lingo5 Posted May 7, 2012 Share Posted May 7, 2012 Hi, I am uploading an image to a folder and I need to display the uploaded image in a new page. This is my code for the upload: <?php //define a maxim size for the uploaded images in Kb define ("MAX_SIZE","2000000"); //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['upload'])) { //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>Tipo de imagen no permitido.</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>La imagen es demasiado grande.</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) $newname="banners/".$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>Error al subir la imagen.</h1>'; $errors=1; }}} //If no errors registred, print the success message if(isset($_POST['Submit']) && !$errors) { echo "<h1>La imagen se ha cargado correctamente</h1>"; } $query = "INSERT INTO banner_rotator.t_banners (banner_path) ". "VALUES ('$newname')"; mysql_query($query) or die('Error, query failed : ' . mysql_error()); //echo "<br>Files uploaded<br>"; header("Location: PC_cropbanner.php"); } Sowhat I need to do is to display the uploaded image in PC_cropbanner.php. How can I do this? Link to comment https://forums.phpfreaks.com/topic/262211-please-help-displaying-image-after-upload/ Share on other sites More sharing options...
wigwambam Posted May 7, 2012 Share Posted May 7, 2012 You could pass the image name using your redirect: header("Location: PC_cropbanner.php?image=".$image_name); Then, at the top of PC_copbanner.php: $image_name = $_GET['image']; You will need to check that the image variable has a value... Link to comment https://forums.phpfreaks.com/topic/262211-please-help-displaying-image-after-upload/#findComment-1343793 Share on other sites More sharing options...
lingo5 Posted May 8, 2012 Author Share Posted May 8, 2012 Thanks wigwambam !!! I did that and it works perfect. Link to comment https://forums.phpfreaks.com/topic/262211-please-help-displaying-image-after-upload/#findComment-1343909 Share on other sites More sharing options...
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