Hey my php/mysql/html people,
I have this variable that holds a path to an image and then insert it into a db (yes, I am using mysql_real_escape_string) which works perfectly.
$file_name = $_FILES["file"]["name"];
$path='images\ '. $file_name;
insert into...blah blah blah
The path is later pulled from the db and stored in said variable.
$path = $row['file_path'];
I am trying to display it with:
// the contents of $path in this case is: images\ cats.jpg
echo "<img src=" . $path . ">";
However the image breaks because the only thing that src pics up is: images\
and not the actualname+extension of the image. I know this probably has to do with the slashes, but I am a newbie and could use some help. Thanks in advance!
your
$path='images\ '. $file_name;
should be
$path='images\'. $file_name;
Print $path variable & see the values.
echo $path;exit;
You will get value on web browser.
Copy the value displayed on browser & paste it in the URL, if you see the image, your saved path is correct, if not, you have gone wrong in specifying the relative path too the resource.
Related
So I have a few images in the server (public_html/img/profile_pictures/).
This is how I currently set the image:
echo "<img src='img/profile_pictures/main_photo.png'/>";
The main_photo can change each day, but if it changes to main_photo.jpg insted, it wont show (because the extension is hardcoded on that line(.png)). Is it possible to display the photo without knowing the extension for the image file?
If you want a PHP code, then try this. This code will look for main_photo.* inside your folder and automatically set the extension upon finding one.
Remember to set the path properly
<?php
$yourPhotoPath = "img/profile_pictures/";
foreach (glob($yourPhotoPath.'main_photo.*') as $filename) {
$pathInfo = pathinfo($filename);
$extension = $pathInfo['extension'];
$fileName = chop($pathInfo['basename'], $extension);
echo "<img src='".$yourPhotoPath.$fileName.$extension."'/>";
}
?>
if a Photo isn't loaded, it's width and size is null.
Although I would advise you to write a class that checks and loads images, I get a feeling you want a simple solution. so, given by the premise that the photo is either
<img src='img/profile_pictures/main_photo.png'/>
or
<img src='img/profile_pictures/main_photo.jpg'/>
and that neither this path nor this filename ever changes and in the folder is only one picture,
you could simply echo both.
The img of the one that is empty will not be shown.
A better way was to write a class that loads your photo and checks if the photo is really there, like
$path = 'img/profile_pictures/main_photo.png';
if(!file_exists('img/profile_pictures/main_photo.png'))
{
//use the jpg path
$path = 'img/profile_pictures/main_photo.jpg';
}
You can ofc just inline this if case, but it's bad practise to intermix buisinesslogic and format logic, so I advice you to write a class for it.
I'm basically a designer who hand codes HTML, but am a hack when it comes to PHP. I've been asked to add image icons within a product search results table on a PHP/MySQL site that has had many programmers over a decade, causing a sloppy mess of code.
I have the following code at the top of a search results page that calls out the name of the image:
$image2 = $row['item'] . ".jpg";
$imagefile2 = $_SERVER['DOCUMENT_ROOT'] . "/product_images/$imagefile2";
if(file_exists($imagefile2)){
Within the table itself I hacked this to get the correct image to show:
print "<td><center><img src=/product_images/$image2 width=80><br>
Of course, if there is no image, there is a broken image link. I dod not know the proper syntax to tell the server that IF there is no image, THEN show noimage.jpg (located in same folder). This is probably a couple lines of added code at best, but after a couple hours of searches and attempts I surrender.
Check to see if the file exists and if it does, set the image2 variable to point to it. If it doesn't exist, set the image2 variable to point to the "no image" image.
if(file_exists($_SERVER['DOCUMENT_ROOT'] . "/product_images/".$row['item'].".jpg")){
$image2=$row['item'].".jpg";
}else{
$image2="/product_images/noimage.png";
}
<?php
if(file_exists($_SERVER['DOCUMENT_ROOT'] . "/product_images/".$row['item'].".jpg")){
$image2=$row['item'].".jpg";
}else{
// Specify No Image File Path to Variable
$image2="/product_images/noimage.png";
}
?>
<td><center><img src="<?php echo $image2; ?>" width="80"><br>
I have been working on a little PHP script that dynamically creates & saves images in a folder tmp (which already exists!) from a base64 string. This is my save.php file.
// get base-64 string from form
$filteredData = substr($_POST['img_val'], strpos($_POST['img_val'], ",")+1);
// decode string
$unencodedData = base64_decode($filteredData);
// create unique filename
$a = uniqid();
// name, location and file extension
$compfile = '/tmp/' . $a . '.png';
// save image
file_put_contents($compfile, $unencodedData);
// print image
echo '<img src="' . $compfile . '" />';
Strangely enough: It will neither create nor render the image on the page, because of the /tmp/ in my $compfile variable. When I remove it, everything works like a charm and the image is being created in the same folder.
Unfortunately, I really want the image to be created in the /tmp/ folder. Before $compfile was a randomly generated filename, and instead was called /tmp/img.png I was able to save to create an image by the name img.png and save it to the tmp.
