PHP - Want To Store A File Directly In The Right Folder
Hi all,
i want to download a file from the server but instead of storing it in the downloads i want it to store it directly in the folder i want and i also dont want to show any download window that appears while we download any file. Friend please help..... Similar TutorialsHey, 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 Hi,
I need to create a landing page with a form. That form needs to be recorded somewhere instead of sent to email. I know I can write it to a SQL database, and then to an excel file. But I only need a temporary solution so I figured I'd just go straight to CSV.
Is this bad practice? What potential problems might I encounter other than security issues?
Hi there, Can anybody help me to write php/javascript code which will allow users to open files directly from web browser into desktop application? Here is the specification: I have a photo editing business with many people working in Photoshop. I am currently developing a web based application (joblist) using Javascript and PHP which should allow the photoshop designers to browse and open files/images directly from joblist/web browser into photoshop. The reason I want this instead of browsing folder is that I have a database where I store who worked on which file, when and how long it took. The concept is that, designers will select a file and click on start, as soon as they click on start the original file will open in Photoshop and there will be an entry into database (using PHP). Once they finish the task they will close the file and click on Finish button. My joblist application will be published in a local server and the file will be open on a local network, so when they save the file it will be saved where the source file is located in (local server). The application should work in both PC and Mac. I have already done all other part of the application except file opening directly from browser to desktop application functionality. Anybody can help me to write the code (PHP or Javascript) which can open the file from browser (local server) directly into desktop application e.g. PHotoshop or Illustrator? Thank you very much I look forward to someone's real help! Best regards Mr. Sumon I have a page using forms to help build listing templates for eBay. I have a folder where I have hundreds of logos stored. I know the logo names but not their extensions. . . . I have to test each potential (jpg, jpeg, gif, png, etc.) until I guess right. Here is an example code for the web form:
<form action="extension_test2.php" method="post"> <p>Logo: <input name="e" value="" type="text" size="15" maxlength="30" /><p> <input name="Submit" type="Submit"/> </form> and the form's result: <? $e =$_POST['e']; if(!empty($e)) { echo '<img src="http://www.gbamedica...ebayimg/logos/'.$e.'">'; }; ?> Here is a link to the example: http://www.gbamedica...ension_test.php Use "olympus.jpg" for test. I am looking for code that can determine the file type and dynamically add the extension. Can it be done? Hi Guys, I need help for in storing data from PHP from array in mysql. I'm very new to PHP/Mysql and have started learing it just few weeks back. I'm tryting to build a website for myself. I'm having tough time trying to insert the data from Form in mysql. I have a form where user update the company name and insert there product name with there image. I want to rename the image with the corresponding text field value and insert the upload path in the table. my FORM 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=utf-8" /> <title>Untitled Document</title> <style type="text/css"> body,td,th { color: #000; font-family: Tahoma, Geneva, sans-serif; font-size: 80%; } body { background-color: #9CF; } table { background: #CCF; } th { font: bold normal 16px/normal "Times New Roman", Times, serif; text-transform: capitalize; color: #00F; background: #FFC; } td { font: bold 14px Georgia, "Times New Roman", Times, serif; text-transform: capitalize; color: #3A00FF; background: #FF9; } </style> </head> <body> <form action="upload-file1.php" method="post" enctype="multipart/form-data" name="form"> <strong>Company Name:</strong> <input name="product" type="text" /> <br /> <table> <tr bgcolor="#FF9900"> <th colspan="3" bgcolor="#CCFFFF">filter </th> <th colspan="5" bgcolor="#CCFFFF">heater</th> </tr> <tr><td>filter1</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater1: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter2</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater2: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter3</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater3: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter4</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater4: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter5</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater5: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> </table> <br /> <br /> <input name="submit" type="submit" value="submit" /> </form> </body> </html> my PHP Code Code: [Select] <?php require("connect.php"); ?> <?php $product = $_POST['product']; echo $product; if(isset($_POST["submit"])){ $sql = "INSERT INTO company (product) VALUES ('$product')"; $query = mysql_query($sql) OR DIE(mysql_error()); $comp_id = mysql_insert_id(); echo $sql; } ?> <?php function filter() { if(isset($_POST['submit'])) { foreach($_POST['filter'] as $key => $val) { if(trim($val) != '') $filter[] = $val; } $total_records_filter = count($filter); for($i = 0; $i < $total_records_filter; $i++) { $prod_val = $filter[$i]; $sql_prod = "INSERT INTO filter(filter_id, filter) VALUES('', '$prod_val')"; echo $sql_prod.'<br>'; mysql_query($sql_prod) OR die(mysql_error()); } } } filter(); function heater() { if(isset($_POST['submit'])) { foreach($_POST['heater'] as $key => $val) { if(trim($val) != '') $heater[] = $val; } $total_records_heater = count($heater); for($i = 0; $i < $total_records_heater; $i++) { $dir_val = $heater[$i]; $sql_dir = "INSERT INTO heater(heater_id, heater) VALUES('', '$dir_val')"; echo $sql_dir.'<br>'; mysql_query($sql_dir) OR die(mysql_error()); } } } heater(); ?> <?php mysql_close($link)?> My Table structure company comp_id int(5) PK product varchar(50) No filter filter_id int(5) PK filter varchar(25) No filter_path varchar(100) No heater heater_id int(5) PK heater varchar(25) No heater_path varchar(100) No company_filter id int(5) PK comp_id int(5) PK FK (From company table) filter_id int(5) PK FK (From filter table) company_heater id int(5) PK comp_id int(5) PK FK (From company table) heater_id int(5) PK FK (From heater table) Regards BW 0
