PHP - This Beautiful Script Needs Some Help!
Okay, I need to make it so it shows " ago " after the time it states.
It shows "2minutes 20 seconds" but I can't figure out how to make it say "2minutes 20 seconds ago" (simply just need to add "ago" after) This uses alot of ternary operators, and I am not very knowledgable in that area, so if You can help, it would be great. Code: [Select] function timeAgo($tm,$rcs = 1) { $cur_tm = time(); $dif = $cur_tm-$tm; $pds = array('second','minute','hour','day','week','month','year','decade'); $lngh = array(1,60,3600,86400,604800,2630880,31570560,315705600); for($v = sizeof($lngh)-1; ($v >= 0)&&(($no = $dif/$lngh[$v])<=1); $v--); if($v < 0) $v = 0; $_tm = $cur_tm-($dif%$lngh[$v]); $no = floor($no); if($no <> 1) $pds[$v] .='s'; $x=sprintf("%d %s ",$no,$pds[$v]); if(($rcs > 0)&&($v >= 1)&&(($cur_tm-$_tm) > 0)) $x .= $this->timeAgo($_tm, --$rcs); return $x; } Simply, just need to add "ago" somewhere, but I can't figure it out, this uses unixtimestamp also. Similar TutorialsExperienced, skilled and perfectionist coded needed for a small, minimal application.
Skills
- Skilled using raw PHP (no framework), utilising PHP5, in OOP structure
- Experienced designing optimised and clean MYSQL schemas
- Concious of both performance and security
- Understanding of Jquery
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- Knowledge of stripe
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- Basic HTML / CSS
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Edited by jameswillson, 03 September 2014 - 08:47 AM. 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 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 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 error Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/nrkgamin/public_html/perp/gameserver/fetch_users5.php on line 7 Code: [Select] <?php mysql_connect("xxxxxxx", "xxxxxxxx", "xxxxxxx"); mysql_select_db("phpbb"); $query = mysql_query("SELECT phpbb_profile_fields_data.pf_steamid, phpbb_users.user_rank FROM user, phpbb_profile_fields_data WHERE (phpbb_users.user_rank='16' OR phpbb_users.user_rank='10' OR phpbb_users.user_rank='9' OR phpbb_users.user_rank='14' OR phpbb_users.user_rank='11' OR phpbb_users.user_rank='15' OR phpbb_users.user_rank='13' OR phpbb_users.user_rank='12' OR phpbb_users.user_rank='44' OR phpbb_users.user_rank='47') AND phpbb_users.user_id=phpbb_profile_fields_data.user_id"); while ($row = mysql_fetch_assoc($query)){ if ($row['pf_steamid'] != "" && $row['user_rank'] != ""){ echo $row['pf_steamid'] . "\t" . $row['user_rank'] . "\n"; } } ?>srr for my bad english Kind of inherited this project and could use some assistance. The script below worked on a windows server running IIS and php 4. Upgraded server and PHP to version 5.5.3 and the script no longer works. Keep getting an error of undefined variable for the following lines of code:
$name=strtoupper($name); This does not work correctly and I need help with it. If $rating is equal to zero it displays 4 stars on this page http://rwdev.whekle.com/view.php?id=4571 what I am doing wrong, it is probably a simple mistake. Code: [Select] if($rating <= '1.5'){ $star1 = "yes"; }elseif($rating <= '2.5'){ $star2 = "yes"; }elseif($rating <= '3.5'){ $star3 = "yes"; }elseif($rating <= '4.5'){ $star4 = "yes"; }elseif($rating <= '5.0'){ $star5 = "yes"; } i want to help with my php scrip coz every hackd my script so please anyone help me for make secure script. I need help with my script! I search the database and post the thumbnail pictures, I have an href statement that I want the link I put in the database to link to the picture. When I look at the html code the href is blank. I don't know what wrong, please help. Below is the code: Thanks for your help. <?php include 'config1.php'; include 'form.php'; $search=$_GET["search"]; // Connect to server and select database. mysql_connect($dbhost, $dbuser, $dbpass)or die("cannot connect"); mysql_select_db("xxxxxx")or die("cannot select DB"); $result = mysql_query("SELECT * FROM $dbname WHERE year LIKE '%$search%'") or die(mysql_error()); // store the record of the "" table into $row //$current = ''; // keeps getting the next row until there are no more to get if($result && mysql_num_rows($result) > 0) { $i = 0; $max_columns = 6; echo "<table>"; echo "<br>"; while($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><a href="<?php echo $link; ?>"><img src="<?php echo $tn; ?>"></a></div> <div><?php echo $title; ?></div> </div> <?php echo "</td>"; // increment counter - if counter = max columns, reset counter and close row if(++$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(); ?