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PHP - Help: Copying Files Using Results From A Db
Ok fellas....this one has me stumped.
I am trying to copy a whole bunch of pictures from one directory into a new directory. However, the new directory will store the images based upon the user's id. Initially I stored all the images in one directory. HUGE mistake.. That single directory is over 5GB and takes and eternity to open and edit the files for 1 single user. So I need to separate them out. Anyways, here is what I am trying to do in theory. All the information about the user and the file they uploaded is stored in my database. But how do I create a script that is somewhat dynamic? One that will refresh the copy file script for each result. Here is the general idea....or a conversation starter code i am working with so far. Code: [Select] <?php include 'db_connect.php'; $userId=2; // Make the directory that will store the users images with the desired folder structure $structure = './images/' .$userId. ''; // create the directory and set permissions if (!mkdir($structure, 0777, true)) { die('Failed to create folder'); chmod($structure, 0777); } //Query the database for the images $query_image = "SELECT * FROM images WHERE angler=$userId ORDER BY submit_id DESC"; $image_result = mysql_query($query_image) or die(mysql_error()); // get count of how many rows in case we need that info $rowCount = mysql_num_rows($image_result); //get the results into an array by setting each result as a variable while($row = mysql_fetch_assoc($image_result)) $galleryresults[]= $row; //now move files $file = './submitted_pics/{$galleryresults[0]['image']}'; $newfile = './images/' .$userId. '/{$galleryresults[0]['image']}'; if (!copy($file, $newfile)) { echo "failed to copy $file...\n"; } ?> Similar TutorialsI want to copy everything in templates/blue to the folder code/ However: shell_exec("cp -r 'templates/blue' 'code'"); Creates a folder called blue inside code. I tried cp -r 'templates/blue/*' 'code', but that didn't do anything. Any ideas? What I'm trying to do is copy all files from one server to another folder on another server. Here is what I have have so far.. <?PHP //connection settings $ftp_server = "server"; $ftp_user_name = "user"; $ftp_user_pass = "pass"; $dir = "/var/test/"; $destination_file = "/test/"; // set up basic connection $conn_id = ftp_connect($ftp_server); // login with username and password $login_result = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass); // check connection if ((!$conn_id) || (!$login_result)) { echo "FTP connection has failed!"; echo "Attempted to connect to $ftp_server for user $ftp_user_name"; exit; } else { echo "Connected to $ftp_server, for user $ftp_user_name"; } if (ftp_chdir($conn_id, $dir)) { echo " <br/>Current directory is now: " . ftp_pwd($conn_id) . "\n<p/>"; } else { echo "Couldn't change directory\n<p/>"; } $buff = ftp_rawlist($conn_id, $dir); foreach($buff as $files) { echo $files. "<br/>"; if (ftp_get($conn_id, $destination_file."test.file", $dir."test.txt", FTP_BINARY)) { echo "<br/>Successfully written to $destination_file\n"; } else { echo "There was a problem\n"; } ?> That doesn't work. Any ideas? Thanks, Sean Hi all, Maybe someone can help me with this, I am looking to copy specific content from one website to another automatically. So when ever one website updates the certain content the other site is automatically updated with it to. I have tried using file_get_contents but this wasnt working. Does anyone have any ideas? Im looking to copy https://annablais.scentsy.us/Home starting at line 270, the end is determined by how many Parties there are. Thanks all Ok, I would just like some advice on how to go about doing something. A website that alot of my friends use to make their builds in-game goes down frequently causing them not to be able to pull up their build information. I would like to be able to use php to copy not only the html source, but