superchrisc

Good Evening, If anyone could help with this little problem be much appreciated. what i'm trying to do is using a drop down menu select a folder name which is stored in a database table.. then taking that folder name and upload various files to that specified folder. There's probably an easier way to do this but i've searched google argggh.. The first part of code is the php to grab the folder names and the file selection form <?php include("connection.php"); $sql="SELECT id, name FROM gallery"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $id=$row["id"]; $thing=$row["name"]; $options.="<OPTION VALUE=\"$id\">".$thing ; } ?> <table width="500" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC"> <tr> <form action="postFiles.php" method="post" enctype="multipart/form-data" name="form1" id="form1"> <td> <table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF"> <tr> <td><strong>multiple Files Upload </strong></td> </tr> <td><SELECT NAME=thing> <OPTION VALUE=0>Choose <?=$options?> </SELECT> </td> <tr> <td>Select file <input name="ufile[]" type="file" id="ufile[]" size="50" /></td> </tr> <tr> <td>Select file <input name="ufile[]" type="file" id="ufile[]" size="50" /></td> </tr> <tr> <td>Select file <input name="ufile[]" type="file" id="ufile[]" size="50" /></td> </tr> <tr> <td align="center"><input type="submit" name="Submit" value="Upload" /></td> </tr> </table> </td> </form> </tr> </table> The second part of the code below is the php code to upload those files, the problem being how do i specify the upload destination using the drop down in the previous form (instead of UPLOAD) which it is default set to <?php //set where you want to store files //in this example we keep file in folder upload //$HTTP_POST_FILES['ufile']['name']; = upload file name //for example upload file name cartoon.gif . $path will be upload/cartoon.gif $path1= "upload".$HTTP_POST_FILES['ufile']['name'][0]; $path2= "upload".$HTTP_POST_FILES['ufile']['name'][1]; $path3= "upload".$HTTP_POST_FILES['ufile']['name'][2]; //copy file to where you want to store file copy($HTTP_POST_FILES['ufile']['tmp_name'][0], $path1); copy($HTTP_POST_FILES['ufile']['tmp_name'][1], $path2); copy($HTTP_POST_FILES['ufile']['tmp_name'][2], $path3); //$HTTP_POST_FILES['ufile']['name'] = file name //$HTTP_POST_FILES['ufile']['size'] = file size //$HTTP_POST_FILES['ufile']['type'] = type of file echo "File Name :".$HTTP_POST_FILES['ufile']['name'][0]."<BR/>"; echo "File Size :".$HTTP_POST_FILES['ufile']['size'][0]."<BR/>"; echo "File Type :".$HTTP_POST_FILES['ufile']['type'][0]."<BR/>"; echo "<img src=\"$path1\" width=\"150\" height=\"150\">"; echo "<P>"; echo "File Name :".$HTTP_POST_FILES['ufile']['name'][1]."<BR/>"; echo "File Size :".$HTTP_POST_FILES['ufile']['size'][1]."<BR/>"; echo "File Type :".$HTTP_POST_FILES['ufile']['type'][1]."<BR/>"; echo "<img src=\"$path2\" width=\"150\" height=\"150\">"; echo "<P>"; echo "File Name :".$HTTP_POST_FILES['ufile']['name'][2]."<BR/>"; echo "File Size :".$HTTP_POST_FILES['ufile']['size'][2]."<BR/>"; echo "File Type :".$HTTP_POST_FILES['ufile']['type'][2]."<BR/>"; echo "<img src=\"$path3\" width=\"150\" height=\"150\">"; /////////////////////////////////////////////////////// // Use this code to display the error or success. $filesize1=$HTTP_POST_FILES['ufile']['size'][0]; $filesize2=$HTTP_POST_FILES['ufile']['size'][1]; $filesize3=$HTTP_POST_FILES['ufile']['size'][2]; if($filesize1 && $filesize2 && $filesize3 != 0) { echo "We have recieved your files"; } else { echo "ERROR....."; } ////////////////////////////////////////////// // What files that have a problem? (if found) if($filesize1==0) { echo "There're something error in your first file"; echo "<BR />"; } if($filesize2==0) { echo "There're something error in your second file"; echo "<BR />"; } if($filesize3==0) { echo "There're something error in your third file"; echo "<BR />"; } ?> Thanks in advance for any help if you need to see a live version of these files i can provide a link

Thats just a silly error on my part.. Thank so you so much for your help on a basic problem!! which i should of realised

Thanks for your quick reply, it half working... it's updating the database but replacing the fields with no content there are 2 errors left in the code which am confused about. error below: Notice: Undefined index: header in E:\Domains\o\onlinegurusites.co.uk\user\htdocs\garageRus\admin\update_ac.php on line 9 Notice: Undefined index: content in E:\Domains\o\onlinegurusites.co.uk\user\htdocs\garageRus\admin\update_ac.php on line 10 Successful View result these are the problem lines of code; $header = $_POST['header']; $content = $_POST['content']; Thanks and thanks in advance for your help!!

