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PHP - Php Script To Prompt For Yes/no
I have a PHP script with no HTML in it. The script runs through and updates a bunch of fields in our Active Directory LDAP accounts. Well, I don't want it to just update every account automatically. I want it to prompt me for for each user. Ex. "Would you like to update John Doe's position to Director?" "Yes/No" I'm guessing I'll have to use Javascript for this. But I don't want to add any HTML/JS if possible. But I'm guessing this isn't possible without adding some.
Thanks Mike Similar TutorialsI need to process a CSV file in database (MySql using LOAD DATA INFILE)), while the code is working fine but it works only with file that is already on my apache server (where php is installed). Is there any way (Java/Ajax or anything) that I can prompt user to select a file from any loaction they prefer including there desktop?? Thanks in advance!! I previously asked this question in the Apache Server forum with no reply so I'm thinking maybe that was the wrong area to post it in. I have a directory with media files in it (mp3's, videos etc.) that I want to secure from being downloaded yet be able to be played by a local player on the site. I was thinking of trying to secure the directory using .htaccess and .htpasswd files. I would then like to have the page the player is on pass the login credentials to allow the media files to be played. I do not want the user to be prompted for login credentials but if they tried to access the directory to download a file they would be denied. The server is (of course) Apache 2.2.17 and the scripting languages are php (version 5.2) and jQuery. Is this possible or am I going about this the wrong way? Thanks in advance. P.S. Yes, I have extensively searched for answers before posting but could not find the answer I was looking for. Does anyone know what the source code for the PHP MyAdmin login prompt (the pop out box) looks like?
Hello! This bit of code actually does work, I just want to check that this is the best/proper/most efficiant way of doing it. Basically i'm calling a PHP function that uses the javascripts prompt function, then returning the value that the user types in back to PHP. Most probably easier to show you. Code: [Select] <?php //prompt function function prompt($prompt_msg){ echo("<script type='text/javascript'> var answer = prompt('".$prompt_msg."'); </script>"); $answer = "<script type='text/javascript'> document.write(answer); </script>"; return($answer); } //program $prompt_msg = "Please type your name."; $name = prompt($prompt_msg); $output_msg = "Hello there ".$name."!"; echo($output_msg); ?> That is the entire code, I just want to know if this is the proper way to do something like this (using javascript and PHP together) thanks guys! Need help urgent. the website sudden prompt out error below: Warning: require_once() [function.require-once]: open_basedir restriction in effect. File(/var/www/vhosts/db2020.co.uk/application/application/config/admin/config.live.php) is not within the allowed path(s): (/var/www/vhosts/db2020.co.uk/subdomains/admin/httpdocs:/tmp) in /var/www/vhosts/db2020.co.uk/subdomains/admin/httpdocs/index.php on line 11 Warning: require_once(/var/www/vhosts/db2020.co.uk/application/application/config/admin/config.live.php) [function.require-once]: failed to open stream: Operation not permitted in /var/www/vhosts/db2020.co.uk/subdomains/admin/httpdocs/index.php on line 11 Fatal error: require_once() [function.require]: Failed opening required '/var/www/vhosts/db2020.co.uk/application/application/config/admin/config.live.php' (include_path='.:') in /var/www/vhosts/db2020.co.uk/subdomains/admin/httpdocs/index.php on line 11 and also after this problem occur, trafic of this domain go overlimit. how can i solve this problem? urgent.... website = https://db2020.co.uk/ thanks. OK so I am working on a default response command prompt that will display X:\XOS(root)>user entered text 'user entered text' is not recognized as an internal or external command, operable program or batch file. However so far when i do this it distorts the response. See pictures. Code: [Select] <?php function stripslashes_deep($value) { $value = is_array($value) ? array_map('stripslashes_deep', $value) : stripslashes($value); return $value; } if (get_magic_quotes_gpc()) { $_GET = stripslashes_deep($_GET); } if (!isset($_GET['idx'])) { $idx = 115; $text = "XOS [Version 0.2.13]\n"; $text .= "Copyright (c) 2012 RedLine Services, LLC. All rights reserved.\n"; } else { $idx = $_GET['idx'] + 3; $input = $_GET['text']; $text = substr($input, 0, $idx); $user_input = substr($input, $idx); $text .= "\n '$user_input' is not recognized as an internal or external command,\noperable program or batch file.