the images as well to my websvr to be able to view when their site is offline. Now I know I could simply just click page -> save as complete html in internet explorer and then upload that... However I don't want to have to do that for everyone stuff.. its time consuming.. I was hoping to automate the process with a script. I know how to get the html source from a website, but I don't know how to pull the content "pictures ect." to make it actually look like their site and have more than just text. I'll appreciate any advice you can give, I think some simple direction should be enough to get me going. I'm trying to compare 2 files, each with data separated by ::. I'm trying to make a comparison that if the data to the left of the separator from one file is not found in the same column of the other file, then it is copied to an array. For example: file1.txt abc::def ggg::hhh mmm::ppp file2.txt abc::def zzz::aaa bbb::ccc So... ggg::hhh mmm::ppp is copied from file1.txt because ggg and mmm was not found on the same side in file2.txt ...but I'm not getting any error message with this. The operation just hangs. Code: [Select] <?php $dic1 = file('file1.txt', FILE_IGNORE_NEW_LINES | FILE_SKIP_EMPTY_LINES); $dic2 = file('file2.txt', FILE_IGNORE_NEW_LINES | FILE_SKIP_EMPTY_LINES); $titleText = array(); // Storing results into this array. foreach ($dic1 as $dic1line){ list($dic1field1, $dic1field2) = explode('::', $dic1line); foreach ($dic2 as $dic2line){ list($dic2field1, $dic2field2) = explode('::', $dic2line); if ($dic1field1 != $dic2field1) { // Found a non match in DB. array_push($titleText, "$dic1field1 :: $dic1field2\n"); // Store all finds in array. } } } // Finish outputting anything left over. if (empty($titleText)) { // $value was not found -- array is empty. echo 'All matched, or something else.'; } else { $arrayOut = implode("", $titleText); // output the results found in the search. echo $arrayOut; } unset($value); ?> Anyone know how to do this? Thanks. I'm trying to allow a user to upload an image, and have that image copied to an Image folder on the server When I run the code I get the success message near the bottom of the code, but the image never gets copied to the folder. I'm not sure what's causing it to not copy the image if($_POST["store"] != "") // if upload was hit { $fileExt = strrchr($_FILES['userfile']['name'], "."); // grab extension if($fileExt != ".jpg" && $fileExt != ".jpeg") // check extension { $_SESSION["badFileType"] = "You cannot upload a file of type ".$fileExt; // set error } else { $fileName = $_FILES['userfile']['name']; // set file name if(!is_uploaded_file($_FILES['userfile']['tmp_name'])) { echo "Problem: possible file upload attack!"; // set error exit(); // end } $upfile = "../Images/lg_".$category.$counter.".jpg"; // set path $newFileName = $category.$counter.".jpg"; // set new file name if(!copy($_FILES['userfile']['tmp_name'], $upfile)) // if not able to copy { echo "Problem: file could not be copied to directory!"; // set error exit(); // end } $_SESSION["badfileType"] = "File uploaded successfully!"; // set success message } } else { $_SESSION["badFileType"] = ""; // clear error } I'm using copy() to copy an image to a different directory. The problem is the image could be in 1 of 3 folders. With out knowing which folder its in how can I do this? This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=313840.0 I've been developing a php application that runs my entire company for the last 4 years. One of the things I never thought of until now is that the server guys or anyone else could copy the source code and db and be able to start up another company which brings up my question to you.... How would you protect your application? My thought is to create one small php file that is encrypted with something that is required to make the entire site run (not sure at this point what it would be that they couldn't just rebuild). Then if this file sees it's on a different domain/ip it requests data from my site which logs the info for me to look at. If I find out it's something not approved, it would then not allow the program to run and will give a error. What is your idea? So far I have managed to create an upload process which uploads a picture, updates the database on file location and then tries to upload the db a 2nd time to update the Thumbnails file location (i tried updating the thumbnails location in one go and for some reason this causes failure) But the main problem is that it doesn't upload some files Here is my upload.php
