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PHP - Need Help Displaying Imagedeleted.jpg When I Delete An Image.
i have an image uploader... that people can link directly to an image. But say for example that image gets deleted. When someone tries to link to it i want it to display a custom imagedeleted.jpg
how can i do this? Similar TutorialsHello
I am new here, and so i am to php.. i have a small question - i think anyone here could help me, im sure :-)
I have a php-gallery running locally on my desktop (using xampp), the gallery just shows images from a directory where a webcam saves its pictures in it (the gallery is named "ubergallery" - maybe anyone knows this gallery).
Now i really would like to have following option:
- A Button on the thumbnail called "delete" - when clicking on it, the picture should get replaced with another, predefined, placeholder-picture. I really hope somebody here can help me ! Thank you so much for your time !
I have attached the files!
Attached Files
UberGallery.php 37.81KB
0 downloads
index.php 2.81KB
0 downloads
index.php 663bytes
0 downloads I have a PHP CRUD application and when i delete a row i need it to delete from the MySQL and the image from the uploads folder as well. I have researched and tried dozens of ways with the unlink option but nothing works. If i take out the unlink from my code it will delete fine from the DB. I am new to coding and PHP so any help would be awesome. The file_path is correct. The uploads is the name of the folder where the image is stored and the $_POST["image"] is the column name in MySQL where the image name is stored. delete.php
<?php
//start PHP session
session_start();
if (!isset($_SESSION['success']))
{
header("Location: login_page.php");
die();
}
// check if value was posted
if($_POST){
// include database and object file
include_once 'config/database.php';
$file_path = 'uploads/' . $_POST["image"];
if(unlink($file_path))
{
// delete query
$query = "DELETE FROM myDBname WHERE id = ?";
$stmt = $con->prepare($query);
$stmt->bindParam(1, $_POST['object_id']);
}
if($stmt->execute()){
// redirect to read records page and
// tell the user record was deleted
echo "Record was deleted.";
}else{
echo "Unable to delete record.";
}
}
?>
this is the delete button code
echo "<a delete-id='{$id}' class='btn btn-danger delete-object'>";
echo "<span class='glyphicon glyphicon-remove'></span> Delete";
echo "</a>";
This is the javascript for the delete button as well.
// delete record
$(document).on('click', '.delete-object', function(){
var id = $(this).attr('delete-id');
bootbox.confirm({
message: "<h4>Are you sure?</h4>",
buttons: {
confirm: {
label: '<span class="glyphicon glyphicon-ok"></span> Yes',
className: 'btn-danger'
},
cancel: {
label: '<span class="glyphicon glyphicon-remove"></span> No',
className: 'btn-primary'
}
},
callback: function (result) {
if(result==true){
$.post('delete.php', {
object_id: id
}, function(data){
location.reload();
}).fail(function() {
alert('Unable to delete.');
});
}
}
});
return false;
});
If you need any other info that would help you help me just let me know and i will get that in here ASAP. Thanks again for any help on this. Help me. What I am doing is editing a form with checkboxes... and updating image .. I uploaded scripts I want to delete old jpeg file from thumb folder and update new filename in db Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Untitled Document</title> <script type="text/javascript"> function show_confirm() { var r=confirm("Delete?"); if (r==true) { //javascript:window.location='view.php'; alert("Deleted!!"); } else { //javascript:window.location='view.php'; alert("You pressed Cancel!"); return false; } } </script> </head> <body> <?php require("connect.php"); if ($_REQUEST['submit']) { $id=$_POST['hide']; for($i=1; $i<=$id; $i++) { $query2 = mysql_query("SELECT * from stu_detail WHERE id = $id "); while($file = mysql_fetch_array($query2)) { $unlink = unlink("up_images/thumb/" . $file['filename']); } } $new_name=$_POST['name']; $new_email=$_POST['email']; $new_phone=$_POST['phone']; $new_gender=$_POST['rad']; $new_city=$_POST['city']; $new_lang=implode(",",$_POST['chk']); $tar = "c:/wamp/www/rms/vignesh_2/up_images/"; if ($_FILES['efile']['error'] > 0) echo "Error" . $_FILES["efile"]["error"]; else { if (file_exists ("$tar" . $_FILES['efile']['name'])) { echo $_FILES['efile']['name'] . " already exsists"; } else { move_uploaded_file ($_FILES['efile']['tmp_name'],"$tar" . $_FILES['efile']['name']); } } $qry=mysql_query("UPDATE stu_detail SET name='".