\n"; } $text .= "\n"; $text .= "X:\XOS(root)>"; $idx = strlen($text); //echo strlen($text); //$fn = "text.txt"; //$file = fopen($fn, "a"); //fwrite($file, "$msg2"); //fclose($file); ?> <html> <style> body{ background-color:#000000;color:#ffffff } textarea { background-color:#000000; color:#ffffff; width:100%; height:400px; } </style> <body> <form method="GET"> <textarea name="text"><?php echo $text; ?></textarea> <input type="text" name="idx" value=""> <button type="submit" name="submit">Submit</button> </form> </body> </html> I can determine which radio button is selected for each file upload prompt because I have a name and a value. So the array name=image can have one of three values for each file upload prompt. Is there a way to determine which of the three file upload prompts was loaded (clicking browse & select image) with an image and which was not? There doesn't seem to be a value like there is with radio buttons. I want to be able to replace an already uploaded image with another. Thanks. <input type='radio' name='image[image0]' value='keep' checked='checked'/> <input type='radio' name='image[image0]' value='delete' /> <input type='radio' name='image[image0]' value='replace' /> <input type="file" name="image[]" /> <input type='radio' name='image[image1]' value='keep' checked='checked'/> <input type='radio' name='image[image1]' value='delete' /> <input type='radio' name='image[image1]' value='replace' /> <input type="file" name="image[]" /> <input type='radio' name='image[image2]' value='keep' checked='checked'/> <input type='radio' name='image[image2]' value='delete' /> <input type='radio' name='image[image2]' value='replace' /> <input type="file" name="image[]" /> if (isset($_POST['submitted']) && ($image = $_POST['image'])) { foreach($image as $imageKey => $imageValue) { if ($imageValue == 'keep') { do this } } } Hi, I need just quick run down on how to run a command line user interface for program for php using Windows command prompt: Let's say I have a calculator program: Code: [Select] <?php class Calculator { function factorial($num) { if($num==1) return 1; else return $num*factorial($num-1); } //other functions... }//END CLASS Calculator ?> //Test runner: Code: [Select] $calculator = new Calculator; $quit = false; while($quit != true) { echo "Please enter the number to find //how do i pass user's keyboard input to the function?? //so now pass user input to $calculator->factorial(//user input??); echo "Do you want to run program again?' //Again need to take in user input I feel it is similar to Java or C, but I just need some refresher, a little rustry, is it printf('%d', $quit) etc??? I have a script that seems to work well to insert a bookmark into a users database when he/she is logged into the system but I am having a hard time figuring out how I would go about making a work-a-round for having an item selected before being logged in, and inserted after they have logged in or registered. For example, I would like a user to be able to select an Item to add to bookmark whether that user is logged in/registered or not and if they are not, they would be greeted with a login/registration form and after successful login the add bookmark script would be initiated on the item previously selected. What I've got this far: Simple form to add bookmark: <form name="bm_table" action="add_bms.php" method="post"> <input type="text" name="new_url" value="http://" /> <input type="submit" value="Add Bookmark"/> </form> Then I have the add bookmark script: BEGIN php $new_url = $_POST['new_url']; try { check_valid_user(); //cannot get past this part since it ends the script....code below if (!filled_out($_POST)) { throw new Exception('Form not completely filled out.'); } // check URL format if (strstr($new_url, 'http://') === false) { $new_url = 'http://'.$new_url; } // check URL is valid if (!(@fopen($new_url, 'r'))) { throw new Exception('Not a valid URL.'); } // try to add bm add_bm($new_url); echo 'Bookmark added.'; // get the bookmarks this user has saved if ($url_array = get_user_urls($_SESSION['valid_user'])) { display_user_urls($url_array); } } catch (Exception $e) { echo $e->getMessage(); } END php Checking valid user - the portion I cannot get past in the above script: function check_valid_user() { // see if somebody is logged in and notify them if not if (isset($_SESSION['valid_user'])) { echo "Logged in as ".$_SESSION['valid_user'].".<br />"; } else { // they are not logged in do_html_heading('Problem:'); echo 'You are not logged in.