<?php
include 'dbconnect.php';
$statusMsg = '';
$Title = $conn -> real_escape_string($_POST['Title']) ;
$BodyText = $conn -> real_escape_string($_POST['ThreadBody']) ;
// File upload path
$targetDir = "upload/";
$fileName = basename($_FILES["file"]["name"]);
$targetFilePath = $targetDir . $fileName;
$fileType = pathinfo($targetFilePath,PATHINFO_EXTENSION);
$Thumbnail = "upload/Thumbnails/'$fileName'";
if(isset($_POST["submit"]) && !empty($_FILES["file"]["name"])){
// Allow certain file formats
$allowTypes = array('jpg','png','jpeg','gif','pdf', "webm", "mp4");
if(in_array($fileType, $allowTypes)){
// Upload file to server
if(move_uploaded_file($_FILES["file"]["tmp_name"], $targetFilePath)){
// Insert image file name into database
$insert = $conn->query("INSERT into Threads (Title, ThreadBody, filename) VALUES ('$Title', '$BodyText', '$fileName')");
if($insert){
$statusMsg = "The file ".$fileName. " has been uploaded successfully.";
$targetFilePathArg = escapeshellarg($targetFilePath);
$output=null;
$retval=null;
//exec("convert $targetFilePathArg -resize 300x200 ./upload/Thumbnails/'$fileName'", $output, $retval);
exec("convert $targetFilePathArg -resize 200x200 $Thumbnail", $output, $retval);
echo "REturned with status $retval and output:\n" ;
if ($retval == null) {
echo "Retval is null\n" ;
echo "Thumbnail equals $Thumbnail\n" ;
}
}else{
$statusMsg = "File upload failed, please try again.";
}
}else{
$statusMsg = "Sorry, there was an error uploading your file.";
}
}else{
$statusMsg = 'Sorry, only JPG, JPEG, PNG, GIF, mp4, webm & PDF files are allowed to upload.';
}
}else{
$statusMsg = 'Please select a file to upload.';
}
//Update SQL db by setting the thumbnail column to equal $Thumbnail
$update = $conn->query("update Threads set thumbnail = '$Thumbnail' where filename = '$fileName'");
if($update){
$statusMsg = "Updated the thumbnail to sql correctly.";
echo $statusMsg ; }
else
{ echo "\n Failed to update Thumbnail. Thumbnail equals $Thumbnail" ; }
// Display status message
echo $statusMsg;
?>
And this does work on most files however it is not working on a 9.9mb png file which is named "test.png" I tested on another 3.3 mb gif file and that failed too? For some reason it returns the following Updated the thumbnail to sql correctly.Updated the thumbnail to sql correctly. Whereas on the files it works on it returns REturned with status 0 and output: Retval is null Thumbnail equals upload/Thumbnails/'rainbow-trh-stache.gif' Failed to update Thumbnail. Thumbnail equals upload/Thumbnails/'rainbow-trh-stache.gif'The file rainbow-trh-stache.gif has been uploaded successfully. Any idea on why this is? Hello I have a simple question about file handling... Is it possible to list all files in directories / subdirectories, and then read ALL files in those dirs, and put the content of their file into an array? Like this: array: [SomePath/test.php] = "All In this php file is being read by a new smart function!"; [SomePath/Weird/hello.txt = "Hello world. This is me and im just trying to get some help!";and so on, until no further files exists in that rootdir. All my attempts went totally crazy and none of them works... therefore i need to ask you for help. Do you have any ideas how to do this? If so, how can I be able to do it? Thanks in Advance, pros I am using WPSQT plugin in my blog site .I code some files in PHP also.how to add that files in plugin files.
