Zoofu Posted September 13, 2009 Share Posted September 13, 2009 The screen comes up white, even though I think this is probably the wrong way to go about doing it. <?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, otherwize 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>"; $sqlt = "SELECT * FROM `users` WHERE `id`='".$_SESSION['uid']."'"; $rest = mysql_query($sqlt) or die(mysql_error()); $rowt = mysql_fetch_assoc($rest); mysql_query("UPDATE `users` SET avatar='{images/.$_FILES['tmp_name'].}' WHERE `id`=".$_SESSION['uid']."") OR die(mysql_error()); } ?> <!--next comes the form, you must set the enctype to "multipart/frm-data" and use an input type "file" --> <form name="newad" method="post" enctype="multipart/form-data" action=""> <table> <tr><td><input type="file" name="image"></td></tr> <tr><td><input name="Submit" type="submit" value="Upload image"></td></tr> </table> Now I want to know, why doesn't it set that field to what I uploaded? Is there an easier/working/more efficient way to do this? Link to comment https://forums.phpfreaks.com/topic/174127-i-feel-this-is-the-wrong-way-to-do-it/ Share on other sites More sharing options...
Zoofu Posted September 13, 2009 Author Share Posted September 13, 2009 Also... The file directory it's going into is images which is the same directory as that php file. It's uploading the images, but not setting them as the avatar. And how would I get erm... It to be the path. The avatar field would have to be ./images/THEFILESNAME.png Link to comment https://forums.phpfreaks.com/topic/174127-i-feel-this-is-the-wrong-way-to-do-it/#findComment-917867 Share on other sites More sharing options...
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