What am I missing here?
(Thank you for your time.)
Since I guess this was the solution I'll post it as answer.
I think you want "tmp/" . $a . ".png" instead of "/tmp/" . $a . ".png". It's good practice to just always use absolute paths, so: __DIR__ . "/tmp/" . $a . ".png". This takes away any confusion.
Hello dear Overflowers!
I have a little code that so far works so well,
it gets the corresponding Image ID to the corresponding House ID, also Caption and filename are no problem!
I just fail at displaying the image... :(
`
include("functions/config.php");
$images_dir = "/some/url/that/is/correct/cms/houses/";
$query = "SELECT houses.*, gallery_photos.* ".
"FROM houses LEFT JOIN gallery_photos ".
"ON houses.id = gallery_photos.photo_category";
$result = mysql_query($query) or die(mysql_error());
// Print out the contents of each row into a table
while($row = mysql_fetch_array($result)){
$house_id = $row['id'];
$photo_category = $row['photo_category'];
$photo_caption = $row['photo_caption'];
$photo_filename = $row['photo_filename'];
echo "House ID ". $house_id. " - ". $photo_category." - ". $photo_caption." - <img src='".$images_dir.$photo_filename."' />";
echo "<br />";
}
?>`
So as said, everything get selected well, even the image path is correct when you click Image Info in the browser, but I simply do not get the image to display, am I missing something very trivial here?
Thanks in advance!
Check Image Name in Table as well as in folder where you store particular image.
<img src='".$images_dir.$photo_filename."' />
It misses like .jpg or .png whatever the extension of your img is
Check with tool like Firebug what is the path in src attribute of img element. Then copy this link and paste it in browser to see if it opens (with your website name - so if link is '/img/houses/house1.jpg' then you paste to browser 'www.mypage.com/img/houses/house1.jpg').
If it does not then:
Check if path is correct (it will count from web root folder) (your path should be probably like '/cms/houses/5953.jpg')
Check of path has correct letter sizes ('house' vs. 'House')
Check if you can open file on server (maybe file is corrupted)
Check if you have rights to read file
All right guys, finally found the mistake, which I hope that with my anwser, some pople in the future can avoid it!
As it seems some Hosts do not like some of the PHP naming conventions, so what I had to do, so the files would actually be stored in the folder, was change the command to get them from "Tmp/In memory" to the final destination folder.
For better understanding, here what I changed :
copy($photos_uploaded['tmp_name'][$counter],
$images_dir . '/' . $filename);
to
move_uploaded_file($photos_uploaded['tmp_name'][$counter],
$images_dir . '/' . $filename);
I hope no one else ever has to fiddle this long and spend so much money on Coffee ever again.
Thanks to all that tried to help anyway!
1.Hi, I have a internal use only file upload script that uploads the files to a directory. When I upload something from my computer with a spcace in the name i.e example 1.zip it uploads with a space in the name thus killing the link in a email. Is it possible to make apache remove the space when its uploaded or make it a underscore?
The second problem I am having is how would I parse this to make the link an email link with the url of the file as the body of the email amd the email addy anything?
if (move_uploaded_file($_FILES['file']['tmp_name'], $uploaddir . $_FILES['file']['name'])) {
// uploaded file was moved and renamed succesfuly. Display a message.
echo "Link: " . "http://example.org/" . $_FILES["file"]["name"];
You just need to urlencode() your file name and everything is fine:
echo "Link: http://example.org/" . urlencode($_FILES["file"]["name"]);
But if you want to remove the spaces for another reason, you can use str_replace():
$replaced_name = str_replace(' ', '_', $_FILES["file"]["name"]);
rename($uploaddir . '/' . $_FILES['file']['name'], $uploaddir . '/' . $replaced_name);
# You should urlencode() it nonetheless:
echo "Link: http://example.org/" . urlencode($replaced_name);
Try:
$filename = $_FILES['file']['name'];
$filename = preg_replace("/[^a-zA-Z0-9]/", "", $filename);
//then
if (move_uploaded_file($_FILES['file']['tmp_name'], $uploaddir . $filename)) {
// uploaded file was moved and renamed succesfuly. Display a message.
echo "Link: " . "http://example.org/" . $filename;
As a side note : with the code you are using, what is happening if two files with the same name are uploaded ? If you don't do a check (like "is there a file that already has that name in $uploaddir ?") the second file will replace the first one.
That might not be something you want... is it ?
If not, to solve that (potential) problem, one solution is to always rename uploaded files, with names you control. (A simple counter would probably to the trick)
Another thing is : $_FILES["file"]["name"] is sent by the client, and, as such, can probably be forged to contains whatever someone would want. If it contains something like "../../index.php" (or something like this - you get the idea), this could allow someone to put any file they want on your server.
To prevent this from happening, you shoud be sure the file name/path used as destination of move_uploaded_file does not contain anything "dangerous". A solution could be to use basename. (see, for instance, example #2 on POST method uploads)
You might also want to check the mimetype of the uploaded file, so you don't get executables, for instance -- and you should make sure files uploaded are not executable by the webserver.