naveenbj Posted December 29, 2007 Share Posted December 29, 2007 Hello I have a very small qustion . jst have a look on this link>>http://ice.host-care.com/~com/n/upload.html and Im trying to get the image link instantly and automatically on the very next page which should comes after uploading. did you get what i mean?? here is the code which i used for this and im not getting the out put which i want <HTML> <HEAD> <TITLE>upload</TITLE> <META HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=iso-8859-1"> </HEAD> <BODY BGCOLOR="#FFFFFF" TEXT="#000000"> <H1> </H1><TITLE></TITLE><?php if (($_FILES["file"]["type"] == "image/gif") || ($_FILES["file"]["type"] == "image/jpeg") || ($_FILES["file"]["type"] == "image/pjpeg") && ($_FILES["file"]["size"] < 20000)) { if ($_FILES["file"]["error"] > 0) { echo "Return Code: " . $_FILES["file"]["error"] . " "; } else { echo "Upload: " . $_FILES["file"]["name"] . " "; echo "Type: " . $_FILES["file"]["type"] . " "; echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb "; echo "Temp file: " . $_FILES["file"]["tmp_name"] . " "; if (file_exists("upload/" . $_FILES["file"]["name"])) { echo $_FILES["file"]["name"] . " already exists. "; } else { move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $_FILES["file"]["name"]); echo "Stored in: " . "upload/" . $_FILES["file"]["name"]; } } } else { echo "Invalid file"; } ?> </BODY> </HTML> ------------------------------ Friends did you get what i mean and please try to help me for this asap. Naveen Quote Link to comment Share on other sites More sharing options...
priti Posted December 29, 2007 Share Posted December 29, 2007 Hello, If i am write then you want to upload the image file and see on the same page withot refreshing the whole page. If this is true then try to get help of AJAX to upload the file.Let me remind you cann't do fileuploads with ajx but there is a trick to do that.It will help you to upload the file and then your can read and show the image as soon as it is in the location. try to find or google class called AjaxFileUploader.I had worked with the same but its a big set of file i cann't attach or share here.Hope it work for you also. Happy coding Regards Quote Link to comment Share on other sites More sharing options...
naveenbj Posted December 29, 2007 Author Share Posted December 29, 2007 Hello Priti, thanks for reply let me explain tht if you upload any image file ,the file will be saved on server but wht Im trying to get the image . And for tht purpose i need tht image link instantly and automatically on the very next page. Did you get ?? Regards Quote Link to comment Share on other sites More sharing options...
naveenbj Posted December 29, 2007 Author Share Posted December 29, 2007 Hello Friend I think i got wht i need and its soo simple . echo "<A HREF='upload/" . $_FILES["file"]["name"]."' target='_blank'>View uploaded image</A>"; the above line has to be there ("after the last echo statement") Anyways thanks for watching my post and to advice me . if there is still any other way to get the same output thn plz let me know Your reply is highly appriciated Regards, Quote Link to comment Share on other sites More sharing options...
naveenbj Posted December 29, 2007 Author Share Posted December 29, 2007 Hello I have a very small qustion . jst have a look on this link>>http://ice.host-care.com/~com/n/upload.html and Im trying to get the image link instantly and automatically on the very next page which should comes after uploading. did you get what i mean?? here is the code which i used for this and im not getting the out put which i want <HTML> <HEAD> <TITLE>upload</TITLE> <META HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=iso-8859-1"> </HEAD> <BODY BGCOLOR="#FFFFFF" TEXT="#000000"> <H1> </H1><TITLE></TITLE><?php if (($_FILES["file"]["type"] == "image/gif") || ($_FILES["file"]["type"] == "image/jpeg") || ($_FILES["file"]["type"] == "image/pjpeg") && ($_FILES["file"]["size"] < 20000)) { if ($_FILES["file"]["error"] > 0) { echo "Return Code: " . $_FILES["file"]["error"] . " "; } else { echo "Upload: " . $_FILES["file"]["name"] . " "; echo "Type: " . $_FILES["file"]["type"] . " "; echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb "; echo "Temp file: " . $_FILES["file"]["tmp_name"] . " "; if (file_exists("upload/" . $_FILES["file"]["name"])) { echo $_FILES["file"]["name"] . " already exists. "; } else { move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $_FILES["file"]["name"]); echo "Stored in: " . "upload/" . $_FILES["file"]["name"]; } } } else { echo "Invalid file"; } ?> </BODY> </HTML> ------------------------------ Friends did you get what i mean and please try to help me for this asap. Naveen Quote Link to comment Share on other sites More sharing options...
naveenbj Posted December 29, 2007 Author Share Posted December 29, 2007 Hello Friend I think i got wht i need and its soo simple . echo "<A HREF='upload/" . $_FILES["file"]["name"]."' target='_blank'>View uploaded image[/url]"; the above line has to be there ("after the last echo statement") Anyways thanks for watching my post and to advice me . if there is still any other way to get the same output thn please let me know Your reply is highly appriciated Regards, Quote Link to comment Share on other sites More sharing options...
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