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  • PHP Parse Error unexpected '{'

    - by Laxmidi
    Hi, I'm getting a "Parse error: syntax error, unexpected '{' in line 2". And I don't see the problem. <?php class pointLocation {     var $pointOnVertex = true; // Check if the point sits exactly on one of the vertices     function pointLocation() {     }                   function pointInPolygon($point, $polygon, $pointOnVertex = true) {         $this->pointOnVertex = $pointOnVertex;                  // Transform string coordinates into arrays with x and y values         $point = $this->pointStringToCoordinates($point);         $vertices = array();          foreach ($polygon as $vertex) {             $vertices[] = $this->pointStringToCoordinates($vertex);          }                  // Check if the point sits exactly on a vertex         if ($this->pointOnVertex == true and $this->pointOnVertex($point, $vertices) == true) {             return "vertex";         }                  // Check if the point is inside the polygon or on the boundary         $intersections = 0;          $vertices_count = count($vertices);              for ($i=1; $i < $vertices_count; $i++) {             $vertex1 = $vertices[$i-1];              $vertex2 = $vertices[$i];             if ($vertex1['y'] == $vertex2['y'] and $vertex1['y'] == $point['y'] and $point['x'] > min($vertex1['x'], $vertex2['x']) and $point['x'] < max($vertex1['x'], $vertex2['x'])) { // Check if point is on an horizontal polygon boundary                 return "boundary";             }             if ($point['y'] > min($vertex1['y'], $vertex2['y']) and $point['y'] <= max($vertex1['y'], $vertex2['y']) and $point['x'] <= max($vertex1['x'], $vertex2['x']) and $vertex1['y'] != $vertex2['y']) {                  $xinters = ($point['y'] - $vertex1['y']) * ($vertex2['x'] - $vertex1['x']) / ($vertex2['y'] - $vertex1['y']) + $vertex1['x'];                  if ($xinters == $point['x']) { // Check if point is on the polygon boundary (other than horizontal)                     return "boundary";                 }                 if ($vertex1['x'] == $vertex2['x'] || $point['x'] <= $xinters) {                     $intersections++;                  }             }          }          // If the number of edges we passed through is even, then it's in the polygon.          if ($intersections % 2 != 0) {             return "inside";         } else {             return "outside";         }     }               function pointOnVertex($point, $vertices) {         foreach($vertices as $vertex) {             if ($point == $vertex) {                 return true;             }         }          }                   function pointStringToCoordinates($pointString) {         $coordinates = explode(" ", $pointString);         return array("x" => $coordinates[0], "y" => $coordinates[1]);     }           } $pointLocation = new pointLocation(); $points = array("30 19", "0 0", "10 0", "30 20", "11 0", "0 11", "0 10", "30 22", "20 20"); $polygon = array("10 0", "20 0", "30 10", "30 20", "20 30", "10 30", "0 20", "0 10", "10 0"); foreach($points as $key => $point) { echo "$key ($point) is " . $pointLocation->pointInPolygon($point, $polygon) . "<br>"; } ?> Does anyone see the problem? Thanks, -Laxmidi

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  • Parsing CSV File to MySQL DB in PHP

    - by Austin
    I have a some 350-lined CSV File with all sorts of vendors that fall into Clothes, Tools, Entertainment, etc.. categories. Using the following code I have been able to print out my CSV File. <?php $fp = fopen('promo_catalog_expanded.csv', 'r'); echo '<tr><td>'; echo implode('</td><td>', fgetcsv($fp, 4096, ',')); echo '</td></tr>'; while(!feof($fp)) { list($cat, $var, $name, $var2, $web, $var3, $phone,$var4, $kw,$var5, $desc) = fgetcsv($fp, 4096); echo '<tr><td>'; echo $cat. '</td><td>' . $name . '</td><td><a href="http://www.' . $web .'" target="_blank">' .$web.'</a></td><td>'.$phone.'</td><td>'.$kw.'</td><td>'.$desc.'</td>' ; echo '</td></tr>'; } fclose($file_handle); show_source(__FILE__); ?> First thing you will probably notice is the extraneous vars within the list(). this is because of how the excel spreadsheet/csv file: Category,,Company Name,,Website,,Phone,,Keywords,,Description ,,,,,,,,,, Clothes,,4imprint,,4imprint.com,,877-466-7746,,"polos, jackets, coats, workwear, sweatshirts, hoodies, long sleeve, pullovers, t-shirts, tees, tshirts,",,An embroidery and apparel company based in Wisconsin. ,,Apollo Embroidery,,apolloemb.com,,1-800-982-2146,,"hats, caps, headwear, bags, totes, backpacks, blankets, embroidery",,An embroidery sales company based in California. One thing to note is that the last line starts with two commas as it is also listed within "Clothes" category. My concern is that I am going about the CSV output wrong. Should I be using a foreach loop instead of this list way? Should I first get rid of any unnecessary blank columns? Please advise any flaws you may find, improvements I can use so I can be ready to import this data to a MySQL DB.