Having trouble with this upload script, it grabs and displays the content from the database fine. It won't however update the database although the code is throwing up no errors; this is the script that displays the content ############### Code <?php include("includes/connection.php"); $sql="SELECT * FROM mot"; $result=mysql_query($sql); ?> <table width="400" border="0" cellspacing="1" cellpadding="0"> <tr> <td> <table width="400" border="1" cellspacing="0" cellpadding="3"> <tr> <td colspan="4"><strong>List data from mysql </strong> </td> </tr> <tr> <td align="center"><strong>Name</strong></td> <td align="center"><strong>Lastname</strong></td> <td align="center"><strong>Update</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr> <td><? echo $rows['header']; ?></td> <td><? echo $rows['content']; ?></td> // link to update.php and send value of id <td align="center"><a href="update_content.php?id=<? echo $rows['id']; ?>">update</a></td> </tr> <?php } ?> </table> </td> </tr> </table> this next bit of code is the code that user updates the content; <?php include("includes/connection.php"); $id=$_GET['id']; $sql="SELECT * FROM mot WHERE id='$id'"; $result=mysql_query($sql); $rows=mysql_fetch_array($result); ?> <table width="400" border="0" cellspacing="1" cellpadding="0"> <tr> <form name="form1" method="post" action="update_ac.php"> <td> <table width="100%" border="0" cellspacing="1" cellpadding="0"> <tr> <td> </td> <td colspan="3"><strong>Update data in mysql</strong> </td> </tr> <tr> <td align="center"> </td> <td align="center"> </td> <td align="center"> </td> <td align="center"> </td> </tr> <tr> <td align="center"> </td> <td align="center"><strong>Name</strong></td> <td align="center"><strong>Lastname</strong></td> </tr> <tr> <td> </td> <td align="center"><input name="name" type="text" id="name" value="<? echo $rows['header']; ?>"></td> <td align="center"><input name="lastname" type="text" id="lastname" value="<? echo $rows['content']; ?>" size="15"></td> </tr> <tr> <td> </td> <td><input name="id" type="hidden" id="id" value="<? echo $rows['id']; ?>"></td> <td align="center"><input type="submit" name="Submit" value="Submit"></td> <td> </td> </tr> </table> </td> </form> </tr> </table> this final bit is the update code for the database which isnt working but not throwing up any errors <?php include("includes/connection.php"); if(isset($_POST['update'])){ $id = $_GET['id']; $header = $_POST['header']; $content = $_POST['content']; $sql="UPDATE mot SET header = '$header', content = '$content' WHERE id='$id'"; $result=mysql_query($sql); if($result){ echo "Successful"; echo "<BR>"; echo "<a href='mot_edit.php'> View result</a>"; } else { echo "ERROR"; } } ?> Any help would be much appreciated

the form code is <form enctype=”multipart/form-data” action=”uploader.php” method=”post”> <input type="hidden" name="MAX_FILE_SIZE" value="100000" /> ProductName: <input type="text" name="productname" /> ProductDescription: <input type="text" name="productdes"/> ChooseImage: <input name="uploadedfile" type="file" /><br /> <input type="submit" value="Submit" /> </form>

another small problem, it won't insert or upload the image, was wondering if someone could hint or tell me where am going wrong with this code <?php // Where the file is going to be placed $target_path = "upload /"; /* Add the original filename to our target path. Result is "uploads/filename.extension" */ $target_path = $target_path . basename( $_FILES['uploadedfile']['name']); $target_path = "upload/"; $target_path = $target_path . basename( $_FILES['uploadedfile']['name']); if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) { echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded"; } else{ echo "There was an error uploading the file, please try again!"; } ?> <?php $sql="INSERT INTO products (ProductName, ProductDescription, ChooseImage) VALUES ('$_POST[produtname]','$_POST[produtdes]','$_POST[uploadedfile]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; ?> Click <a href="products.php">HERE</a> to go back to form

i'll try that thanks very much, will that insert the image url to? within a varchar record type

Sorry, my problem is. i have a form with four inputs, product name product price product image product description at the moment, my code will upload the image to the upload folder, but i need help or a point in the direction of how i'd store that information including the image URL in my db table. Hope this is a little clearer sorry for the previous post

if anyone needs to see any more of the code thats no problem, and if anyone could solve my problem i'd be very grateful

Hey, i need help storing an image in my database via the URL(image location) at the moment my php code is storing the image in a folder on the directory called upload. here is the code: <?php // Where the file is going to be placed $target_path = "upload /"; /* Add the original filename to our target path. Result is "uploads/filename.extension" */ $target_path = $target_path . basename( $_FILES['uploadedfile']['name']); $target_path = "upload/"; $target_path = $target_path . basename( $_FILES['uploadedfile']['name']); if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) { echo "The file ". basename $_FILES['uploadedfile']['name']). " has been uploaded"; } else{ echo "There was an error uploading the file, please try again!"; } ?> Click <a href="products.php">HERE</a> to go back to form if someone could help me i'd be very grateful

yes am making a bespoke cms and i want to allow the user to upload an image and do both store the image in a folder and the that url in the database and to maybe echo out the image on the front end of the site(but would this just be a basic select from and echo?

Hey, i need help storing an image in my database via the URL(image location) at the moment my php code is storing the image in a folder on the directory called upload. here is the code: <?php // Where the file is going to be placed $target_path = "upload /"; /* Add the original filename to our target path. Result is "uploads/filename.extension" */ $target_path = $target_path . basename( $_FILES['uploadedfile']['name']); $target_path = "upload/"; $target_path = $target_path . basename( $_FILES['uploadedfile']['name']); if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) { echo "The file ". basename $_FILES['uploadedfile']['name']). " has been uploaded"; } else{ echo "There was an error uploading the file, please try again!"; } ?> Click <a href="products.php">HERE</a> to go back to form if someone could help me i'd be very grateful [attachment deleted by admin]