<br />'; do_html_url('login.php', 'Login'); do_html_footer(); exit; } } How would I go about modifying the script so that a user could fill in the form (later it would be a link...obviously they probably wouldn't be filling in a form that is log-in specific - but same concept I think) Thanks in advance for the help! tec4 Hi everyone! I've been working on a php script to replace links that contain a query with direct links to the files they would redirect to. Hi i have this upload script which works fine it uploads image to a specified folder and sends the the details to the database. but now i am trying to instead make a modify script which is Update set so i tried to change insert to update but didnt work can someone help me out please this my insert image script which works fine but want to change to modify instead Code: [Select] <?php mysql_connect("localhost", "root", "") or die(mysql_error()) ; mysql_select_db("upload") or die(mysql_error()) ; // my file the name of the input area on the form type is the extension of the file //echo $_FILES["myfile"]["type"]; //myfile is the name of the input area on the form $name = $_FILES["image"] ["name"]; // name of the file $type = $_FILES["image"]["type"]; //type of the file $size = $_FILES["image"]["size"]; //the size of the file $temp = $_FILES["image"]["tmp_name"];//temporary file location when click upload it temporary stores on the computer and gives it a temporary name $error =array(); // this an empty array where you can then call on all of the error messages $allowed_exts = array('jpg', 'jpeg', 'png', 'gif'); // array with the following extension name values $image_type = array('image/jpg', 'image/jpeg', 'image/png', 'image/gif'); // array with the following image type values $location = 'images/'; //location of the file or directory where the file will be stored $appendic_name = "news".$name;//this append the word [news] before the name so the image would be news[nameofimage].gif // substr counts the number of carachters and then you the specify how how many you letters you want to cut off from the beginning of the word example drivers.jpg it would cut off dri, and would display vers.jpg //echo $extension = substr($name, 3); //using both substr and strpos, strpos it will delete anything before the dot in this case it finds the dot on the $name file deletes and + 1 says read after the last letter you delete because you want to display the letters after the dot. if remove the +1 it will display .gif which what we want is just gif $extension = strtolower(substr($name, strpos ($name, '.') +1));//strlower turn the extension non capital in case extension is capital example JPG will strtolower will make jpg // another way of doing is with explode // $image_ext strtolower(end(explode('.',$name))); will explode from where you want in this case from the dot adn end will display from the end after the explode $myfile = $_POST["myfile"]; if (isset($image)) // if you choose a file name do the if bellow { // if extension is not equal to any of the variables in the array $allowed_exts error appears if(in_array($extension, $allowed_exts) === false ) { $error[] = 'Extension not allowed! gif, jpg, jpeg, png only<br />'; // if no errror read next if line } // if file type is not equal to any of the variables in array $image_type error appears if(in_array($type, $image_type) === false) { $error[] = 'Type of file not allowed! only images allowed<br />'; } // if file bigger than the number bellow error message if($size > 2097152) { $error[] = 'File size must be under 2MB!'; } // check if folder exist in the server if(!file_exists ($location)) { $error[] = 'No directory ' . $location. ' on the server Please create a folder ' .$location; } } // if no error found do the move upload function if (empty($error)){ if (move_uploaded_file($temp, $location .$appendic_name)) { // insert data into database first are the field name teh values are the variables you want to insert into those fields appendic is the new name of the image mysql_query("INSERT INTO image (myfile ,image) VALUES ('$myfile', '$appendic_name')") ; exit(); } } else { foreach ($error as $error) { echo $error; } } //echo $type; ?> I'm having trouble echoing $year in my script. Listed below is the script, just below ,$result = mysql_query("SELECT * FROM $dbname WHERE class LIKE '%$search%'") or die(mysql_error());, in the script I try to echo $year. It doesn't show up in the table on the webpage. Everything else works fine. Any help wold be appreciated greatly. Thanks in advance. <?php include 'config2.php'; $search=$_GET["search"]; // Connect to server and select database. mysql_connect($dbhost, $dbuser, $dbpass)or die("cannot connect"); mysql_select_db("vetman")or die("cannot select DB"); $result = mysql_query("SELECT * FROM $dbname WHERE class LIKE '%$search%'") or die(mysql_error()); // store the record of the "" table into $row //$current = ''; echo "<table align=center border=1>"; echo "<br>"; echo "<tr>"; echo "<td align=center>"; ?