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An7hony

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Everything posted by An7hony

  1. thanks!!
  2. I'm trying to use the js below to check whether lunch is inputted as 0.25, 0.5, 0.75 or 1. Below is the code. It seams to be working as: if ((day_object[dayname].lunch) != "0.25"){ but as if (((((day_object[dayname].lunch) != "0.25") || (day_object[dayname].lunch) != "0.5") || (day_object[dayname].lunch) != "0.75") || (day_object[dayname].lunch) != "1") { it pulls the alert whatever is inputted. else if ((day_object[dayname].start_hhmm != "") && (day_object[dayname].finish_hhmm != "") && (day_object[dayname].lunch != "")) { if (((((day_object[dayname].lunch) != "0.25") || (day_object[dayname].lunch) != "0.5") || (day_object[dayname].lunch) != "0.75") || (day_object[dayname].lunch) != "1") { alert("Check lunch hour 0.25 is 1/4 of an hour. 0.5 is 1/2 an hour. 0.75 is 3/4 of an hour. 1 is 1 hour"); day_object[dayname] = new oDay(day_object[dayname].dayname, day_object[dayname].start_hhmm, day_object[dayname].finish_hhmm, 0); } }
  3. thanks, you are right. What a pain..
  4. $largedate has: 25/10/2011 $smalldate has: 02/10/2011
  5. Can anyone help? the date is pulling through without a value shown in bold below. `Timesheet_start_date` >= '2011-02-10 00:00:00' AND `Timesheet_end_date` <= '1970-01-01 00:00:00' ORDER BY id DESC here is my code: $smalldate = $_POST['small']; $largedate = $_POST['large']; $types[] = (isset($_POST['small']) && strlen(trim($_POST['small'])) > 0) ? "`Timesheet_start_date` >= '".date('Y-m-d',strtotime($smalldate))." 00:00:00'" : ''; $types[] = (isset($_POST['large']) && strlen(trim($_POST['large'])) > 0) ? "`Timesheet_end_date` <= '".date('Y-m-d',strtotime($largedate))." 00:00:00'" : ''; i've echo the $largedate and there is a value in there?
  6. i fixed it by putting it all into one function function checkLogin(){ global $database; //The database connection /* Check if user has been remembered */ if(isset($_COOKIE['cookname']) && isset($_COOKIE['cookid'])){ $this->username = $_SESSION['username'] = $_COOKIE['cookname']; $this->userid = $_SESSION['userid'] = $_COOKIE['cookid']; } /* Username and userid have been set and not guest */ if(isset($_SESSION['username']) && isset($_SESSION['userid']) && $_SESSION['username'] != GUEST_NAME){ /* Confirm that username and userid are valid */ if($database->confirmUserID($_SESSION['username'], $_SESSION['userid']) != 0){ /* Variables are incorrect, user not logged in */ // unset($_SESSION['username']); // unset($_SESSION['userid']); return false; } /* User is logged in, set class variables */ $this->userinfo = $database->getUserInfo($_SESSION['username']); $this->username = $this->userinfo['username']; $this->userid = $this->userinfo['userid']; $this->userlevel = $this->userinfo['userlevel']; return true; } /* User not logged in */ else{ if(isset($_COOKIE['cookname']) && isset($_COOKIE['cookid'])){ $this->clientname = $_SESSION['username'] = $_COOKIE['cookname']; $this->clientid = $_SESSION['userid'] = $_COOKIE['cookid']; } /* Username and userid have been set and not guest */ if(isset($_SESSION['username']) && isset($_SESSION['userid']) && $_SESSION['username'] != GUEST_NAME){ /* Confirm that username and userid are valid */ if($database->confirmClientID($_SESSION['username'], $_SESSION['userid']) != 0){ /* Variables are incorrect, user not logged in */ unset($_SESSION['username']); unset($_SESSION['userid']); return false; } /* User is logged in, set class variables */ $this->clientinfo = $database->getUserClient($_SESSION['username']); $this->clientname = $this->clientinfo['username']; $this->clientid = $this->clientinfo['userid']; $this->clientlevel = $this->clientinfo['userlevel']; return true; } /* User not logged in */ else{ return false; } } }
  7. function checkLogin(){ global $database; //The database connection /* Check if user has been remembered */ if(isset($_COOKIE['cookname']) && isset($_COOKIE['cookid'])){ $this->username = $_SESSION['username'] = $_COOKIE['cookname']; $this->userid = $_SESSION['userid'] = $_COOKIE['cookid']; } /* Username and userid have been set and not guest */ if(isset($_SESSION['username']) && isset($_SESSION['userid']) && $_SESSION['username'] != GUEST_NAME){ /* Confirm that username and userid are valid */ if($database->confirmUserID($_SESSION['username'], $_SESSION['userid']) != 0){ /* Variables are incorrect, user not logged in */ unset($_SESSION['username']); unset($_SESSION['userid']); return false; } /* User is logged in, set class variables */ $this->userinfo = $database->getUserInfo($_SESSION['username']); $this->username = $this->userinfo['username']; $this->userid = $this->userinfo['userid']; $this->userlevel = $this->userinfo['userlevel']; return true; } /* User not logged in */ else{ return false; } }
  8. How can i switch between the 2 lines below in my function. At the moment the bottom line is ignored: /* Determine if user is logged in */ $this->logged_in = $this->checkLogin(); /* Determine if client is logged in */ $this->logged_in_Client = $this->checkLoginClient(); function startSession(){ global $database; //The database connection session_start(); //Tell PHP to start the session /* Determine if user is logged in */ $this->logged_in = $this->checkLogin(); /* Determine if client is logged in */ $this->logged_in_Client = $this->checkLoginClient(); if($this->logged_in){ $database->addActiveUser($this->username, $this->time); } if($this->logged_in_Client){ $database->addActiveClient($this->clientname, $this->time); } /* Set referrer page */ if(isset($_SESSION['url'])){ $this->referrer = $_SESSION['url']; }else{ $this->referrer = "/"; } /* Set current url */ $this->url = $_SESSION['url'] = $_SERVER['REQUEST_URI']; }
  9. thank you!! what a stupid mistake, now i feel silly
  10. I'm having the strangest thing happing. When posting a form i'm not getting the option value. Instead its posting my text field ie: <option name="1">text</option> <?php $query = "SELECT id, name FROM locations ORDER BY name"; $res = mysql_query($query, $cid); while($a = mysql_fetch_array($res)) { echo "<option name=".$a["id"].">".$a["name"]."</option>"; } ?> <?php The code thats grapping it: if($_POST['location']==""){ $location = ""; echo '<div class="message error">Please enter a <strong>Location</strong></div>'; }else{ $location = $_POST['location']; } ?> Can anyone see why? I have also tried it in html to see if its my php, but i get the same result
  11. ok i added: $content2 = file_get_contents($file); i have the attachment although when i open the file it is reading code and symbols?
