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PHP - Transfering From Directory Folder To Database
I have been looking around and have not been able to find any useful information on this. I am trying to transfer files from a folder in the directory to a table in a database. Eventually they will be images but for right now I am just trying to get text files to transfer. Any help/code/ideas would be a lot of help. This is going to be triggered a button on a page by the way.
Similar TutorialsI have a PHP script that reads the image files in a folder and displays them for viewing. When originally tested with three images, it seemed to be displaying them in alphabetical order by name (as desired). Now, with more image files added, it is displaying the pictures in a seemingly random order. Can I use a SORT command similar to the method for organizing data from a database in my script (to organize the files by name)?
if ($d = opendir("uploads")) {
while ($file = readdir($d)) {
if ($file != "." && $file != ".." )
// if( is_file($file) )
{
$files[] = $file;
}
}
closedir($d);
}
if ($file != "." && $file != ".." $file != ???????? )
just a bit stuck here
I've tried this code and a few others, but it only displays 0. I'd like the simplest working way to display the number of files in a directory. Thanks for the help! $directory = "../images/team/harry/"; $filecount = count(glob("" . $directory . "*.*")); echo $filecount; Okay, so I'm kind of a PHP noob.
But out of context, for this site that I'm designing, it's easiest if I make a directory in the temporary folder PHP uses. In my case, /tmp/ because I am on Linux.
I want to use rand() to generate a random name for the page. But I then realized something, rand() could produce duplicates. How do I prevent PHP from trying to make the same directory at the same time? I know there are functions that will check if a file exists but I'm assuming it'll fail if that directory is currently being created, right?
How do I assure thread safety?
I willing to change my idea and not use rand(). Is there a way to get a unique key for each anonymous user on my site?
Hi.. I've seen code for back up database but when I run the code the database was backup outside the folder. I want to put the back up database inside the folder here is the code: Code: [Select] <?php include 'config.php'; backup_tables('localhost','root','','payroll'); /* backup the db OR just a table */ function backup_tables($host,$user,$pass,$name,$tables = '*') { $link = mysql_connect($host,$user,$pass); mysql_select_db($name,$link); //get all of the tables if($tables == '*') { $tables = array(); $result = mysql_query('SHOW TABLES'); while($row = mysql_fetch_row($result)) { $tables[] = $row[0]; } } else { $tables = is_array($tables) ? $tables : explode(',',$tables); } //cycle through foreach($tables as $table) { $result = mysql_query('SELECT * FROM '.$table); $num_fields = mysql_num_fields($result); $return.= 'DROP TABLE '.$table.';'; $row2 = mysql_fetch_row(mysql_query('SHOW CREATE TABLE '.$table)); $return.= "\n\n".$row2[1].";\n\n"; for ($i = 0; $i < $num_fields; $i++) { while($row = mysql_fetch_row($result)) { $return.= 'INSERT INTO '.$table.' VALUES('; for($j=0; $j<$num_fields; $j++) { $row[$j] = addslashes($row[$j]); $row[$j] = ereg_replace("\n","\\n",$row[$j]); if (isset($row[$j])) { $return.= '"'.$row[$j].'"' ; } else { $return.= '""'; } if ($j<($num_fields-1)) { $return.= ','; } } $return.= ");\n"; } } $return.="\n\n\n"; } //save file $myfoldername = "backup_DBPayroll";//your folders name $handle = fopen(getcwd().$myfoldername.'db-backup-'.time().'-'.(md5(implode(',',$tables))).'.sql','w+'); // $handle = fopen('db-backup-'.date('m-d-Y').'-'.(md5(implode(',',$tables))).'.sql','w+'); fwrite($handle,$return); fclose($handle); } $smarty->display('header_cat.tpl'); $smarty->display('backup.tpl'); $smarty->display('footer.tpl'); ?> Thank you in advance 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. hi all i need a script that will allow me upload images to 2 dif folders on my server and then add the info to my database along with some other form data. iv been looking all over for code or scripts for days now and have been playing with cut and copyed code but no look again any help i will be greatful for as im a noob to php but al learning quick here is my html form Code: [Select] <html> <body> <form action="add.php" method="post"> Project Name: <input type="text" name="pro_name" /><br> Thumbnail: <input type="file" name="thumbnail" /><br> ////// this image to ../thum Short Details: <input type="text" name="short_details" /><br> Full Details: <input type="text" name="full_details" /><br> Category: <input type="text" name="cat" /><br> Image1: <input type="file" name="image1" /><br>//// and image1,2,3,4 to ../images Image2: <input type="file" name="image2" /><br> Image3: <input type="file" name="image3" /><br> Image4: <input type="file" name="image4" /><br> <input type="submit" /> </form></body></html> here is my code for add.php witch only adds the info to the DB Code: [Select] <?php error_reporting(E_ALL); include ("../includes/db_config.php"); $con = mysql_connect($db_hostname,$db_username,$db_password); @mysql_select_db($db_database) or die( "Unable to select database"); $sql="INSERT INTO $db_table (pro_name, thumbnail, short_details, full_details, cat, image1, image2, image3, image4) VALUES ('$_POST[pro_name]','$_POST[thumbnail]','$_POST[short_details]','$_POST[full_details]','$_POST[cat]','$_POST[image1]','$_POST[image2]','$_POST[image3]','$_POST[image4]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; mysql_close($con) ?