What is the best way to store this data coming from the api into a csv file to later put into the db. Output: rank, level, xp, rank, level, xp, etc. This api produces about 60 rows of data per name ran and x that by about roughly 300 names that equals a lot of data. Pretty much with my current code I am pretty much creating a endless loop almost that would take a long time to execute so updating said data would be a nightmare i would think. Is there a way to accomplish the same thing without the loop or I am not sure how to go about this. My current code is a mess that much I know I am gonna be told. This code works, just not efficiently. I am thinking there may be a better way to do this. $query = $conn->prepare("SELECT name FROM users LIMIT 1"); $query->execute(); while($row = $query->fetch(PDO::FETCH_ASSOC)){ $name = $row['name']; $url = 'https://secure.runescape.com/m=hiscore/index_lite.ws?player='. $name . '';//api that the csv data is coming from $highscores = file_get_contents($url); $fields = array("Name", "Rank", "Level", "Xp");//this is to add the headers // for the csv files $implode1 = implode($fields, ",");//turn into csv format, not sure this is //even needed $implode1 .= "\n";/*this is to add a line break so that the explode below will properly put it into its own element*/ //otherwise data starts joining togather $extra_data = array("$name");/*This is to add the name that the data pertains too*/ $implode2 = implode($extra_data, ",");//turn into csv format $highscores = $implode1 . $implode2 . $highscores;//join as one array $highscores = explode("\n", $highscores);//turn each csv into an element of //its own element a bunch of unsets to remove unwanted data. Omitted them to condense the code $i = 1; header('Content-Type: text/csv'); header('Content-Disposition: attachment; filename="name.csv"'); $data = $highscores; $fp = fopen('highscores/' . $name . '.csv', 'wb'); foreach ( $data as $line ) { $val = explode(",", $line); fputcsv($fp, $val); } fclose($fp); The pdo part I was gonna include but the way I have it setup now wouldn't work. I was thinking that I would use mysql's LOAD DATA INFILE to insert the csv data to database. The end goal here is to be able to search for player stats and to use said data to keep track of xp earned during a xp competition. I PRAY that i have included enough info... I would really appreciate any feedback if possible. Hello! I have a 2 questions: Q1: How do you store answers from a php file(see code below) into another file?(php preferably) I know it would probably be easier using a database, but I would like to know how to do it with a text/php file. Below you can see a piece of code for a page students would go on to fill in online exercices. They would first have to input their names. These names would have to be sent to a php file called "StudentAnswers.php". As well as their answers I also have another file that is called "QandA.php" which is called on with include in the piece of code below so the students can't see the answers before they submitted their answers. Q2: But how do I complete the if code(see question marks) so the answer only shows IF(as soon as) they clicked on the submit button? Code: [Select] <html> <body> <h2>Exercices</h2> <p> <form name="input" action="StudentAnswers.php" method="get"> Student name: <input type="text" name="user" /> </form> </p> <p> What is the gradient of T(x,y)=2x+3? <form name="input" action="StudentAnswers.php" method="get"> <input type="text" name="1b" /> <input type="submit" value="Input answer"/> </form> if ( ????????? ) { ???????"; } <?php include("QandA.php"); echo $Question["1b"];?> <p> </body> </html> B Hello! I'd like to know how one can save inputted data using a form to store it into another php file. The user should just input their name in the php webpage called "MainPage.php". When they write their names and hit submit, they would just stay on the same page. Their submitted names should be stored permanently into a "Answers.php" file. What do I need to change in my "MainPage.php code? Code: [Select] <html> <body> <form method="post" action="Answers.php" > Name:<input type="text" name="Name" /> <input type="submit" value="Submit name" /> </form> [Some php code not relevant to question] </body> </html> Hi, I want to store files to variable array using glob() like $files[0] = xyz.txt $files[1] = pqr.txt . . . $files[n] = nfile.txt I know how to list files from directory using glob() Code: [Select] <?php foreach (glob("*.txt") as $filename) { echo "$filename size " . filesize($filename) . "\n"; } ?> How can I do that ? Hi, I have a form where a user selects a file to attach to the email. At the moment when you select a file it uploads from the user device. How do i change this so that a user can attach a file from a folder on the server. For example the folder name is uploadinvoice so when the user selects the browse button to attach a file it opens up the uploadinvoice folder on the server so the user can select the file from there ?
