I create php script that shows the Id and the image stored in mysql database, but i can just see the image without the Id, how can i show the image and the id in this php script ?
<?php
$con = mysqli_connect("localhost","root","","othmane") or die(mysqli_error($con));
if($_SERVER['REQUEST_METHOD']=='GET'){
$id = $_GET['id'];
$sql = "SELECT image,image_type FROM images where id = $id";
$r = mysqli_query($con,$sql) or die(mysqli_error($con));;
$result=mysqli_fetch_array($r);
header('Content-Type:image/jpeg');
$ss=$result['image_type'];
if ($ss == 'php') {
echo 'Type of image ' . $result['image_type'];
echo ( $result['image']);
} else if ($ss == 'android') {
echo 'Type of image ' . $result['image_type'];
echo base64_decode( $result['image'] );
}
mysqli_close($con);
}
?>
this is what i get when i execute script :
note : when i remove echo 'Type of image ' . $result['image_type']; from the code the image is properly displayed. but i want to show the image and its type too.
Related
I am trying to render uploaded image from Mysql to frontend in a file called result.php
This is the code that uploaded the image to Mysql
include ('connect.php');
//if button with the name uploadfilesub has been clicked
if(isset($_POST['uploadfilesub'])) {
//declaring variables
$filename = $_FILES['uploadfile']['name'];
$filetmpname = $_FILES['uploadfile']['tmp_name'];
//folder where images will be uploaded
$folder = 'imagesuploadedf/';
//function for saving the uploaded images in a specific folder
move_uploaded_file($filetmpname, $folder.$filename);
//inserting image details (ie image name) in the database
$sql = "INSERT INTO `uploadedimage` (`imagename`) VALUES ('$filename')";
$qry = mysqli_query($conn, $sql);
if( $qry) {
echo "</br>image uploaded";
}
}
?>
The uploaded images are saved on a folder on my server called imagesuploadedf
I used this code below to render the image on the result.php but it's showing the file name only, I need the image to be rendered.
$conn = new mysqli($servername, $username, $password, $dbname);
$sql = 'SELECT * from uploadedimage';
if (mysqli_query($conn, $sql)) {
echo "";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
$count=1;
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) { ?>
<tbody>
<tr>
<td>
<?php echo $row['imagename']; ?>
</td>
</tr>
</tbody>
<?php
$count++;
}
} else {
echo '0 results';
}
?>
</table>
What do I need to write on result.php to show/render the uploaded image on the page?
You should write an HTML page containing <img> elements referring to the images you want to display.
It might look like this:
<td>
<?php
$imgTag = '<img src="/imagesuploadedf/' . $row['imagename'] . '" />';
echo $imgTag; ?>
</td>
This puts out stuff like this into your web page.
<td>
<img src="/imagesuploadedf/image_file_name"/>
</td>
If your web page is coming from https://images.example.com this will make the browser download the image at https://images.example.com/imagesuploadedf/image_file_name from your server and display it in the page.
I am trying to show the blop images i have in my mysql database. But all i get is a whitebox with no picture in it.
PHP code:
<?php
require_once "include/config.php";
session_start();
$sql = "SELECT * FROM docenten";
$result = $conn->query($sql);
if($result->num_rows > 0){
while($row = $result->fetch_assoc()){
echo "<dt><strong>Foto:</strong></dt><dd>" .
'<img src="data:image/jpeg;base64,'.
base64_encode($row['foto']).
'" width="290" height="290">' . "</dd>";
}
}else{
echo "0 result";
}
?>
But when the code runs all i get is this: http://imgur.com/nhIO2LQ
Anyone know a solution?
use addslashes before insert
$img = addslashes(file_get_contents($_FILES['images']['tmp_name']));
$query = "INSERT INTO tableName (id,image) VALUES('','$image')";
i created this form to add my images to mysql database,and i think i did it right-cause it saves something :D - but it wont show me the image, what should i do to "SEE" the image from mysql?!