> I am looking to resize the images (hight and width) after I have uploaded to a folder with php. What the bast way to do this. I am unable to resize them before I upload as I use the same picture in other areas. I'm using this code to view the image and other infomation on the database can some one point me in the right direction or show me how to change the code below. Thanks Code: [Select] <?php mysql_connect("localhost", "", "") or die(mysql_error()) ; mysql_select_db("stafflog") or die(mysql_error()) ; $data = mysql_query("SELECT * FROM ********") or die(mysql_error()); while($info = mysql_fetch_array( $data )) { echo "<img src=http://www.web.com/productimages/" .$info['image'] . "> <br>"; echo "<b>Name:</b> ".$info['yourname'] . "<br> "; echo "<b>price:</b> ".$info['price'] . " <br>"; echo "<b>shortdes:</b> ".$info['shortdes'] . " <hr>"; } ?> 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 $_FILES['uploadedfile']['name']). " has been uploaded"; } else{ echo "There was an error uploading the file, please try again!"; } ?> Click <a href="products.php">HERE</a> to go back to form if someone could help me i'd be very grateful Basically i have a folder with 100+ images they are NOT all the same extension, what im wanting to do is use PHP to find all the images and put them all in a database. how would i go about doing this? thanks Hey guys, i'm new to this site and would need some help with coding. So i'm making a car part website which should has brand/model search. It's a dropdown search which will get the brand / model from database and should display all the parts for that exact model, from folder. Database structure which i have is id, master, name. ( Here's some pictures to clear out what i'm doing. http://imgur.com/a/7XwVd ) So the problem is that it does not get the images from folder named ex: *_audi_a3.jpg. And link to codes what have been written already. Parts.php: http://pastebin.com/q6vdypge Update.php: http://pastebin.com/DymhGQ17 Search_images.php: http://pastebin.com/LF5Q0i8f Core.js: http://pastebin.com/bgc0y4TS I don't know what's wrong with it sadly. One thing i noticed when i used firebug it gives error on category when searching error:true. Hope you guys understand what i mean here, and also all help is appreciated. And move this post if it's in wrong place. Thanks! Edited by aeonius, 17 October 2014 - 12:15 PM. images are uploaded to a directory and the sql database contains the filename. I added 'url' field and am trying to link the two I have used on of the posts to help me adapt some script however I'm not sure I have the syntax right. The image in DW live view looks ok and I get the link finger ok but when previewed in browser the image is not shown at all or any link. Any ideas <? echo !empty($row['pic']) ? "<a href=\"{$row['url']}\" <img src=\"js/images/{$row['pic']}\">" : ''; ?> it echos out to a table Mike Hello All, I have a contact directory database. It has all the employees of my company (name, phone, email, department, building, etc). Say on one page I have the Marketing Department, and I want to say: "The Marketing Department Director is ________" How would I assign that value from the database? Do I want to put in a unique "keyword" field in the database, but then how would I store all the values automatically on the page? I see pages where I would want to list the Marketing Director, and his secretary, then another page with the Sales Director, and his secretary, etc.... all with being able to change the values in the database, and it changing across all the pages instantly. Do I need to say on every page "select * from database where keyword = marketingdirector" and then store that result as a variable? It seems unpractical to repeat that a few times for each different person I want to list. Is there a better way to do this then I'm thinking? Thanks all! if i got some homepage named index.php and got a $_GET function which calls index.php?strana=2, how do i define so my index.php or just www.example.com/folder/ is same like index.php?strana. Cuz Ive got 2 forms on my website and when i put index.php?strana - i get the first one, when i put index.php?strana=2 - i get the second one but when i put index.php only or blank as i said above im still getting second form Hello everybody!
