natie769

I have a form that submits the data to mysql. Then on the page where it displays all the rows in the table (items) I want it to only show the form fields (or columns) that have been filled out. So I'm trying to check if it is empty. I'm using a code that does that now -- but it only works for the first entry in the database, and then applies that to each and every entry after it (even though it isn't true)**. Should I be using a for loop so it checks every database table entry, and not just the first entry? ** When I say it applies to each and every entry I mean if the first entries "Category" is empty, it thinks the rest of the table entries "Category" is empty, when in fact it isn't. Hope that makes sense... Any help is appreciated! $result = mysql_query("SELECT * FROM `Equipment Listings` WHERE id = $id"); while($row = mysql_fetch_assoc($result)){ echo '<table>'; //CATEGORY if (isset($_POST['category'])) { echo '<tr><td><span class="equipmentheader">Category:</span> '; echo $row['category']; echo '</td></tr>'; } else { echo ''; } //YEAR if (isset($_POST['year'])) { echo '<tr><td><span class="equipmentheader">Year:</span> '; echo $row['year']; echo '</td></tr>'; } else { echo ''; } //SERIAL NUMBER if (isset($_POST['serialnum'])) { echo '<tr><td><span class="equipmentheader">Serial Number:</span> '; echo $row['serialnum']; echo '</td></tr>'; } else { echo ''; }

Yes I did, I had gotten a solution from this forum: http://codingforums.com/showthread.php?t=160693

I have a form that allows the user to upload multiple images (5 max). This code works for me for uploading one image, but it doesn't work now that I'm trying to upload multiple images. It uploads the file names to the MySQL table perfectly, but it doesn't upload the image to the server. Can anyone tell me where I've gone wrong -- thanks!!! I'm assuming its something small I'm missing since it work for uploading just one image. Here is my form: <form action="nextpage.php" method="post" enctype="multipart/form-data"> <table> <tr> <td valign="top">Upload Image: </td> <td> <input type="hidden" name="MAX_FILE_SIZE" value="2000000"><input name="photo" type="file" size="30"><br /> <input type="hidden" name="MAX_FILE_SIZE" value="2000000"><input name="photo2" type="file" size="30"><br /> <input type="hidden" name="MAX_FILE_SIZE" value="2000000"><input name="photo3" type="file" size="30"><br /> <input type="hidden" name="MAX_FILE_SIZE" value="2000000"><input name="photo4" type="file" size="30"><br /> <input type="hidden" name="MAX_FILE_SIZE" value="2000000"><input name="photo5" type="file" size="30"></td> </tr> </table> Here is the nextpage.php code: <?php $photo=($_FILES['photo']['name']); $photo2=($_FILES['photo2']['name']); $photo3=($_FILES['photo3']['name']); $photo4=($_FILES['photo4']['name']); $photo5=($_FILES['photo5']['name']); //This is the directory where images will be saved $target = "images/uploaded/"; $target = $target . basename( $_FILES['photo']['name']['photo2']['name']['photo3']['name']['photo4']['name']['photo5']['name']); //Writes the photo to the server if(move_uploaded_file($_FILES['photo']['tmp_name']['photo2']['tmp_name']['photo3']['tmp_name']['photo4']['tmp_name']['photo5']['tmp_name'], $target)) { //Tells you if its all ok echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory"; } else { //Gives and error if its not echo "Sorry, there was a problem uploading your file."; } $database info mysql_connect("$hostname", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $sql="INSERT INTO `Listings` (`photo`, `photo2`, `photo3`, `photo4`, `photo5`) VALUES('".$photo."', '".$photo2."', '".$photo3."', '".$photo4."', '".$photo5."')"; $result=mysql_query($sql);

That worked perfectly! Thank you! But for some reason though it doesn't work with the site I'm trying to link....bummer! Must be something to do with the site, because it works great with any other site.

That loads, but when a link is clicked it goes to a 404 again. The url is then: http://www.myaddress.com/theendoftheexternaladdress

I'm not very good at PHP, so I'm sorry if I have this wrong! This is what i have for my script now: <?php function get_data($url) { $ch = curl_init(); $timeout = 5; curl_setopt($ch,CURLOPT_URL,$url); curl_setopt($ch,CURLOPT_RETURNTRANSFER,1); curl_setopt($ch,CURLOPT_CONNECTTIMEOUT,$timeout); $data = curl_exec($ch); curl_close($ch); return $data; } $returned_content = get_data('http://www.externalsitehere.com'); echo str_replace("</head>", '<base href="http://www.yoursite.com/script.php?url=http://www.externalsitehere.com"></head>', $returned_content); ?> And it says address not found.

Thanks for the quick response! That did allow me to click a link, but once I click the link the content doesn't remain inside the div, but just opens up the external page.

I'm running this script inside a div. What I'm trying to accomplish is loading content from an external site, and also styling it from my own style sheet. It loads and styles the content perfectly, but if I click a link inside the external page, it takes me to a 404 page. Is there anyway to navigate through the external site? The reason I need this is because I'm loading an equipment list for my client, and there is lots of click-through. Here is the code: <div style="border:1px solid black; color:red" id="content area"> <?php function get_data($url) { $ch = curl_init(); $timeout = 5; curl_setopt($ch,CURLOPT_URL,$url); curl_setopt($ch,CURLOPT_RETURNTRANSFER,1); curl_setopt($ch,CURLOPT_CONNECTTIMEOUT,$timeout); $data = curl_exec($ch); curl_close($ch); return $data; } $returned_content = get_data('http://www.externalsitehere.com'); echo $returned_content; ?> </div> Any help appreciated!