this is my form php:
$tmp_name=$_FILES['file']['tmp_name'];
if (isset($_POST['submit'])) {
if ((($_FILES['file']['type']) == "image/jpeg")
|| ($_FILES['file']['type']) == "image/gif"
|| ($_FILES['file']['type']) == "image/pjpeg"
&& ($_FILES['file']['size']) > 200000) {
$tmp_name=$_FILES['file']['tmp_name'];
// i also tried addslasheds
$image = mysql_real_escape_string(file_get_contents($_FILES['file']['tmp_name']));
if ($_FILES['file']['error'] > 0) {
echo "return code : " . $_FILES['FILES']['error'];
}else{
if (file_exists($_FILES['file']['name'])) {
echo "your file is already exists!";
}else{
Query("INSERT INTO image(image) VALUES ('".$tmp_name."')");
echo "FILES has been stored";
}
}
}
}else{
echo "invalid file";
}?>
and my code to show the image is:
<?php
require 'lib.php';
$request=Query('SELECT * FROM image');
while ($row = mysql_fetch_array($request)) {
echo $row['image'];
}?>
Inserting the temporary name into your database won't accomplish anything because it is just that, temporary. You need to save the image somewhere on your server using move_uploaded_file() and then save the new permanent name to your database so that you can use it in html image tags later.
$filename = "myimage.jpg";
$path = "/var/www/images/".$filename;
$link = "http://domain.com/images/".$filename;
move_uploaded_file($FILES['file']['tmp_name'], $path);
$image = mysql_real_escape_string($link);
if ($_FILES['file']['error'] > 0) {
echo "return code : " . $_FILES['FILES']['error'];
}else{
Query("INSERT INTO image(image) VALUES ('".$image."')");
echo "FILES has been stored";
}
Then when you retrieve your image:
<?php
require 'lib.php';
$request=Query('SELECT * FROM image');
while ($row = mysql_fetch_array($request)) {
echo '<img src="'.$row['image'].'" />";
}?>
How can I show my all "blog post" with images? Example:
Mysql table:
post_id | user_id | subject | message | image | img_name.
What is the php code to display all my posts with images in the index page? I used the following code but it doesn't display images, it shows only data. I would like to see something like this:
image | message is here
image | message is here
image | message is here
I used 3 pages
add_post.php(Html form)
add_post_process.php(process the add_post.php)
index.php (which shows my all post)
add_post_process.php:
<?php
include "db.php";
#$file = addslashes($_FILES['image']['tmp_name']);
$lastid= mysql_insert_id();
#$img = addslashes(file_get_contents($_FILES['image']['tmp_name']));
$img_name = addslashes($_FILES['image']['name']);
#$img_size = getimagesize($_FILES['image']['tmp_name']);
$upload_path = './blogpostimg/';
$post_id = addslashes($_POST['post_id']);
$user_id = addslashes($_POST['user_id']);
$subject = addslashes($_POST['subject']);
$message = addslashes($_POST['message']);
$cat_id = addslashes($_POST['cat_id']);
$cat_name = addslashes($_POST['cat_name']);
$comment_count = addslashes($_POST['comment_count']);
$ch_img = addslashes($_FILES['image']['name']);
$query = "SELECT img_name FROM blog_post WHERE img_name = '$ch_img';";
$result = mysql_query($query) or die (mysql_error());
if(isset($subject, $message))
{
$errors = array();
if(empty($subject) && empty($message) && empty($file))
{
$errors[] ='all field required';
}
else
{
if(empty($subject))
$errors[] = 'Subejct requried';
if (empty($message))
$errors[] = 'message required';
if(empty($file))
$errors[] ="SORRY, you have to be upload a image";
elseif($img_size == FALSE)
{
$errors[] ="That's not an image";
}
else
{
if(mysql_num_rows($result) != 0)
$errors[] = " You have to change this image name, already exit in our database";
}
}
if(!empty($errors))
{
foreach($errors as $error)
{
echo "<ul>";
echo "<strong><font color=red><li>$error</li></font></strong><br/>";
echo "</ul>";
}
}
else
{
if(!move_uploaded_file($_FILES['image']['tmp_name'],$upload_path . $img_name))
{
die('File Not Uploading');
}
else
{
$lastid = mysql_insert_id();
$query = mysql_query("INSERT INTO blog_post VALUES ('', '', '$subject',
'$img','$img_name','$message','$cat_id','$cat_name','',NOW() )");
if($query)
echo "Successfully uploaeded your post";
else
{
echo "Something is wrong to Upload";
}
}
}
}
?>
index.php
<?php
include "db/db.php";
$upload_path = "/secure/content/blogpostimg";
$sql= mysql_query("SELECT * FROM blog_post");
while ($rel = mysql_fetch_assoc($sql))
{
$id = $rel['post_id'];
$sub = $rel['subject'];
$imgname = $rel['img_name'];
$img = $rel ['image'];
$msg = $rel['message'];
$date = $rel['date'];
echo "<h1>". "$sub" ."</h1>". "<br/>";
echo "$imgname" ."<br/>";
echo '<img src="$upload_path " />';
echo "$msg" . "<br/>";
echo "$date" . "<br/>";
echo "<hr/>";
echo "<br/>";
}
?>
The mysql table structure is
post_id(int)
User_id(int)
subject(varchar)
image(blob)
img_name(varchar)
message(text)
change
echo '<img src="$upload_path " />';
to
echo '<img src="' . $upload_path . '/' . $img . '" />';
that should do the trick..