Hi guys im in the middle of optimizing code.. Code: [Select] $sql = "SELECT * FROM sa_enemystats WHERE username='$username'"; $res = mysql_query($sql); while ($row = mysql_fetch_assoc($res)) { $enemy['current_health'] = $row['current_health']; $enemy['current_skill'] = $row['current_skill']; $enemy['level'] = $row['level']; $enemy['damage'] = $row['damage']; $enemy['evade'] = $row['evade']; $enemy['accuracy'] = $row['accuracy']; $enemy['speed'] = $row['speed']; $enemy['luck'] = $row['luck']; echo 'debug: variables synced with db table'; }I know that the setting of these variables are messy and can be done in a better way... but how? I have tried foreach and copying other's code for an array copy but it lead no where. Then again my syntax could be wrong... I used something like Code: [Select] foreach ( $enemy[$value] as $row => $value) { $enemy[$value] = $row[$value]; } I am testing live and the data is downloading to the database but when it goes to the next page it gets this message. Warning: Wrong parameter count for mysql_result() in /home/ebermy5/public_html/html/login.php on line 25 Could not execute query THIS IS MY CODING ANY IDEAS WHERE I AM GOING WRONG. YES I HAVE THE OPEN AND CLOSE PHP? if (@$_SESSION['auth'] != "yes") @include('Connections/connect_to_mysql.php'); $query = "SELECT firstName, lastName FROM `Members` WHERE email='{$_SESSION['id']}'"; $result = mysql_result($query) or die("Could not execute query"); echo "<html> <head><title>New Member Welcome</title></head> <body> <h2 style='margin-top: .7in; text-align: center'> Welcome $firstName </h2>\n"; hi all, Okay, after successfully getting a form to input data to a MySQL database, I'm now trying to get the data back out by searching. Search form code: Code: [Select] <form name="search" method="post" action="process_search.php"> Seach for: <input type="text" name="find" /> in <select name="columns"> <option value="">Please choose one:</option> <option value="status">Status</option> <option value="date">Date</option> <option value="species">Species</option> <option value="breed">Breed</option> <option value="sex">Sex</option> <option value="primary_colour">Primary Colour</option> <option value="colour">Colour</option> <option value="distinctive_traits">Distinctive Traits</option> <option value="fur_length">Fur Length</option> <option value="age">Age</option> <option value="desexed">Desexed</option> <option value="microchipped">Microchipped</option> <option value="suburb">Suburb</option> <option value="pound_area">Pound Area</option> <option value="contact">Contact</option> <option value="link">Link</option> </Select> <input type="hidden" name="searching" value="yes" /> <input type="submit" name="search" value="Search" /> </form> All good there! Form displays how I want it to, and submits correctly. The bit I'm struggling with is the process_search.php side of things. This is what I have (note that i'm trying to get the results into a formatted table and I'd like more results if the match is in more than one column): Code: [Select] <? //Here we display stuff if they have submitted the form if ($searching =="yes") { echo "<h2>Results</h2><p>"; //If they stuffed up and didn't search for anything, we show them this if ($find == "") { echo "<p>Oh, Bianca, you need to enter SOMETHING to search!"; exit; } //If everything is all good, we connect to the database mysql_connect("localhost", "******", "******") or die(mysql_error()); mysql_select_db("******") or die(mysql_error()); //Let's not forget the register globals off crap $status = $_POST['status']; $date = $_POST['date']; $species = $_POST['species']; $breed = $_POST['breed']; $sex = $_POST['sex']; $primary_colour = $_POST['primary_colour']; $colour = $_POST['colour']; $distinctive_traits = $_POST['distinctive_traits']; $fur_length = $_POST['fur_length']; $age = $_POST['age']; $desexed = $_POST['desexed']; $microchipped = $_POST['microchipped']; $suburb = $_POST['suburb']; $pound_area = $_POST['pound_area']; $contact = $_POST['contact']; $link = $_POST['link']; $columns = $_POST['columns']; // We perform a bit of filtering $find = strtoupper($find); $find = strip_tags($find); $find = trim ($find); //Now we search for our search term, in the field the user specified $data = mysql_query("SELECT * FROM animal_info WHERE upper($columns) LIKE'%$find%'"); //And we display the results where($result = mysql_fetch_array( $data )) { echo "<tr>"; echo "<td>".$row['status']."</td>"; echo "<td>".$row['date']."</td>"; echo "<td>".$row['species']."</td>"; echo "<td>".$row['breed']."</td>"; echo "<td>".$row['sex']."</td>"; echo "<td>".$row['primary_colour']."</td>"; echo "<td>".$row['colour']."</td>"; echo "<td>".