How do you display an image from a SQL database in php.
The file name is stored in the database and the images itself is saved in my images file.
The image should change accordingly and correspond to the information I want.
Tried using this codes but failed
echo"<tr><td> <img src='mysql_query('SELECT imagefile FROM recipe WHERE recipeid = $recipeid')'> </td>";
Create a php file called image.php and an extra parameter which corresponds to the path.
Like, for example image.php?v=007
Hence, image.php should be able to look up the database using 007 as a reference and find it's path.
Path can be stored in a column in the database.
EDIT:
Create a php file called image.php
Next create a database with with columns like ID, REFERENCE AND PATH.
Use the following code to get the reference:
$reference = $_GET["v"];
Next look up the database for the path:
$stat = $conn -> prepare ("SELECT PATH FROM TABLE_NAME WHERE REFERENCE = :reference");
$stat -> bindValue(':reference', $reference);
$stat -> execute();
$path = $stat -> fetchColumn();
// Now you can use this code to display the image.
echo '<img src="' . $path . '">';
It will take the user to the image's path.
For this, you need to insert reference and path into the database everytime where a new image is uploaded.
Related
I have a hosted website and i want to upload my images to image directory and link the image to mysql database .
What i did:
Creating form as :
<form method=POST align=center enctype="multipart/form-data">
<input type=file name=image>
Php code :
$uploaddir = 'images/';
// echo $uploaddir;
$target = "$uploaddir".basename($_FILES['image']['name']);
echo $target;
$image = $_FILES['image']['name'];
$sql="insert into News2(image_News2)
values($image)";
$res=mysqli_query($con,$sql);
copy($_FILES['tmp_name']['name'],$target,true);
What this code actually do is only inserting the image into the database not copying the image into the images folder.
What i searched for and found :
You shouldn't specify a http:// URL as $uploaddir but a path relative to the path where the php script is running from.
So i what i really wanted to do is :
1- Copying the image into image folder directory on website.
2- Linking the copied image to the database.
you're missing the name ïmage"
not:
copy($_FILES['tmp_name']['name'],$target,true);
but:
move_uploaded_file($_FILES['image']['tmp_name'], $target);
And maybe you should try the move first, before inserting it in the database, should that action fail (or filesize = 0) etc.
do check on both the file actions and the db actions, should they fail: rollback
extra:
is only 1 column present in your database table?
is it wise to use the uploaded file's name? Duplicates can happen.
I would after checking the file insert the original name into the database. Get the autoincrement id of that record. rename the file to ./1234.jpg
through the database you can then later use the original name. But this numbering might go too var for a simple application
PHP Code :
if (isset($_FILES['image']['name']) && $_FILES['image']['error']
== 0) {
$uploads_dir = '/uploads';
$temp_file = $_FILES['image']['tmp_name'];
$name = basename($_FILES["image"]["name"]);
if(move_uploaded_file($temp_file, "$uploads_dir/$name")){
//Fire Insert Query For Insert image name in your database
}
else
{
echo "Error !! , Your image not uploaded.";
}
}
?>
as per your given data you must close Form tag then give form action ="" attribute to specify url . and if your oprating system is linux then check folder permission in set or not , also check your database table if field name is set unique or not ?
then write below code in given url file
Am saving a file as a long blob in php by first saving it in a folder then to a db, the problem is that the server has write permissions so i would like to save it directly
This is what i have tried (This works perfectly):
if(isset($_POST['image'])){
$id = 0;
$image = $_POST['image'];
$tmp_image = date('YmdHisu').'.jpg';
file_put_contents($tmp_image, base64_decode($image));
$sql = "INSERT INTO fingerprint(template)
VALUES ('".addslashes(file_get_contents($tmp_image))."')";
try
{
$connection=Yii::app()->db;
$command=$connection->createCommand($sql);
$rowCount=$command->execute(); // execute the non-query SQL
echo "saved successifully";
unlink($tmp_image);
}
catch(Exception $ex)
{
echo 'Query failed' , $ex->getMessage();
unlink($tmp_image);
}
}
How can i save this in a blob field in mysql without first having to save it in a folder then saving to db
Let's assume you've sent the file via Android using POST method to your server running Yii1.
