showoff Posted July 27, 2007 Share Posted July 27, 2007 hi, i just wanted to say first that some of the stuff on this site has helped me thus far. ok my question is i followed a tut about making a image upload script to add to my site. i got that working, my only main question is what do i need to do to get it to give me the url of the image. when it is uploaded, somewhat like photobucket or other image hosting sites do. what i am doing is i do graphic design work mostly signatures and wallpapers, and i put together a site to show my stuff, but i have a forum( a free one for now) and some of my users we want to make a flash gallery but for now i want it to were they can host stuff to share between us for now. i hope my question makes sense, i know enough about html but am very new to php, i also have a mysql database on my server. if using that is better or easier i noticed some of the ones i have seen do that but i really don't want all the stuff that the one i found has on it. thanks in advance. Quote Link to comment Share on other sites More sharing options...
btherl Posted July 27, 2007 Share Posted July 27, 2007 Please post your current code Quote Link to comment Share on other sites More sharing options...
showoff Posted July 27, 2007 Author Share Posted July 27, 2007 this what i have and it will upload but i just want to add were it will show the link back to the image. <form enctype="multipart/form-data" action="upload.php" method="POST"> Please choose a file: <input name="uploaded" type="file" /><br /> <input type="submit" value="Upload" /> </form> <?php $target = "upload/"; $target = $target . basename( $_FILES['uploaded']['name']) ; $ok=1; //This is our size condition if ($uploaded_size > 1000000) { echo "Your file is too large.<br>"; $ok=0; } //This is our limit file type condition if ($uploaded_type =="text/php") { echo "No PHP files<br>"; $ok=0; } //Here we check that $ok was not set to 0 by an error if ($ok==0) { Echo "Sorry your file was not uploaded"; } //If everything is ok we try to upload it else { if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target)) { echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded"; } else { echo "Sorry, there was a problem uploading your file."; } } ?> Quote Link to comment Share on other sites More sharing options...
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