$new_name."', email='".$new_email."', phone='".$new_phone."', gender='".$new_gender."', city='".$new_city."', lang='".$new_lang."', filename='".$_FILES['efile']['name']."' WHERE id=$id ") or die(mysql_error()); require_once("create_thumb.php"); echo generateThumbs(); if (!$qry) { echo "not updated"; } else { echo "<table align='center' border='1' width='800' height='600'>"; ?><tr> <th scope="col">Name</th> <th scope="col">E-mail Address</th> <th scope="col">Phone</th> <th scope="col">Gender</th> <th scope="col">City</th> <th scope="col">Language</th> </tr> <?php $recyc=mysql_query("SELECT * FROM stu_detail where status = 1") or die(mysql_error()); $del = mysql_num_rows($recyc); if ($del != 0) echo "<a href='recycled.php'>Recycle bin</a>"; if ($qry2=mysql_query("SELECT * FROM stu_detail where status = 0 order by id desc") ) { while($res=mysql_fetch_array($qry2)) { ?> <tr> <td><?php echo $res['name']; ?></td> <td><?php echo $res['email']; ?></td> <td><?php echo $res['phone']; ?></td> <td><?php echo $res['gender']; ?></td> <td><?php echo $res['city']; ?></td> <td><?php echo $res['lang']; ?></td> <td> <img src="up_images/thumb/<?php echo $res['filename']; ?>" /> </td> <td> <a href="edit.php?uid=<?php echo $res['id']; ?>">edit</a></td> <!--<td><button type="button" onClick="return show_confirm();">Delete</button> </td>--> <td> <a href="delete.php?uid=<?php echo $res['id']; ?>" onClick="return show_confirm();">delete</a></td> </tr> <?php } } ?> </table> <?php } } ?> </body> </html> Hi all. Here is my scripts which allow user to check multiple rows of data and delete it , but it require select data and click for twice to delete the rows , what should be the error? Code: [Select] <form name="frmSearch" method="post" action="insert-add.php"> <table width="600" border="1"> <tr> <th width="50"> <div align="center">#</div></th> <th width="91"> <div align="center">ID </div></th> <th width="198"> <div align="center">First Name </div></th> <th width="198"> <div align="center">Last Name </div></th> <th width="250"> <div align="center">Mobile Company </div></th> <th width="100"> <div align="center">Cell </div></th> <th width="100"> <div align="center">Workphone </div></th> <th width="100"> <div align="center">Group </div></th> </tr> </form> <? echo "<form name='form1' method='post' action=''>"; while($objResult = mysql_fetch_array($objQuery)) { echo "<tr>"; echo "<td align='center'><input name=\"checkbox[]\" type=\"checkbox\" id=\"checkbox[]\" value=\"$objResult[addedrec_ID]\"></td>"; echo "<td>$objResult[addedrec_ID] </td>"; echo "<td>$objResult[FirstName]</td>"; echo "<td>$objResult[LastName] </td>"; echo "<td>$objResult[MobileCompany] </td>"; echo "<td>$objResult[Cell] </td>"; echo "<td>$objResult[WorkPhone] </td>"; echo "<td>$objResult[Custgroup] </td>"; echo "</tr>"; } echo "<td colspan='7' align='center'><input name=\"delete\" type=\"submit\" id=\"delete\" value=\"Delete\">"; if (isset($_POST['delete']) && isset($_POST['checkbox'])) // from button name="delete" { $checkbox = ($_POST['checkbox']); //from name="checkbox[]" $countCheck = count($_POST['checkbox']); for($d=0;$d<$countCheck;$d++) { $del_id = $checkbox[$d]; $sql = "DELETE from UserAddedRecord where addedrec_ID = $del_id"; $result2=mysql_query($sql) or trigger_error(mysql_error());;; } if($result2) { $fgmembersite->GetSelfScript(); } else { echo "Error: ".mysql_error(); } } echo "</form>"; Thanks for every reply. <body> <?php
include 'sql.php';
$query = "SELECT * FROM validation";
$result = mysqli_query($con , $query);
$rows = mysqli_fetch_assoc($result) ;
$totals = mysqli_num_rows($result) ;
?>
<div id="css">
<form >
<table width="80%" border="0" cellpadding="2" cellspacing="2" >
<caption><h2>Personal Details of Customers</h2></caption>
<tr class="white">
<td bgcolor="#330033"> </td>
<td bgcolor="#330033"> Id Number </td>
<td bgcolor="#330033"> Full Name </td>
<td bgcolor="#330033"> Email Address </td>
<td bgcolor="#330033"> Website </td>
<td bgcolor="#330033"> Comment </td>