> <div style="float: center;"><a><h1><?php echo $year; ?></h1></a></div> <?php echo "</td>"; echo "</tr>"; echo "</table>"; // keeps getting the next row until there are no more to get if($result && mysql_num_rows($result) > 0) { $i = 0; $max_columns = 2; echo "<table align=center>"; echo "<br>"; while($row = mysql_fetch_array($result)) { // make the variables easy to deal with extract($row); // open row if counter is zero if($i == 0) echo "<tr>"; echo "<td align=center>"; ?> <div style="float: left;"> <div><img src="<?php echo $image1; ?>"></div> </div> <?php echo "</td>"; // increment counter - if counter = max columns, reset counter and close row if(++$i == $max_columns) { echo "</tr>"; $i=0; } // end if } // end while } // end if results // clean up table - makes your code valid! if($i > 0) { for($j=$i; $j<$max_columns;$j++) echo "<td> </td>"; echo '</tr>'; } mysql_close(); ?> </table> I'm trying to use this script known as SimpleImage.php that can be found here <a href="http://www.white-hat-web-design.co.uk/articles/php-image-resizing.php">link</a> I'm trying to include what is on the bottom of the page to my existing script can anyone help me I've tried several ways but its not working. Code: [Select] <?php session_start(); error_reporting(E_ALL); ini_set('display_errors','On'); //error_reporting(E_ALL); // image upload folder $image_folder = 'images/classified/'; // fieldnames in form $all_file_fields = array('image1', 'image2' ,'image3', 'image4'); // allowed filetypes $file_types = array('jpg','gif','png'); // max filesize 5mb $max_size = 5000000; //echo'<pre>';print_r($_FILES);exit; $time = time(); $count = 1; foreach($all_file_fields as $fieldname){ if($_FILES[$fieldname]['name'] != ''){ $type = substr($_FILES[$fieldname]['name'], -3, 3); // check filetype if(in_array(strtolower($type), $file_types)){ //check filesize if($_FILES[$fieldname]['size']>$max_size){ $error = "File too big. Max filesize is ".$max_size." MB"; }else{ // new filename $filename = str_replace(' ','',$myusername).'_'.$time.'_'.$count.'.'.$type; // move/upload file $target_path = $image_folder.basename($filename); move_uploaded_file($_FILES[$fieldname]['tmp_name'], $target_path); //save array with filenames $images[$count] = $image_folder.$filename; $count = $count+1; }//end if }else{ $error = "Please use jpg, gif, png files"; }//end if }//end if }//end foreach if($error != ''){ echo $error; }else{ /* -------------------------------------------------------------------------------------------------- SAVE TO DATABASE ------------------------------------------------------------------------------------ -------------------------------------------------------------------------------------------------- */ ?> Hello, I stored a fsockopen function in a separate "called.php" file, in order to run it as another thread when it needs. The called script should return results to the "master.php" script. I'm able to run the script to get the socket working, and I'm able to get results from the called script. I tried for hours but I can't do the twice both My master.php script (with socket working): Code: [Select] <?php $command = "(/mnt/opt/www/called.php $_SERVER[REMOTE_ADDR] &) > /dev/null"; $result = exec($command); echo ("result = $result\r\n"); ?> and my called.php script Code: [Select] #!/mnt/opt/usr/bin/php-cli -q <?php $device = $_SERVER['argv'][1]; $port = "8080"; $fp = fsockopen($device, $port, $errno, $errstr, 5); fwrite($fp, "test"); fclose($fp); echo ("normal end of the called.php script"); ?> In the master script, if I use Code: [Select] $command = "(/mnt/opt/www/called.php $_SERVER[REMOTE_ADDR] &) > /dev/null"; the socket works, but I have nothing in $result (note also that I don't anderstand why the ( ... &) are needed!?) and if I use Code: [Select] $command = "/mnt/opt/www/called.php $_SERVER[REMOTE_ADDR]"; I have the correct text "normal end of the called.php script" in $result but the socket connection is not performed (no errors in php logs) Could you help me to find a way to let's work the two features correctly together? Thank you. hey guys im really just after a bit of help/information on 2 things (hope its in the right forum).
1. basically I'm wanting to make payments from one account to another online...like paypal does...im wondering what I would need to do to be able to do this if anyone can shine some light please?
2.as seen on google you type in a query in the search bar and it generates sentences/keywords from a database
example:
so if product "chair" was in the database
whilst typing "ch" it would show "chair" for a possible match
I know it would in tale sql & json but im after a good tutorial/script of some sort.
if anyone can help with some information/sites it would be much appreciated.