  12. this is similar to what i'm getting instead of the file attachment: This is the body of the message. --frontier Content-Type: application/octet-stream Content-Transfer-Encoding: base64 PGh0bWw+CiAgPGhlYWQ+CiAgPC9oZWFkPgogIDxib2R5PgogICAgPHA+VGhpcyBpcyB0aGUg Ym9keSBvZiB0aGUgbWVzc2FnZS48L3A+CiAgPC9ib2R5Pgo8L2h0bWw+Cg== --frontier-- http://en.wikipedia.org/wiki/MIME
  13. ok i am not getting any errors now. This is what i have: function sendNewCandidate($user, $email, $agent, $jobtitle, $useremail, $cv) { $file = $_SERVER['DOCUMENT_ROOT']."/elements/".$cv; $file_size = filesize($file); $handle = fopen($file, "r"); $content = fread($handle, $file_size); fclose($handle); $content = chunk_split(base64_encode($content)); $uid = md5(uniqid(time())); $name = basename($file); $subject = "New applicant"; $header = "From: ".$user." <".$useremail.">\r\n"; $header .= "Reply-To: ".$useremail."\r\n"; $header .= "MIME-Version: 1.0\r\n"; $header .= "Content-Type: multipart/mixed; boundary=\"".$uid."\"\r\n\r\n"; $header .= "This is a multi-part message in MIME format.\r\n"; $header .= "--".$uid."\r\n"; $header .= "Content-type:text/plain; charset=iso-8859-1\r\n"; $header .= "Content-Transfer-Encoding: 7bit\r\n\r\n"; $header .= $agent.",\n\n" ."".$user." has applied for your job ".$jobtitle."." ."We suggest you go to your admin panel and view the CV" ."- Web Site. "; $header .= "--".$uid."\r\n"; $header .= "Content-Type: application/octet-stream; name=\"".$name."\"\r\n"; // use different content types here $header .= "Content-Transfer-Encoding: base64\r\n"; $header .= "Content-Disposition: attachment; filename=\"".$name."\"\r\n\r\n"; $header .= $content."\r\n\r\n"; $header .= "--".$uid."--"; return mail($email, $subject, "", $header); } }; Although instead of the file being attached to the email i am getting 0M8R4KGxGuEAAAAAAAAAAAAAAAAAAAAAPgADAP7/CQAGAAAAAAAAAAAAAAABAAAAIgAAAAAAAAAA etc etc
  14. ah it may be because $cv reads path/filename.doc instead of filename.doc
  15. At the moment i am getting: Warning: filesize() [function.filesize]: stat failed for http://www.website.co.uk/elements/$cv in /home/*****/public_html/elements/mailer.php on line 93 Warning: fread() [function.fread]: Length parameter must be greater than 0 in /home/******/public_html/elements/mailer.php on line 95 function sendNewCandidate($user, $email, $agent, $jobtitle, $useremail, $cv) { $file = 'http://www.website.co.uk/elements/$cv'; $file_size = filesize($file); $handle = fopen($file, "r"); $content = fread($handle, $file_size); fclose($handle); $content = chunk_split(base64_encode($content)); $uid = md5(uniqid(time())); $name = basename($file); $subject = "New applicant"; $header = "From: ".$user." <".$useremail.">\r\n"; $header .= "Reply-To: ".$useremail."\r\n"; $header .= "MIME-Version: 1.0\r\n"; $header .= "Content-Type: multipart/mixed; boundary=\"".$uid."\"\r\n\r\n"; $header .= "This is a multi-part message in MIME format.\r\n"; $header .= "--".$uid."\r\n"; $header .= "Content-type:text/plain; charset=iso-8859-1\r\n"; $header .= "Content-Transfer-Encoding: 7bit\r\n\r\n"; $header .= $agent.",\n\n" ."".$user." has applied for your job ".$jobtitle."." ."We suggest you go to your admin panel and view the CV" ."- Website. "; $header .= "--".$uid."\r\n"; $header .= "Content-Type: application/octet-stream; name=\"".$cv."\"\r\n"; // use different content types here $header .= "Content-Transfer-Encoding: base64\r\n"; $header .= "Content-Disposition: attachment; filename=\"".$cv."\"\r\n\r\n"; $header .= $content."\r\n\r\n"; $header .= "--".$uid."--"; return mail($email, $subject, "", $header); } }; can anyone explain? Thanks