<td bgcolor="#330033"> Time </td>
</tr>
<?php while($rows=mysqli_fetch_assoc($result)
{
<tr>
<input type="raido" name="ID" value="<?php echo $rows['ID']; ?>" />
<td bgcolor="#FFFFCC"><?php echo $rows['ID'];?></td>
<td bgcolor="#FFFFCC"><?php echo $rows['Name'];?> </td>
<td bgcolor="#FFFFCC"><?php echo $rows['Email'];?></td>
<td bgcolor="#FFFFCC"><?php echo $rows['Website'];?></td>
<td bgcolor="#FFFFCC"><?php echo $rows['Comment'];?></td>
<td bgcolor="#FFFFCC"><?php echo $rows['Time'];?></td>
<td> </td>
<td> <a href="delete.php? ID= "$rows[ID]" /"> <input type="submit" name="del" value="Delete" /> </a> <input type="button" name= "edit" value="Edit" /> </td>
</tr>
}
?></table> </form> </div> </body> Hi,
I wish to find out is there any possible that I can delete some data inside my php website but inside my sql database, the record will still at there?
Thank you.
I am using this code Code: [Select] <?php echo IMAGES_HEADER . "header_02.jpg"; ?> Instead of displaying the actual image on my site I am getting the path of the image. It is displaying "images/header/header_02.jpg" instead. Thank You in advance i have a db which store jpg image thru the upload prg code in php.But the image is not displayed properly in the <img> and also when i echo the blob data. how to correct this code.i am pasting the code below $result = mysql_query("select cover from Movies where movie_id=4"); $row = mysql_fetch_row($result); $data = base64_decode($row[0]); $im = imagecreatefromstring($row[0]); imagejpeg($im); header('Content-type: ' . $mime); // 'image/jpeg' for JPEG images echo $data; <img src=<?php echo $data;?>> Hi guys, I have followed a tutorial and made a members only area using sessions. The user can upload an image and which gets renamed as their username. I was hoping to display all the users images that are logged in. I know how to do it with a single image by just setting the img src as the session username but I don't know how I would display multiple images if more than one person were logged in. Is it even possible? I must be doing something incredibly daft, because I'm incredibly new at this. I have an image stored in a DB under a table called 'images' and I want to display it on my website but instead of that image I get the error: Warning: Cannot modify header information - headers already sent by (output started at /home/... This is how I'm trying to achieve it. Any ideas where I'm doing wrong? Thanks. Code: [Select] <?php $user="###"; $password="###"; $database="###"; $con = mysql_connect(localhost,$user,$password); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db($database, $con); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>MySite</title> </head> <body> <div id="container"> <?php include("../navbar.php"); ?> <div id="left-content"></div> <div id="right-content"> <?php $item = $_GET['item']; $query = "SELECT * FROM main WHERE Ref='$item'"; $result = mysql_query($query) or die("Oops" .mysql_error()); $row = mysql_fetch_array($result,MYSQL_BOTH) or die("Oops" .mysql_error()); extract($row); $query2 = "SELECT image FROM main WHERE Ref='$item'"; // the result of the query $result2 = mysql_query($query2) or die("Invalid query: " . mysql_error()); header("Content-type: image/jpg"); echo mysql_result($result2, 0,'image'); echo "<p><strong>Name: </strong>".$FirstName." - ".$SecondName."</p>"; mysql_close($con); ?> </div> <br /> <?php include("../footer.php"); ?> </div> </div> </body> </html> MOD EDIT: [code] . . . [/code] BBCode tags added. Hi everyone, i am just trying to learn php for a bit of fun really and started making a sort of 'facebook' website. I am having trouble however trying to display different users images, for example when trying to find a correct 'friend' only the image of the last result is being shown for all people with the same name... here is my code below, if anyone can help me out that would be great file 1 $count=1; while ($numids>=$count){ echo "<form method=\"post\" action=\"friendadded.php\">"; $frienduserid=$_SESSION["passedid[$ii]"]; $friendfirstname=$_SESSION["passedfirstname[$ff]"]; $friendlastname=$_SESSION["passedlastname[$ll]"]; $_SESSION['friendsuserpicid'] = $frienduserid; echo "<table width=\"700\" height=\"50\" border=\"1\" align=\"center\">"; echo "<tr>"; echo "<th></th>"; echo "<th>First Name</th>"; echo "<th>Last Name</th>"; echo "</tr>"; echo "<tr>"; echo "<td><center>"; echo "<img border=\'0\' src=\"frienduserpic.php\" width=\"80\" height=\"80\" align=\"middle\"/>"; echo "</center></td>"; echo "<td><center>"; echo $friendfirstname; echo "</center></td>"; echo "<td><center>"; echo $friendlastname; echo "</center></td>"; echo "</tr>"; echo "</table>"; echo "<center><input type=\"submit\" value=\"Add this friend\" name=\"Add Friend\"></center><br/>"; $ii=$ii+1; $ff=$ff+1; $ll=$ll+1; $count=$count+1; echo "</form>"; } file 2 session_start(); $passeduserid=$_SESSION['friendsuserpicid']; $timespost=$_SESSION['postednum']; $host= $username= $password= $db_name= $tbl_name= mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $query = mysql_query("SELECT * FROM $tbl_name WHERE picid='".$passeduserid."'"); $row = mysql_fetch_array($query); $content = $row['image']; header("Content-type: image/jpeg"); echo $content; Thanks in advance Now I am trying to display the images from my table and I have almost got it working, except for one small thing --- it seems to be displaying all the records, but instead of displaying the right image for each record, it's displaying the same image across all the records. Can anyone tell me what I have done wrong? Code: [Select] this is the gallery <?php $dbhost = 'localhost'; $dbuser = 'webdes17_lizkula'; $dbpass = 'minimoon'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'webdes17_jewelry'; mysql_select_db($dbname); $all_records = "SELECT * FROM gallery"; $all_records_res = mysql_query($all_records); $image = mysql_result($all_records_res, 0, 'image'); while($nt=mysql_fetch_array($all_records_res)){ echo "<img src=http://www.webdesignsbyliz.com/wdbl_wordpress/wp-content/themes/twentyten_2/upload/$image>"; } ?> I'm guessing I have the $image variable in the wrong place, but when I tried placing it within the while statement, the page never loaded, and instead acted like it was loading forever. What am I doing wrong? Here is the link to the page so you can see what is happening: http://webdesignsbyliz.com/wdbl_wordpress/test-submit/ I just can't get anything right today! Now I am trying to have a list of images, and when the user clicks on the image, the next page will display the image along with other fields for that record. I am sending the id through the hyperlink to the next page, and I have echoed it to ensure it's comign through, but I cannot get anythign to display ont he next page. What am I doin wrong? Here is the link to the page. If you click on one of the images, you 'll see that the next page is empty: http://webdesignsbyliz.com/wdbl_wordpress/test-submit/ Here is my code: Code: [Select] this is the gallery <?php $dbhost = 'localhost'; $dbuser = 'user'; $dbpass = 'pass'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'jewelry'; mysql_select_db($dbname); $all_records = "SELECT * FROM gallery"; $all_records_res = mysql_query($all_records); $image = mysql_result($all_records_res, 0, 'image'); $id = mysql_result($all_records_res, 0, 'id'); while($nt=mysql_fetch_array($all_records_res)){ echo "<a href=http://webdesignsbyliz.com/wdbl_wordpress/test-display/?id=" .$nt['id']." ><img src=http://www.webdesignsbyliz.com/wdbl_wordpress/wp-content/themes/twentyten_2/upload/".$nt['image']." width=133 height=86></a>"; } ?> display page: Code: [Select] <?php $id = $_GET['id']; $dbhost = 'localhost'; $dbuser = 'user'; $dbpass = 'pass'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'jewelry'; mysql_select_db($dbname); $all_records = "SELECT * FROM gallery WHERE id = $_GET[id]"; $all_records_res = mysql_query($all_records); $image = mysql_result($all_records_res, 0, 'image'); $id = mysql_result($all_records_res, 0, 'id'); while($nt=mysql_fetch_array($all_records_res)){ echo "<img src=http://www.webdesignsbyliz.com/wdbl_wordpress/wp-content/themes/twentyten_2/upload/".$nt['image']." width=133 height=86></a>"; } ?> What an I doing wrong this time?? :-( I am having trouble with displaying an image from a url. In the following code if I echo out $test I get the url of the image. However, in the current code I am trying to display the image, but the only thing that gets displayed is a small broken image in the upper left.
