josephman1988 Posted August 15, 2008 Share Posted August 15, 2008 Hey guys, I have this form, which uploads data to a table, then uploads a file to the server, and also stores that information into another table. What happens is it creates a page that shows both of these inputted files/data. My DB tables: operation_data -------------------- | oid | name | date | -------------------- screenshot_operation --------------------------------------- | sid | name | title | operation_dataoid | --------------------------------------- Form: <form enctype="multipart/form-data" method="post" action="<?php $_SERVER['PHP_SELF'];?>"> Case Name:<br /> <div id="operationupload"> <input type="text" title="casename" name="casename" /></div><br /><br /> Briefing:<br /> <div id="operationupload"> <textarea cols="50" rows="10" name="briefing" title="briefing"></textarea></div><br /><br /> Upload Screenshot:<br /> <div id="operationupload"> <input type="text" name="pictitle" title="pictitle" /><br /> <input type="file" name="imageupload" title="imageupload" /> </div> <br /> <input type="submit" value="Submit" name="submit" /> <input type="reset" value="Reset" name="reset" /> </form> PHP Code: <?php if (isset($_POST['casename'], $_POST['briefing'])) { $casename = $_POST['casename']; $briefing = $_POST['briefing']; $pictitle = $_POST['pictitle']; $picname = $_FILES['imageupload']['name']; $sql = ("INSERT INTO operation_data SET name = '$casename', briefing = '$briefing', date = CURDATE()"); if (@mysql_query($sql)) { }else{ echo "Cannot Be Added, Something Went Wrong!"; } $sql2 = @mysql_query("INSERT INTO screenshot_operation SET name = '$picname', title = '$pictitle', date = CURDATE()"); if ( isset( $_FILES['imageupload'] ) ) { $_FILES['imageupload']['name']; $_FILES['imageupload']['size']; $_FILES['imageupload']['tmp_name']; $_FILES['imageupload']['type']; $_FILES['imageupload']['error']; basename($_FILES['imageupload']['name']); $source = $_FILES['imageupload']['tmp_name']; $target = "../images/operations/screenshots/".$_FILES['imageupload']['name']; if($_FILES['imageupload']['type'] == "image/pjpeg" || $_FILES['imageupload']['type']=="image/jpeg") { }else{ exit('Permission Denied, Please Only Upload mages'); } if (file_exists($target)) { exit( "File Name Exists, Please Alter" ); } else { move_uploaded_file( $source, $target ); echo "News Added. Thankyou!!"; } } } ?> However, what I want to do, is associate this uploaded image to this very 'Operation'. How would I go about it? I thought putting the 'operation_dataoid in the screenshot_operation table, i would be able to grab that operations id i was adding so in the template i could call the operations id from screenshot_operation table and display it. Am i correct? If not, how do I do this? Thanks in advance, Joe. Link to comment https://forums.phpfreaks.com/topic/119830-associating-images-to-operation-id-in-one-form/ Share on other sites More sharing options...
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