {$image_location="images/$your_photo";} // if there is any image in database
?>
<img src="<?php echo base_url(); ?><?php echo $image_location ;?>" width="150" height="170" />
解決方案4
0 2012-01-04 13:47:54
首先檢查您是否確實能夠成功將文件上傳到服務器。 遵循本教程：
http://www.tizag.com/phpT/fileupload.php
這將幫助您排除整個旅程的一半，但我強烈建議您檢查日志中是否存在實際錯誤，如果需要其他幫助，請明確指出它們。
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