Thank you
Well the subject line is pretty explicit. I found this script that uploads a picture onto a folder on the server called images, then inserts the the path of the image on the images folder onto a VACHAR field in a database table. Code: [Select] <?php //This file inserts the main image into the images table. //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user //Start session session_start(); //Connect to database require ('config.php'); //Check whether the session variable id is present or not. If not, deny access. if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) { header("location: access_denied.php"); exit(); } else{ // Check to see if the type of file uploaded is a valid image type function is_valid_type($file) { // This is an array that holds all the valid image MIME types $valid_types = array("image/jpg", "image/jpeg", "image/bmp", "image/gif"); if (in_array($file['type'], $valid_types)) return 1; return 0; } // Just a short function that prints out the contents of an array in a manner that's easy to read // I used this function during debugging but it serves no purpose at run time for this example function showContents($array) { echo "<pre>"; print_r($array); echo "</pre>"; } // Set some constants // This variable is the path to the image folder where all the images are going to be stored // Note that there is a trailing forward slash $TARGET_PATH = "images/"; // Get our POSTed variable $image = $_FILES['image']; // Sanitize our input $image['name'] = mysql_real_escape_string($image['name']); // Build our target path full string. This is where the file will be moved to // i.e. images/picture.jpg $TARGET_PATH .= $image['name']; // Make sure all the fields from the form have inputs if ( $image['name'] == "" ) { $_SESSION['error'] = "All fields are required"; header("Location: member.php"); exit; } // Check to make sure that our file is actually an image // You check the file type instead of the extension because the extension can easily be faked if (!is_valid_type($image)) { $_SESSION['error'] = "You must upload a jpeg, gif, or bmp"; header("Location: member.php"); exit; } // Here we check to see if a file with that name already exists // You could get past filename problems by appending a timestamp to the filename and then continuing if (file_exists($TARGET_PATH)) { $_SESSION['error'] = "A file with that name already exists"; header("Location: member.php"); exit; } // Lets attempt to move the file from its temporary directory to its new home if (move_uploaded_file($image['tmp_name'], $TARGET_PATH)) { // NOTE: This is where a lot of people make mistakes. // We are *not* putting the image into the database; we are putting a reference to the file's location on the server $sql = "insert into images (member_id, image_cartegory, image_date, image) values ('{$_SESSION['id']}', 'main', NOW(), '" . $image['name'] . "')"; $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error()); header("Location: images.php"); echo "File uploaded"; exit; } else { // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to // Make sure you chmod the directory to be writeable $_SESSION['error'] = "Could not upload file. Check read/write persmissions on the directory"; header("Location: member.php"); exit; } } //End of if session variable id is not present. ?> The script seems to work fine because I managed to upload a picture which was successfully inserted into my images folder and into the database. Now the problem is, I can't figure out exactly how to write the script that displays the image on an html page. I used the following script which didn't work. Code: [Select] //authenticate user //Start session session_start(); //Connect to database require ('config.php'); $sql = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main' "); $row = mysql_fetch_assoc($sql); $imagebytes = $row['image']; header("Content-type: image/jpeg"); print $imagebytes; Seems to me like I need to alter some variables to match the variables used in the insert script, just can't figure out which. Can anyone help?? How would I go about making it to where,, I can tell the script to use a certain extension of php in the script like curl.. ? Thanks I have an application which runs on more than one server and need to launch one PHP script from another PHP script. Since this is different than a function call I'm not sure how it's done. I plan to include parameters in the URL I send and use GETs to pick up parameters in the "called" PHP script. Thanks for sugestions Hi, I am trying to run two scripts on one page. When I use just one script on the page they work however when I place both scripts on the same page one of them disrupts the other script. This script prevents the other following script from working: Code: [Select] ini_set('display_errors', 1); error_reporting(-1); { $query = "SELECT * FROM answers ORDER BY `aid` DESC LIMIT 0, 11"; } $result = mysql_query($query); while($row = mysql_fetch_assoc($result)) { $answer = $row['answer']; $aid = $row['aid']; echo " <div class='questionboxquestion'> <a href= 'http://www.domain.co.uk/test/easy/answer.php?aid=$aid' class='questionlink'>$answer</a> </div> <div class='questionboxnotes'> </br> </div> <div class='questionboxlinks'> <div class='questionboxcategory'> <div class='questionboxcategorytitle'> Category: </div> <a href= 'http://www.domain.co.uk/test/easy/furniture-category.php' class='questionanswerlink'></a> </div> <div class='questionboxanswerlink'> <a href= 'http://www.domain.co.uk/test/oeasy/index.php' class='questionanswerlink'>Answer</a> </div> </div> "; } Code: [Select] <?php if($error) echo "<span style=\"color:#ff0000;\">".$error."</span><br /><br />"; ?> <label for="username">Username: </label> <input type="text" name="username" value="<?php if($_POST['username']) echo $_POST['username']; ?>" /><br /> <label for="password">Password: </label> <input type="password" name="password" value="<?php if($_POST['password']) echo $_POST['password']; ?>" /><br /> <label for="password2">Retype Password: </label> <input type="password" name="password2" value="<?php if($_POST['password2']) echo $_POST['password2']; ?>" /><br /> <label for="email">Email: </label> <input type="text" name="email" value="<?php if($_POST['email']) echo $_POST['email']; ?>" /><br /><br /> <input type="submit" name="submit" value="Register" /> i want to help with my php scrip coz every hackd my script so please anyone help me for make secure script. |
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