I'm here today with a mystery i can't solve...
The problem i'm having is that i need to be able to transfer mysql data from one server to another one.
Retrieving data from the one database is easy, but making it transfer to another server is not that easy for me.
I have it printing out the data ATM
This is how i retrieve data from one server:
<?php
$dbhost = "";
$dbname = "eg_xenforo";
$dbuser = "";
$dbpass = "";
$conn = mysqli_connect("$dbhost","$dbuser","$dbpass","$dbname") or die("Error " . mysqli_error($conn));
$sql= "SELECT username
FROM xf_user
WHERE FIND_IN_SET('9',secondary_group_ids)
UNION
SELECT username
FROM xf_user
WHERE FIND_IN_SET('10',secondary_group_ids)
UNION
SELECT username
FROM xf_user
WHERE FIND_IN_SET('12',secondary_group_ids)";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
$name = $row['username'];
echo "$name <br>";
}
} else {
echo "0 results";
}
?>
This is what i get as result:
CosmonautBob ooglebrain JamieKG Quinnter16 Jordan Zoehamp skynet1123 misscupcake1306 Sir CrimsonThese are usernames that should be added onto another server in a database table. Any suggestions? Thanks in advance!
I have a form on which the filepond plugin send the file manually
Look here
if(empty($_POST['image'])) {
echo 'add file!';
}
$myimage = $_POST['image'];
$myimage = str_replace('data:image/png;base64,', '', $myimage);
$myimage = str_replace(' ', '+', $myimage);
$decode = base64_decode($myimage);
$myfile = $_SERVER['DOCUMENT_ROOT'].'/mages/' . uniqid() . '.png';
//now put the file
file_put_contents($myfile, $decode);
From the beginning, the user accesses my website and uploads an image. Currently I'm using SimpleImage.php available at http://www.white-hat-web-design.co.uk/articles/php-image-resizing.php to resize uploaded pictures. I create two copies, a regular size and a thumbnail size. Then I transfer the converted files through FTP to my storage server. The regular size works great, but the thumbnail transfers as a blank file. I don't want to save the files on the non-storage server at all, so I'm trying to work with temporary files. include('./SimpleImage.php'); // A few variables $file_name = basename($_FILES['photo1']['name']); $random_digit=mt_rand(111111111,2147483647); // New file name $new_file_name = $userInfo['id'].'_'.$random_digit.$random_digit.'_'.substr($file_name, -5); // Paths to temporary files $target_path = $_FILES['photo1']['tmp_name']; $thumb_path = tmpfile(); // Start SimpleImage and load temp file $simImage = new SimpleImage(); $simImage->load($target_path); // Get the temp image dimensions list($width, $height) = getimagesize($target_path); // Resize the file if it's too tall if ($height > $var['maxPhotoHeight']) { $simImage->resizeToHeight($var['maxPhotoHeight']); $simImage->save($target_path); } // Get the newly resized image dimensions list($width, $height) = getimagesize($target_path); // Resize the image if it's too wide if ($width > $var['maxPhotoWidth']) { $simImage->resizeToWidth($var['maxPhotoWidth']); $simImage->save($target_path); } // Resize for thumbnail width $simImage->resizeToWidth($var['thumbWidth']); $simImage->save($thumb_path); // Get thumbnail dimensions list($width, $height) = getimagesize($thumb_path); // Resize the thumbnail if it's too tall if ($height > $var['thumbHeight']) { $simImage->resizeToHeight($var['thumbHeight']); $simImage->save($thumb_path); } // Was