  16. i have just read this, so i'm going to try it out: $my_file = "somefile.zip"; $my_path = $_SERVER['DOCUMENT_ROOT']."/your_path_here/"; $my_name = "Olaf Lederer"; $my_mail = "my@mail.com"; $my_replyto = "my_reply_to@mail.net"; $my_subject = "This is a mail with attachment."; $my_message = "Hallo,\r\ndo you like this script? I hope it will help.\r\n\r\ngr. Olaf"; mail_attachment($my_file, $my_path, "recipient@mail.org", $my_mail, $my_name, $my_replyto, $my_subject, $my_message);
  17. Hi I need to attach a file to an email from my database records So far i have: function sendNewCandidate($user, $email, $agent, $jobtitle, $useremail, $cv) { $from = "From: ".$user." <".$useremail.">"; $subject = "New applicant"; $attachment = chunk_split(base64_encode(file_get_contents('http://www.website.co.uk/$cv'))); $body = $agent.",\n\n" ."".$user." has applied for your job ".$jobtitle."." ."We suggest you go to your admin panel and view the CV" ."- Thank you. "; return mail($email, $subject, $body, $from); } }; $cv holds the location of the file for the user. These files are generally .doc I have read i may need something like this: Content-Type: application/zip; name="attachment.zip" Content-Transfer-Encoding: base64 Content-Disposition: attachment but how would i incorporate this? Thanks guys
  18. cool thanks guys echo "Job inserted!", mysql_insert_id(); worked fine
  19. so just to confirm i would write this echo "Job inserted!", mysql_insert_id();
  20. How would i return the id in the echo of this post: $query = "INSERT INTO jobs SET agent = '".mysql_real_escape_string($agent)."', title = '".mysql_real_escape_string($title)."', location = '".mysql_real_escape_string($location)."', salary_from = '".$salary_from."', salary_to = '".$salary_to."', frequency = '".$frequency."', jobtype='".$jobtype."', description='".$description."', email = '".$email."', created_at = '".$created_at."', closing_date ='".$closing_date."', categories = '".$categories."', sectors = '".$sectors."', ref='".$ref."'"; if(mysql_query($query)) { echo "Job inserted!"; } else { die(mysql_error()); } Thanks for the help guys
  21. An7hony

    joins

    i've added GROUP BY norton_rent.Prop_VebraID and its working ok now Thanks
  22. An7hony

    joins

    Hi i am trying to do a join between 2 tables but it is pulling duplicate results. i.e: $query = "SELECT norton_rent.id, norton_rent.Prop_DisplayAddress, norton_rent.Prop_DisplayAddress, norton_rent.Prop_PropType, norton_rent.Prop_Price, norton_rent.Prop_Bedrooms, norton_rent.Prop_Description, norton_rent.Prop_VebraID, norton_rent_img.File_url, norton_rent_img.Prop_VebraID, norton_rent_img.File_index FROM norton_rent LEFT JOIN norton_rent_img ON (norton_rent.Prop_VebraID = norton_rent_img.Prop_VebraID) WHERE norton_rent.Prop_Bedrooms LIKE '%".$_GET['beds']."%' AND norton_rent.Prop_Price < '".$_GET['maxsaleprice']."' AND norton_rent.Prop_VebraID=norton_rent_img.Prop_VebraID AND norton_rent_img.File_index=0 LIMIT ".$offset.", ".$per_page.""; $result = mysql_query($query) or die(mysql_error()); Its ok if i take out WHERE norton_rent.Prop_Bedrooms LIKE '%".$_GET['beds']."%' AND norton_rent.Prop_Price < '".$_GET['maxsaleprice']."' just wondered if anyone can see why? thanks
  23. Thanks a lot!!! Works perfectly if(isset($_GET['salary_to'])) { $salary = explode(":", $_GET['salary_to']); } $types[] = (isset($_GET['salary_to']) && strlen(trim($_GET['salary_to'])) > 0) ? "`salary_from` <= " . intval($salary[1]) . " AND `salary_to` >= " . intval($salary[0]) ." " : '';
  24. it seemed to work but say my search is 15001 - 20000 it wont show a job which is 15000 - 22000
  25. will it be ok to do: if(isset($_GET['salary_to'])) { $salary = explode(":", $_GET['salary_to']); } $types[] = (isset($_GET['salary_to']) && strlen(trim($_GET['salary_to'])) > 0) ? "`salary_from` >= " . intval($salary[0]) . " AND `salary_to` >= " . intval($salary[1]) ." " : '';
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