$row['distinctive_traits']."</td>"; echo "<td>".$row['fur_length']."</td>"; echo "<td>".$row['age']."</td>"; echo "<td>".$row['desexed']."</td>"; echo "<td>".$row['microchipped']."</td>"; echo "<td>".$row['suburb']."</td>"; echo "<td>".$row['pound_area']."</td>"; echo "<td>".$row['contact']."</td>"; echo "<td>".$row['link']."</td>"; echo "</tr>"; } else { echo "ERROR: ".mysql_error(); } //This counts the number or results - and if there wasn't any it gives them a little message explaining that $anymatches=mysql_num_rows($data); if ($anymatches == 0) { echo "Nope, couldn't find anything here! Maybe refine your search criteria?<br><br>"; } //And we remind them what they searched for echo "<b>Searched For:</b> " .$find; } ?> Trouble with this is that I'm not getting any errors OR results! ANY help would be appreciated HUGELY! Cheers, Dave What I expect: - The function to loop output all of the rows which match the search results. The problems: - Only 1 row is matched. - Notice: Undefined index: id in G:\xampp\htdocs\xampp\dsa\search_func.php on line 57 - The image is not showing Line 57: Code: [Select] $id = $_REQUEST['id']; Entire Code: Code: [Select] <?php /* return the details of an employee parameter empno - employee number eg empHTML.php?empno=7521 This is very basic e.g no error checking */ function search($dbc, $id) { if (isset($_GET['terms']) && ($_GET['terms'] != 'Search...') ) { $terms = $_GET['terms']; $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) or die *('Error connecting to MySQL server'); $query = "SELECT * FROM stick, location, identification WHERE make LIKE '%$terms%' AND stick.sid = location.lid AND identification.stickID = stick.sid AND identification.stickID = $id"; $result=mysqli_query($dbc,$query); $num_rows = mysqli_num_rows($result); $output = Array(); $image = mysqli_fetch_object($result); if ($num_rows > 0){ if($result=mysqli_query($dbc,$query)){ $output[] = '<ul>'; $output[] = '<li> Type: '.$image->make .'<br />Size: '.$image->size .$image->type .'<br />Colour: '.$image->colour . '<br /> Street Missing in: ' .$image->street.'<br />Town missing in: '.$image->town .'<br />City missing in: '.$image->city.'</li>'; $output[] = '</ul>'; } return join('',$output); } else { echo "<h3>Sorry,</h3>"; echo "<p>your search: "" .$terms. "" returned zero results</p>"; } } else { // Tell them to use the search form. echo '<p class="error">Please use the search form at the top of the window to search this site.</p>'; } } // connect $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); // get the id from the URL $id = $_REQUEST['id']; print search($dbc, $id); print "<img src='getPhoto.php?id=$id'/>"; // close the database mysqli_close($dbc); My aim is for this what if the table is empty which means no results will be retrieved in the first query then I want it to repeat the $testArray as "TBD" as well. Code: [Select] function getTop5() { $this->db->select('character1_id, character2_id, character3_id, character4_id, character5_id'); $this->db->from('site_top5'); $this->db->where('status_id', '1'); $this->db->order_by('id', 'desc'); $query = $this->db->get(); $row = $query->row(); $ids = array( $row->character1_id, $row->character2_id, $row->character3_id, $row->character4_id, $row->character5_id ); $testArray = array(); foreach ($ids as $id) { if($id !== "0") { $this->db->select('character_name'); $this->db->from('characters'); $this->db->where('id', $id); $query = $this->db->get(); $row = $query->row(); $testArray[] = $row->character_name; } else { $testArray[] = "TBA"; } } return $testArray; } hi friends. to fetch results from the database i made a function and how i call the function on a page like $results = GetResults($someID); when i print_r the $results i get all the results, but now im confused on how do i print the rows individually? say i got Title row how will i only get the Title row from the $results VAR? I have a variable called clientid being passed via a url. I place that in a variable called $clientid I have two tables, client and web_info I want to select all from client and web_info for the appropriate clientid. I may not have my tables set up correctly. The primary key on the table client is auto genereated and titled client_id the primary key on web_info is e_mail and I want to pull information from the table web_info where the e_mail columns match. Below is my PHP: $query = "SELECT * FROM client WHERE $clientid = client_id and FROM web_info WHERE client.e_mail = web_info.e_mail"; $result = mysqli_query($connection,$query); $row = mysqli_fetch_array($result); I know I probably screwed all this up but please help |
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