Thanks
coding i have at moment function ValidateEmail($email) { $pattern = '/^([0-9a-z]([-.\w]*[0-9a-z])*@(([0-9a-z])+([-\w]*[0-9a-z])*\.)+[a-z]{2,6})$/i'; return preg_match($pattern, $email); } if ($_SERVER['REQUEST_METHOD'] == 'POST' && isset($_POST['formid']) && $_POST['formid'] == 'form1') { $mailto = $_POST['youremail']; $mailfrom = isset($_POST['myemail']) ? $_POST['myemail'] : $mailto; $subject = 'Message'; $message = 'Message'; $success_url = './test.php'; $error_url = ''; $eol = "\n"; $error = ''; $internalfields = array ("submit", "reset", "send", "filesize", "formid", "captcha_code", "recaptcha_challenge_field", "recaptcha_response_field", "g-recaptcha-response"); $boundary = md5(uniqid(time())); $header = 'From: '.$mailfrom.$eol; $header .= 'Reply-To: '.$mailfrom.$eol; $header .= 'MIME-Version: 1.0'.$eol; $header .= 'Content-Type: multipart/mixed; boundary="'.$boundary.'"'.$eol; $header .= 'X-Mailer: PHP v'.phpversion().$eol; try { if (!ValidateEmail($mailfrom)) { $error .= "The specified email address (" . $mailfrom . ") is invalid!\n<br>"; throw new Exception($error); } $message .= $eol; foreach ($_POST as $key => $value) { if (!in_array(strtolower($key), $internalfields)) { if (!is_array($value)) { $message .= ucwords(str_replace("_", " ", $key)) . " : " . $value . $eol; } else { $message .= ucwords(str_replace("_", " ", $key)) . " : " . implode(",", $value) . $eol; } } } $body = 'This is a multi-part message in MIME format.'.$eol.$eol; $body .= '--'.$boundary.$eol; $body .= 'Content-Type: text/plain; charset=ISO-8859-1'.$eol; $body .= 'Content-Transfer-Encoding: 8bit'.$eol; $body .= $eol.stripslashes($message).$eol; if (!empty($_FILES)) { foreach ($_FILES as $key => $value) { if ($_FILES[$key]['error'] == 0) { $body .= '--'.$boundary.$eol; $body .= 'Content-Type: '.$_FILES[$key]['type'].'; name='.$_FILES[$key]['name'].$eol; $body .= 'Content-Transfer-Encoding: base64'.$eol; $body .= 'Content-Disposition: attachment; filename='.$_FILES[$key]['name'].$eol; $body .= $eol.chunk_split(base64_encode(file_get_contents($_FILES[$key]['tmp_name']))).$eol; } } } $body .= '--'.$boundary.'--'.$eol; if ($mailto != '') { mail($mailto, $subject, $body, $header); } header('Location: '.$success_url); } catch (Exception $e) { $errorcode = file_get_contents($error_url); $replace = "##error##"; $errorcode = str_replace($replace, $e->getMessage(), $errorcode); echo $errorcode; } exit;
} I'm trying to extract the contents of a zip file to a folder. I found the ZipArchive class and followed the examples to get it to work for the most part. But I want to extract the files in the folder inside the zip file but leave the folder out. So it should extract just the files to my given destination. I found this on php.net. Code: [Select] If you want to copy one file at a time and remove the folder name that is stored in the ZIP file, so you don't have to create directories from the ZIP itself, then use this snippet (basically collapses the ZIP file into one Folder). <?php $path = 'zipfile.zip' $zip = new ZipArchive; if ($zip->open($path) === true) { for($i = 0; $i < $zip->numFiles; $i++) { $filename = $zip->getNameIndex($i); $fileinfo = pathinfo($filename); copy("zip://".$path."#".$filename, "/your/new/destination/".$fileinfo['basename']); } $zip->close(); } ?> For some reason that 'copy' line is not working for me. Obviosly I've changed the variables in the line to the correct variables. Can someone help me out. Thanks Mike When I use require('../config.php'); It does not works on my machine but it works on shared hosting Can someone help me what must be the issue? Thanks in advance CSJakharia The root directory:
header.php
stylesheet.css
In the following example I am trying to include the header.php file in a sub folder.