> Hi i have 2 days setup so far but when i go to add more days the script does not work. can you please tell me how to add a extra 5 days to the below script please. <html> <head> <meta http-equiv="refresh" content="180"> </head> <body> <?php //set timezone putenv ('GMT=Europe/London'); mktime(0,0,0,1,1,1970); $day = date("n"); $time = date("Hi"); if ($day == 7) { switch($time) { case ($time >= '0000' && $time <= '0459'): echo"<img src='http://lcradio.co.uk/presenters/blazey.png' class='onair'/>"; break; case ($time >= '0500' && $time <= '0600'): echo"<img src='http://lcradio.co.uk/presenters/gilly.png' class='onair'/>"; break; case ($time >= '0600' && $time <= '0900'): echo"<img src='http://lcradio.co.uk/presenters/stephen.png' class='onair'/>"; break; case ($time >= '0900' && $time <= '1200'): echo"<img src='http://lcradio.co.uk/presenters/homer.gif'class='onair'/>"; break; case ($time >= '1200' && $time <= '1400'): echo"<img src='http://lcradio.co.uk/presenters/gilly.png' class='onair'/>"; break; case ($time >= '1400' && $time <= '1600'): echo"<img src='http://lcradio.co.uk/presenters/rich.png'class='onair'/>"; break; case ($time >= '1600' && $time <= '1800'): echo"<img src='http://lcradio.co.uk/presenters/blazey.png' class='onair'/>"; break; case ($time >= '1800' && $time <= '2000'): echo"<img src='http://lcradio.co.uk/presenters/martan.png'class='onair'/>"; break; case ($time >= '2000' && $time <= '2400'): echo"<img src='http://lcradio.co.uk/presenters/gilly.png'class='onair'/>"; break; default: echo"<img src='http://lcradio.co.uk/presenters/blazey.png' alt='On-Air Now' class='onair'/>"; break; } } elseif($day == 6) { echo"<img src='http://lcradio.co.uk/presenters/rich.png'alt='KNAS' class='onair'/>"; } else { switch($time) { case ($time >= '0000' && $time <= '0459'): echo"<img src='http://lcradio.co.uk/presenters/gilly.png' class='onair'/>"; break; case ($time >= '0500' && $time <= '0600'): echo"<img src='http://lcradio.co.uk/presenters/stephen.png'class='onair'/>"; break; case ($time >= '0600' && $time <= '0900'): echo"<img src='http://lcradio.co.uk/presenters/blazey.png' class='onair'/>"; break; case ($time >= '0900' && $time <= '1200'): echo"<img src='http://lcradio.co.uk/presenters/rich.png'class='onair'/>"; break; case ($time >= '1200' && $time <= '1400'): echo"<img src='http://lcradio.co.uk/presenters/stephen.png'class='onair'/>"; break; case ($time >= '1400' && $time <= '1600'): echo"<img src='http://lcradio.co.uk/presenters/homer.gif'class='onair'/>"; break; case ($time >= '1600' && $time <= '1800'): echo"<img src='http://lcradio.co.uk/presenters/stephen.png'class='onair'/>"; break; case ($time >= '1800' && $time <= '2000'): echo"<img src='http://lcradio.co.uk/presenters/martan.png'class='onair'/>"; break; case ($time >= '2000' && $time <= '2400'): echo"<img src='http://lcradio.co.uk/presenters/homer.gif' class='onair'/>"; break; } } ?> </body> </html> I'm working on this application that allows the user to submit up to 10 files into the server and database. The files can be either images or video files.. What i'm looking for is for people in this forum to see if they can look it over and see if they find anything that can be fixed that might screw things up. Also I have yet to test it, i know i should test first and them come here for help, but i currently don't have a server to test it until tomorrow..so all i need is to see if you guys can look it over and let me know if i'm doing something wrong if there would be a better way to do it Thankyou <?php if(isset($_POST['submit'])){ $dir = 'images/content'; $date_sub = date('m/d/y'); $time_sub = date('h:i A'); $ip = $_SERVER['REMOTE_ADDR']; foreach($_FILES['file']['name'] as $key => $file){ $type = mysql_real_escape_string($_POST['type']); $fileupload = $dir.$file; move_uploaded_file($_FILES['file']['tmp_name'][$key], $fileupload); $add_files = mysql_query("INSERT INTO content (contentURL, contentType, profileID, date_submitted, time_submitted, ip_log) VALUES('$file', '$type', '$profileID', '$time_sub', '$date_sub', 'ip')") or die(mysql_error()); } } ?> I cant figure out why my log out script doesnt work all you need is session_start(); session_destroy(); right? here is my script to keep people loged in which works fine. Code: [Select] if ($_SESSION['login_time'] < strtotime('-60 minutes')) { header("Location: signup.php"); exit(); session_destroy(); I need PHP script for create a program i have 1 header.html (which is my header file) 2 menu.html(which contain menu on right of the page) 3 footer.html(which is my footer) 4 content.html(which is my content area of the page) 5 multiple around 10 .php form pages now i want to use them into single page which have header,menu,footer as same & my php forms call on content area. so please help me out |