First of all the physical file is contained in the $_FILES variable and in the $_POST variable, as you said, contains a string of the file encoded in base 64 format (i write this for a clear answer).
$_FILE DOCUMENTATION
Now this is how you can try to upload the file with the standard MVC yii way using yii code:
It's true that you're uploading your file from an external device but Yii came in your help with CUploadedFile Class:
Call getInstance to retrieve the instance of an uploaded file, and then use saveAs to save it on the server. You may also query other information about the file, including name, tempName, type, size and error.
In particular you should use the function getInstancesByName that returns an array of instances starting with specified array name.
$temp = CUploadedFile::getInstanceByName("image"); // $_FILES['image']
and with this you can access to the file data:
$temp->name; // Name of the file
$temp->type; // Type of the file
$temp->size; // Size of the file
// etc etc..
$temp->saveAs("/your/path/" . $tmp_image . $temp->type); // Save your file
for final you can check if the file was saved and execute your query:
if($temp->saveAs("/your/path/" . $tmp_image . $temp->type)) {
// File is saved you can execute the query for save the record in your db
} else {
// Something went wrong.
}
You can also transfer this logic in a model.
check this link for more infos
Hope this will help you.
I have a directory full of images (40,000 +) that I need sorted. I have designed a script to sort them into knew proper directories, however, I am having issues with the file name.
The images urls with the id they belong to are stored in a database, and I am using the database in conjunction with the script to sort the images.
My Problem:
The image url's in the database are shortened. An example of such corresponding images are like this:
dsc_0107-367.jpg
dsc_0107-367-5478-2354-0014.jpg
The first part of the filenames are the same, but the actual file contains more info. I'd like a way to move the file from the database with the known part of the file name.
I have a basic code:
<?php
$sfiles = mysqli_query($dbconn, "SELECT * FROM files WHERE gal_id = '$_GET[id']");
while($file = mysqli_fetch_assoc($sfiles)){
$folder = $file['gal_id'];
$fileToMove = $file['filename'];
$origDir = "mypath/to/dir";
$newDir = "mypath/to/new/dir/$file['gal_id']";
mkdir "$newDir";
mv "$fileToMove" "$newDir";
}
Im just confused on how to select the file based on the small part from the database.
NOTE: It's not as simple as changing the number of chars in the db, because the db was given to me from an external site thats been deleted. So this is all the data I have.
PHP can open files using the function glob() . Glob searches your server, or specified directory, for any files containing a "match" to a pattern you specify.
Using glob() like this will pull your images from a partial name.
Run this query separate from the second:
$update = mysqli($dbconn, "UPDATE files
SET filename = REPLACE(filename, '.info', ''));
filename should be the column in your database that contains the list of images. The reason we are removing the .jpg from the db columns is if your names are partial, the .jpg may not match with the given name in your directory. With it removed, we can search solely for the pattern of the name.
Build the query to select and move the folders:
$sfiles = mysqli_query($dbconn, "SELECT * FROM files");
while($file = mysqli_fetch_assoc($sfiles)){
$fileToMove = $file['filename'];
// because glob outputs the result set into an array,
// we will use foreach to run each result from the array individually.
foreach(glob("$fileToMove*") as filename){
echo "$filename <br>";
// I'm echoing this out to see that the results are being run
// one line at a time and to confirm the photo's are
// matching the pattern.
$folder = $file['gal_id'];
// pulling the id from the db of the gallery the photo belongs to.
// This will specify which folder to move the pic to.
// Replace gal_id with the name of your column.
$newDir = $_SERVER['DOCUMENT_ROOT']."/admin/wysiwyg/kcfinder/upload/images/gallery/old/".$folder;
copy($filename,$newDir."/".$filename);
// I would recommend copy rather than move.
// This will leave the original photo in its place.
// This measure is to ensure the photo made it to the new directory so you don't lose it.