I don't now if it matters in performanceways.. but howcome you use blob instead of a varchar.. 255 characters for a filename should be enough.
In your index.php you have <img src="$upload_path" />. Instead of $upload_path, src needs to be the URL to your image.
I just noticed that you are storing the image itself in the database. There is no need to do this if you move it to a publicly accessible directory, e.g. /var/www/images/myimage.jpg
UPDATE:
I have have ten columns in my SQL table: id, imgid, urlid, image1, image2, image3, image4, image5, and comment. Id, imgid, and urlid are int type. Image[1-5] are mediumblob type. Url and comment are text type. Imgid is the number of images uploaded (which should be 5), urlid is the number of urls submitted (which should be one right now), url holds the url, and comment holds user comments. I'd like to retrieve and output all the image data from the image columns in a row but I'm getting torn page icons. I am using a front.php to output images and a get.php to convert images to viewable format.
Here's my front.php:
<?php
mysql_connect ("","","") or die(mysql_error());
mysql_select_db ("") or die(mysql_error());
$defaultqry = mysql_query ("SELECT * FROM dbp");
while($row = mysql_fetch_assoc($defaultqry))
{
echo ($row['image1'] != NULL) ? '<img src="get.php?id='.$row[id].'&col=1 " width="30" height="30"/> ' : '';
echo ($row['image2'] != NULL) ? '<img src="get.php?id='.$row[id].'&col=2 " width="30" height="30"/> ' : '';
echo ($row['image3'] != NULL) ? '<img src="get.php?id='.$row[id].'&col=3 " width="30" height="30"/> ' : '';
echo ($row['image4'] != NULL) ? '<img src="get.php?id='.$row[id].'&col=4 " width="30" height="30"/> ' : '';
echo ($row['image5'] != NULL) ? '<img src="get.php?id='.$row[id].'&col=5 " width="30" height="30"/> ' : '';
}
?>
Here's my get.php:
<?php
mysql_connect ("","","") or die(mysql_error());
mysql_select_db ("") or die(mysql_error());
$query = mysql_query("SELECT * FROM dbp WHERE id = ".mysql_real_escape_string($_GET['id']));
$row = mysql_fetch_array($query);
$col = intval($_GET['col']);
if(isset($row['image'.$col]) && $row['iamge'.$col] != NULL){
$content = $row['image'.$col];
}else{
exit; // col is not existent, or image is empty
}
header('Content-type: image/jpg');
echo $content;
exit;
?>
first, you use a DATAbase, store data in it not files. (you should store the link/path to the files in the DB not the images itself, this only blows up your database and makes it slow) see Storing Images in DB - Yea or Nay? for more info on this topic.
but if you want to do it this way then you will have to output the id of the row AND a column identifier like:
echo ($row['image1'] != NULL) ? '<img src="get.php?id='.$row['id'].'&col=1 " width="30" height="30"/> ' : '';
in your get.php:
mysql_connect ("localhost","","") or die(mysql_error());
mysql_select_db ("") or die(mysql_error());
$query = mysql_query("SELECT * FROM dbp WHERE id = ".mysql_real_escape_string($_GET['id']));
$row = mysql_fetch_array($query);
$col = intval($_GET['col']);
if(isset($row['image'.$col]) && $row['image'.$col] != NULL){
$content = $row['image'.$col];
}else{
exit; // col is not existent, or image is empty
}
header('Content-type: image/jpg');
echo $content;
exit;