told this was important, but I've tried with and without it and got the same result fseek($thumb_path,0); // Perform File Transfer ftpTransfer($target_path, "$new_file_name"); ftpTransfer($thumb_path, "thumbs/$new_file_name"); // My Simple FTP Transfer Function: function ftpTransfer($source, $name) { global $ftp; $ftpMeta = "ftp://" . $ftp['user_name'] . ":" . $ftp['user_pass'] . "@" . $ftp['server'] . $ftp['dir'] . $name; $ftpData = file_get_contents($source); file_put_contents($ftpMeta, $ftpData); } Like I said, the regular sized image works great!! the thumbnail file is only created on the storage server but it has a 0kb file size. I'm sure it has to do with me working with a temporary file. tmpfile() claims to remove the file after it's closed, so maybe by using the Save feature in SimpleImage, it's closing the file and then the server automatically removes the temporary file? SimpleImage has a feature where I can write the image output [possibly to a file] using $simImage->output() for the data. If that helps with a solution? I don't know. I know it might be complicated, but any help is appreciated! This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=333230.0 I got this script: But it give me error, file_get_contents cannot open stream. I need to add the FTP connection with user/pass paramaters. then look in set http url, to get the file contents(images) and transfer to ftp server location. Can Anyone take alook and tell me if I am going down the right path and how to get there. Please Code: [Select] function postToHost($host, $port, $path, $postdata = array(), $filedata = array()) { $data = ""; $boundary = "---------------------".substr(md5(rand(0,32000)),0,10); $fp = fsockopen($host, $port); fputs($fp, "POST $path HTTP/1.0\n"); fputs($fp, "Host: $host\n"); fputs($fp, "Content-type: multipart/form-data; boundary=".$boundary."\n"); // Ab dieser Stelle sammeln wir erstmal alle Daten in einem String // Sammeln der POST Daten foreach($postdata as $key => $val){ $data .= "--$boundary\n"; $data .= "Content-Disposition: form-data; name=\"".$key."\"\n\n".$val."\n"; } // Sammeln der FILE Daten if($filedata) { $data .= "--$boundary\n"; $data .= "Content-Disposition: form-data; name=\"".$filedata['name']."\"; filename=\"".$filedata['name']."\"\n"; $data .= "Content-Type: ".$filedata['type']."\n"; $data .= "Content-Transfer-Encoding: binary\n\n"; $data .= $filedata['data']."\n"; $data .= "--$boundary--\n"; } // Senden aller Informationen fputs($fp, "Content-length: ".strlen($data)."\n\n"); fputs($fp, $data); // Auslesen der Antwort while(!feof($fp)) { $res .= fread($fp, 1); } fclose($fp); return $res; } $postdata = array('var1'=>'today', 'var2'=>'yesterday'); $filedata = array( 'type' => 'image/png', 'data' => file_get_contents('http://xxx/tdr-images/images/mapping/dynamic/deals/spot_map') ); echo '<pre>'.postToHost ("localhost", 80, "/test3.php", $postdata, $filedata).'</pre>'; hello friends, while clicking the form all the information goes to database, I have one image upload field, when cliking the submit button, i would like 'image name' to go in database and file to go in /upload folder, i have tried this for hours and gave up, if anyone help me in this, i would be very greatful |
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