// You could go back and delete the photos after if you'd prefer.
}
}
Your MySQL query is ripe for SQL Injection, and your GET statement needs to be sanitized, if I went to your page with something similar to :
pagename.php?id=' DROP TABLE; #--
this is going to end extremely badly for you.
So;
OVerall it's much better to use Prepared Statements. THere's LOTS and LOTS of data about how to use them all over SO and the wider internet. What I show below is only a stopgap measure.
$id = (int)$_GET['id'] //This forces the id value to be numeric.
$sfiles = mysqli_query($dbconn, "SELECT * FROM files WHERE gal_id = ".$id);
Also keep note of closing your ' and " quotes as your original doesn't close the array key wrapper quotes.
I never used mysqli_fetch_assoc and always used mysqli_fetch_array so will use that as it fits the same syntax :
while($file = mysqli_fetch_array($sfiles)){
$folder = $id //same thing.
$fileToMove = $file['filename'];
$origDir = "mypath/to/dir/".$fileToMove;
//This directory shold always start with Server['DOCUMENT_ROOT'].
//Please read the manual for it.
$newDir = $_SERVER['DOCUMENT_ROOT']."/mypath/to/new/dir/".$folder;
if(!is_dir($newDir)){
mkdir $newDir;
}
// Now the magic happens, copies the file to the new directory.
// Then (optionally) delete the original.
copy($origDir,$newDir."/".$fileToMove);
unlink($origDir); //removes original.
// Add a flag to your Database to know that this file has been copied,
// ideally you should resave the filepath to the correct new one.
//MySQL update saving the new filepath.
}
Read up on PHP Copy and PHP unlink.
And; please use Prepared Statements for PHP and Database interactions.!
I made a form in which a user can upload images. I am storing the copy of the uploaded image into a folder named "pics"(full path is C:\wamp\www\pics). I am also storing the name of the image in my database table.
I want to display all images in a page. I created a script. Used while and foreach loop to fetch the name of the image and then added the name in the image tag. It's now displaying a blank broken box.
The address of the image is correct. I select "Copy image location" by right clicking on the pic and it's displaying the correct path. eg: C:\wamp\www\pics\abc.JPG
Here's the code:
<?php
//Including SQL Connection Credentials
require_once("sql_connection.php");
$sql_fetch_pics = "SELECT name FROM pics";
$query_fetch_pics = mysql_query($sql_fetch_pics);
while($fetched_pic = mysql_fetch_assoc($query_fetch_pics)){
foreach($fetched_pic as $display_pic){
$full_img_name = dirname(__FILE__)."\pics\\".$display_pic;
echo "<img src='".$full_img_name."'/>";
}
}
?>
The src tag should be URL, not filesystem path:
$full_img_name = "/pics/".$display_pic;
Solution:
$full_img_name = "/pics/".$display_pic;
Explanation goes here: dirname(__FILE__) on localhost
I have a list of files in my html where I have removed the directory ImageFiles/ from the filename. So for example the file names look like below:
tulips.png
koala.png
jellyfish.png
Now if the user clicks on one of these file names, then I want to retrieve the file from the database, but in the database all files contain the directory ImageFiles/ at the start of the file name. So in the database the files look like this below:
ImageFiles/tulips.png
ImageFiles/koala.png
ImageFiles/jellyfish.png
My question is simply how do I finish the WHERE clause in order to retrieve the correct file from the database? For example if I am looking for the file tulips.png, how can I get the query to be able to find ImageFiles/tulips.png and only this file from the database?
Below is the mysqli query:
$getimage = $_GET['filename'];
$imagequery = "SELECT ImageFile FROM Image WHERE (ImageFile = ?)";
if (!$imagestmt = $mysqli->prepare($imagequery)) {
// Handle errors with prepare operation here
}
// Bind parameter for statement
$imagestmt->bind_param("s", $getimage);
// Execute the statement
$imagestmt->execute();
Just prepend the directory name to the $getimage variable:
$getimage = 'ImageFiles/' . $_GET['filename'];