$html = file_get_contents($website.$criteria);
$dom = new DOMDocument;
@$dom->loadHTML($html);
$links = $dom->getElementsByTagName('img');
header('Content-Type: image/png');
foreach ($links as $link){
$test = $link->getAttribute('src');
echo file_get_contents($test);
}
image that gets displayed: http://imgur.com/epfF3wJ
Hi everyone, Sorry not sure if this is a php or html problem. Im using php and a html form to upload images to my site, i have made it so that jpg, jpeg, gif and png images can be uploaded. The problem im having is displaying the image. Code: [Select] <img src="images/<?php echo $name ?>.jpg" /> That works fine if a jpg was uploaded, but what if a png or a gif was uploaded? The $name is going to be unique, there will not be more than one image with the same name, so what do i have to do to display the image regardless of what the extension is? Thanks Hi all, I have this script below where I am trying to display a default image if an image can not be found. For some reason though it is not working. <?php foreach (glob('./aircraft/' . $rowX['reg'] . '[0-8].jpg') as $file) { if (file_exists($file)) { echo "<img src=\"" . $file . "\" /><br />"; } else { echo "><img src=\"aircraft/wrightflyer.jpg\" /><br />";} } ?> I've taken this eg from php manual site for displaying txt on image wonder why it is not working Code: [Select] <?php // Create a 300x150 image $im = imagecreatetruecolor(300, 150); $black = imagecolorallocate($im, 0, 0, 0); $white = imagecolorallocate($im, 255, 255, 255); // Set the background to be white imagefilledrectangle($im, 0, 0, 299, 299, $white); // Path to our font file $font = './arial.ttf'; // First we create our bounding box for the first text $bbox = imagettfbbox(10, 45, $font, 'Powered by PHP ' . phpversion()); // This is our cordinates for X and Y $x = $bbox[0] + (imagesx($im) / 2) - ($bbox[4] / 2) - 25; $y = $bbox[1] + (imagesy($im) / 2) - ($bbox[5] / 2) - 5; // Write it imagettftext($im, 10, 45, $x, $y, $black, $font, 'Powered by PHP ' . phpversion()); // Create the next bounding box for the second text $bbox = imagettfbbox(10, 45, $font, 'and Zend Engine ' . zend_version()); // Set the cordinates so its next to the first text $x = $bbox[0] + (imagesx($im) / 2) - ($bbox[4] / 2) + 10; $y = $bbox[1] + (imagesy($im) / 2) - ($bbox[5] / 2) - 5; // Write it imagettftext($im, 10, 45, $x, $y, $black, $font, 'and Zend Engine ' . zend_version()); // Output to browser header('Content-Type: image/png'); imagepng($im); imagedestroy($im); ?> Good day: Im trying to display an image from a folder and the image name is retrieved by a query. The query is getting the image name all right but the image is not displaying. This is the code im using for the image display Code: [Select] <?php $aid = 1; $connection = mysql_connect("localhost", "username", "password"); mysql_select_db("articles", $connection); $query="SELECT imagename, description FROM articles_description WHERE id='$aid'"; $result=mysql_query($query); $num=mysql_numrows($result); mysql_close(); ?> <table width ="1000" border="1" cellspacing="2" cellpadding="2"> </tr> <?php $j=0; while ($j < $num) { $f8=mysql_result($result,$i,"description"); $f9=mysql_result($result,$i,"imagename") ?> <tr> <td valign="top"> <img src="images/'.$f9.'; ?>" alt="" name="picture" width="100" height="100" border="1" /></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f8; ?></font></td> </tr> <tr> </tr> <?php $j++; } ?> Any help will be appreciated Hi, I am uploading an image to a folder and I need to display the uploaded image in a new page. This is my code for the upload: Code: [Select] <?php //define a maxim size for the uploaded images in Kb define ("MAX_SIZE","2000000"); //This function reads the extension of the file. It is used to determine if the // file is an image by checking the extension. function getExtension($str) { $i = strrpos($str,"."); if (!