When I include the header.php like in the following example then the stylesheet.css file will not work anymo
<?php include("../header.php"); ?>The stylesheet.css file is included in the head tags of the header.php file. Is the above example the right way to do it? If yes, how can I do it so the stylesheet.css file will work too. Hello, I'm trying to have my index.php to open/run another test.php file. I'm having my own server that I play with, that I run Ubuntu on. So the index.php are located at /var/www/ directory, but I want to run a file that are located at /testing/test.php The final test.php is file for showing pictures, and I don't want to out all the pictures under the /var/www/ location. It's alot of photos. I don't know much about php but I have been trying this: <?php header("Location: /var/www/testing/test.php"); //These below are desperat old tries. //header("Location: ./testing/test.php"); //header("Location: ../testing/test.php"); //$handle = fopen("/privat/Web_pictures/test.php", "r"); //"/testing/test.php" // header("Location: ./test.php"); //This one actually works, but I'm still in the wrong folder (/var/www/) echo "test "; // NN4 requires that we output something... exit(); ?> Thankful for help! This topic has been moved to HTML Help. http://www.phpfreaks.com/forums/index.php?topic=308625.0 if (!file_exists('../images/flags/imNum.txt')) { $file1 = fopen('../images/flags/imNum.txt','c'); fclose($file1); } why won't that work =\ it makes no sense to me Hi everyone, I'm still learning, but getting intermediate in PHP now, but it is still challenge to learn. I'm trying to have php check to see if one file inside folder in server, seem I could not get it right, but I tested it on other site, it works, but not this script, I don't understand why it won't work...maybe logical is wrong? here my code: if ($_POST['video']) { $path1 = "UPLOADS/Home/"; $path2 = "UPLOADS/Breakfast/"; $path3 = "UPLOADS/Spider/"; $scan1 = scandir($path1); $scan2 = scandir($path2); $scan3 = scandir($path3); $count1 = count($scan1) - 3; $count2 = count($scan2) - 3; $count3 = count($scan3) - 3; if($count1 > 0) { header('location:exist.html'); }elseif ($count2 > 0) { header('location:exist.html'); }elseif ($count3 > 0) { header('location:exist.html'); } But other site, it works: $scan = scandir($path); $count = count($scan) - 3; echo $count; if($count > 1){ echo "Hello yourself!<br />"; } Anyone will help will be appreciate! Thank you! Gary Edited April 3, 2019 by sigmahokiesI upload an image and put every information inside $_SESSION['tmp'] and $_SESSION['path'] then once user click on button then i use move_uploaded_file($_SESSION['tmp'],$_SESSION['path']) but file uploaded not appeared in my upload folder, and again i try to echo everything but all information still kept well in $_SESSION is there something missing here? thanks Hello all, i am go9090go. Today i made a domains for a jar file people can upload from my website. I made this to make the jar file close source and its easy to update. Now i made a java classloader and everything i made works. The classloader call a php document with the password and username. The pass and name will be checked inside a databse and if its inside i use header() to load the jar file. But when i just go to my main domain i get the index of the site and people can easly download the jar file without have to walk thru the php pass checker. So i want to place the jar file inside a protected folder,and i want that only way you get acces to this jar is by the php file. How can i get a file from a protected folder? here is the php used when the jar file is not inside a protected folder: <?php $DBName = "name";//name database $DBUser = "name";//user $DBPassword = "pass"; //passs $DBHost = "host"; //might be different mysql_connect($DBHost, $DBUser, $DBPassword); mysql_select_db($DBName); $username = $_GET['username']; $password = $_GET['password']; $IP = $_SERVER['REMOTE_ADDR']; $string = "Java"; $pos = strpos($agent, $string); if (!strpos($_SERVER['HTTP_USER_AGENT'], "Java")) { echo("Your Auth has been banned for trying to breach security."); //mysql_query("delete from users where username='$username'"); exit(); } $query = "select * from users where name='$username' and pass='$password'"; mysql_query($query); $num = mysql_affected_rows(); if ($num > 0) { header('Location:script/Script.jar'); } ?> now i want to use the header to a file inside a folder that is protected : so how can i make the header() methode to open script.jar inside a protected folder. The folder haves name and pass: blabla,balbla for exempel thanks for help Hi
I echo out images from a folder. The problem is that I have a html file in the same folder but I don't want to echo that out. How can I write my code to get rid of the html in the output?
Here my code:
<?php $filetype = '*.html'; $dirname = substr('$filetype'); $i=0; if (isset($_POST['submit2'])) { //get the dir from POST $selected_dir = $_POST['myDirs']; //now get the files from within the selected dir and echo them to the screen foreach(glob($selected_dir . DIRECTORY_SEPARATOR . '*') as $dirname) { echo substr($dirname, 0, -4); echo '<img src="'.$dirname.'" />'; echo "<label><div class=\"radiobt\"><input type='radio' name='radio1' value='$i'/></div></label>"; } } ?>P.S. when I echo out : substr($dirname, 0, -4); I want to get ride of the html filename there too. How can I do so something like this? |