Boxerman Posted October 3, 2008 Share Posted October 3, 2008 hi guys im in need of a script that when a user uploads it will display the picture uploaded and add to a page where other users have uploaded too! this is what i got so far.. <?php //define a maxim size for the uploaded images in Kb define ("MAX_SIZE","100"); //This function reads the extension of the file. It is used to determine if the file is an image by checking the extension. function getExtension($str) { $i = strrpos($str,"."); if (!$i) { return ""; } $l = strlen($str) - $i; $ext = substr($str,$i+1,$l); return $ext; } //This variable is used as a flag. The value is initialized with 0 (meaning no error found) //and it will be changed to 1 if an errro occures. //If the error occures the file will not be uploaded. $errors=0; //checks if the form has been submitted if(isset($_POST['Submit'])) { //reads the name of the file the user submitted for uploading $image=$_FILES['image']['name']; //if it is not empty if ($image) { //get the original name of the file from the clients machine $filename = stripslashes($_FILES['image']['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 echo '<h1>Unknown extension!</h1>'; $errors=1; } else { //get the size of the image in bytes //$_FILES['image']['tmp_name'] is the temporary filename of the file //in which the uploaded file was stored on the server $size=filesize($_FILES['image']['tmp_name']); //compare the size with the maxim size we defined and print error if bigger if ($size > MAX_SIZE*1024) { echo '<h1>You have exceeded the size limit!</h1>'; $errors=1; } //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) $newname="images/".$image_name; //we verify if the image has been uploaded, and print error instead $copied = copy($_FILES['image']['tmp_name'], $newname); if (!$copied) { echo '<h1>Copy unsuccessfull!</h1>'; $errors=1; }}}} //If no errors registred, print the success message if(isset($_POST['Submit']) && !$errors) { echo "<h1>File Uploaded Successfully! Try again!</h1>"; } ?> Quote Link to comment https://forums.phpfreaks.com/topic/126879-picture-upload/ Share on other sites More sharing options...
Boxerman Posted October 3, 2008 Author Share Posted October 3, 2008 ? Quote Link to comment https://forums.phpfreaks.com/topic/126879-picture-upload/#findComment-656486 Share on other sites More sharing options...
budimir Posted October 3, 2008 Share Posted October 3, 2008 And your problem is ... ? Quote Link to comment https://forums.phpfreaks.com/topic/126879-picture-upload/#findComment-656489 Share on other sites More sharing options...
Boxerman Posted October 3, 2008 Author Share Posted October 3, 2008 Its not displaying the uploaded image... onto a page called usersimages.php Quote Link to comment https://forums.phpfreaks.com/topic/126879-picture-upload/#findComment-656520 Share on other sites More sharing options...
Boxerman Posted October 3, 2008 Author Share Posted October 3, 2008 hmmm.. still cant work it out :-\ Quote Link to comment https://forums.phpfreaks.com/topic/126879-picture-upload/#findComment-656584 Share on other sites More sharing options...
Flames Posted October 3, 2008 Share Posted October 3, 2008 is it uploading the images then? Quote Link to comment https://forums.phpfreaks.com/topic/126879-picture-upload/#findComment-656641 Share on other sites More sharing options...
Boxerman Posted October 3, 2008 Author Share Posted October 3, 2008 Yes uploads to the folder fine.. how do i display them.. along with the other ones uploaded before? Quote Link to comment https://forums.phpfreaks.com/topic/126879-picture-upload/#findComment-656644 Share on other sites More sharing options...
Boxerman Posted October 5, 2008 Author Share Posted October 5, 2008 Its not displaying the uploaded image... onto a page called usersimages.php Quote Link to comment https://forums.phpfreaks.com/topic/126879-picture-upload/#findComment-657486 Share on other sites More sharing options...
budimir Posted October 5, 2008 Share Posted October 5, 2008 Well, since I haven't seen your code. I assume you could do something like $sql = "SELECT * FROM pictures ORDER BY id DESC LIMIT 1"; In this way you get last (id) picture in your database and that one is shown. Again, I haven't seen your code so I just asume what you could do. Quote Link to comment https://forums.phpfreaks.com/topic/126879-picture-upload/#findComment-657492 Share on other sites More sharing options...
JasonLewis Posted October 5, 2008 Share Posted October 5, 2008 The code you have above, that's not usersimages.php is it? We would need to see the code of usersimages.php because that is the code that is relevant to the displaying of the image. Also, you just upload the image. You don't store it's name somewhere for later reference or anything? How are you supposed to know which image to load up? You need a database or perhaps a flat file which stores the id of the user who is uploading, along with the name of the image. Quote Link to comment https://forums.phpfreaks.com/topic/126879-picture-upload/#findComment-657494 Share on other sites More sharing options...
Boxerman Posted October 5, 2008 Author Share Posted October 5, 2008 ok, well the database i got wont allow upload of pictures to it? Quote Link to comment https://forums.phpfreaks.com/topic/126879-picture-upload/#findComment-657498 Share on other sites More sharing options...
JasonLewis Posted October 5, 2008 Share Posted October 5, 2008 You shouldn't upload pictures to a database. Instead, you store the name of the uploaded image to the database. Assign it a unique ID, and perhaps the ID of the user who uploaded. Then you can call on the image later and get the name of the image from the database. Quote Link to comment https://forums.phpfreaks.com/topic/126879-picture-upload/#findComment-657500 Share on other sites More sharing options...
Boxerman Posted October 5, 2008 Author Share Posted October 5, 2008 ahh... but guests upload on my site... so unique id per one... and call on it later? like the query it to display on a page? Quote Link to comment https://forums.phpfreaks.com/topic/126879-picture-upload/#findComment-657501 Share on other sites More sharing options...
JasonLewis Posted October 5, 2008 Share Posted October 5, 2008 If you want to just display all uploaded images, without caring of the storage of the image information you can run a basic script like this: foreach(glob("images/*.*") as $image){ echo "<img src='" . $image . "' /><br />"; } Of course if you have any .php files in there it'll try and put them in there, so you need to add in more error checking. Quote Link to comment https://forums.phpfreaks.com/topic/126879-picture-upload/#findComment-657503 Share on other sites More sharing options...
Boxerman Posted October 5, 2008 Author Share Posted October 5, 2008 ok, so how do i store file names etc.. in the script i have ad the moment? Quote Link to comment https://forums.phpfreaks.com/topic/126879-picture-upload/#findComment-657505 Share on other sites More sharing options...
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