$i) { return ""; } $l = strlen($str) - $i; $ext = substr($str,$i+1,$l); return $ext; } //This variable is used as a flag. The value is initialized with 0 (meaning no // error found) //and it will be changed to 1 if an errro occures. //If the error occures the file will not be uploaded. $errors=0; //checks if the form has been submitted if(isset($_POST['upload'])) { //reads the name of the file the user submitted for uploading $image=$_FILES['image']['name']; //if it is not empty if ($image) { //get the original name of the file from the clients machine $filename = stripslashes($_FILES['image']['name']); //get the extension of the file in a lower case format $extension = getExtension($filename); $extension = strtolower($extension); //if it is not a known extension, we will suppose it is an error and // will not upload the file, //otherwise we will do more tests if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { //print error message echo '<h1>Tipo de imagen no permitido.</h1>'; $errors=1; } else { //get the size of the image in bytes //$_FILES['image']['tmp_name'] is the temporary filename of the file //in which the uploaded file was stored on the server $size=filesize($_FILES['image']['tmp_name']); //compare the size with the maxim size we defined and print error if bigger if ($size > MAX_SIZE*1024) { echo '<h1>La imagen es demasiado grande.</h1>'; $errors=1; } //we will give an unique name, for example the time in unix time format $image_name=time().'.'.$extension; //the new name will be containing the full path where will be stored (images //folder) $newname="banners/".$image_name; //we verify if the image has been uploaded, and print error instead $copied = copy($_FILES['image']['tmp_name'], $newname); if (!$copied) { echo '<h1>Error al subir la imagen.</h1>'; $errors=1; }}} //If no errors registred, print the success message if(isset($_POST['Submit']) && !$errors) { echo "<h1>La imagen se ha cargado correctamente</h1>"; } $query = "INSERT INTO banner_rotator.t_banners (banner_path) ". "VALUES ('$newname')"; mysql_query($query) or die('Error, query failed : ' . mysql_error()); //echo "<br>Files uploaded<br>"; header("Location: PC_cropbanner.php"); }Sowhat I need to do is to display the uploaded image in PC_cropbanner.php. How can I do this? I'm trying to set it so that it will delete an entire populated directory based upon a value in the database then after finishing that to go back and delete that row in the database. my current code is Code: [Select] <?php $page_title = "Central Valley LLC | Photo Addition" ?> <?php include("header.php"); ?> <?php include("nav.html"); ?> <div id="content"> <form action="delprod.php" method="post" enctype="multipart/form-data"> <label for="which">Choose A Product To Remove:</label> <?php $con = mysql_connect("localhost","phoenixi_cv","centraladmin"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("phoenixi_cvproducts", $con); $result = mysql_query("SELECT * FROM Products"); echo "<select name=\"which\">"; while($row = mysql_fetch_array($result)) { echo "<option "; echo "value=\"" . $row['id'] . "\">"; echo $row['Name'] . "</option>"; } echo "</select>"; mysql_close($con); ?> <br /> <input type="submit" name="submit" value="Submit" /> </form> </div><!--#content--> <?php include("footer.html") ?> and the delete script Code: [Select] <?php $con = mysql_connect("localhost","phoenixi_cv","centraladmin"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("phoenixi_cvproducts", $con); $result = mysql_query("SELECT id FROM Products WHERE id=$_POST['which']"); $row = mysql_fetch_array($result) chdir('assets'); chdir('images'); $mydir = $row . '/'; $d = dir($mydir); while($entry = $d->read()) { if($entry!="." && $entry!="..") { unlink($_POST['which'] . '/' . $entry); } } rmdir($mydir); $result = mysql_query("DELETE * FROM Producs WHERE id=$_POST['which']"); ?> Thank you in advance for all your help. any easier ways of approaching this will be welcome as well |
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