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  • PHP & MySQL Undefined variable problem

    - by comma
    I keep getting the following error Undefined variable: id on line 91 can some one help me correct this problem? The error is on this line. $query2 = "INSERT INTO users_skills (skill_id, user_id, date_created) VALUES ('$id', '$user_id', NOW())"; MySQL database tables. CREATE TABLE tags ( id INT UNSIGNED NOT NULL AUTO_INCREMENT, skill VARCHAR(255) NOT NULL, experience VARCHAR(255) NOT NULL, years VARCHAR(255) NOT NULL, PRIMARY KEY (id) ); CREATE TABLE users_skills ( id INT UNSIGNED NOT NULL AUTO_INCREMENT, skill_id INT UNSIGNED NOT NULL, user_id INT UNSIGNED NOT NULL, date_created DATETIME UNSIGNED NOT NULL, PRIMARY KEY (id) ); Here is the PHP & MySQL code. if (isset($_POST['info_submitted'])) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT learned_skills.*, users_skills.* FROM learned_skills INNER JOIN users_skills ON learned_skills.id = users_skills.skill_id WHERE user_id='$user_id'"); if (!$dbc) { print mysqli_error($mysqli); return; } $user_id = '5'; $skill = $_POST['skill']; $experience = $_POST['experience']; $years = $_POST['years']; $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT learned_skills.*, users_skills.* FROM learned_skills INNER JOIN users_skills ON users_skills.skill_id = learned_skills.id WHERE users_skills.user_id='$user_id'"); if (mysqli_num_rows($dbc) == 0) { if (isset($_POST['skill']) && trim($_POST['skill'])!=='') { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $query1 = mysqli_query($mysqli,"INSERT INTO learned_skills (skill, experience, years) VALUES ('" . $skill . "', '" . $experience . "', '" . $years . "')"); if (mysqli_query($mysqli, $query1)) { print mysqli_error($mysqli); return; } $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT id FROM learned_skills WHERE id='" . $skill . "' AND experience='" . $experience . "' AND years='" . $years . "'"); if (!$dbc) { print mysqli_error($mysqli); } else { while($row = mysqli_fetch_array($dbc)){ $id = $row["id"]; } } $query2 = "INSERT INTO users_skills (skill_id, user_id, date_created) VALUES ('$id', '$user_id', NOW())"; } }

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  • php zencart mod - having problems with attributes array

    - by user80151
    I inherited a zencart mod and can't figure out what's wrong. The customer selects a product and an attribute (model#). This is then sent to another form that they complete. When they submit the form, the product and the attribute should be included in the email sent. At this time, only the product is coming through. The attribute just says "array." The interesting part is, when I delete the line that prints the attribute, the products_options_names will print out. So I know that both the product and the products_options_names are working. The attribute is the only thing that is not working right. Here's what I believe to be the significant code. This is the page that has the form, so the attribute should already be passed to the form. //Begin Adding of New features //$productsimage = $product['productsImage']; $productsname = $product['productsName']; $attributes = $product['attributes']; $products_options_name = $value['products_options_name']; $arr_product_list[] = "<strong>Product Name:</strong> $productsname <br />"; $arr_product_list[] .= "<strong>Attributes:</strong> $attributes <br />"; $arr_product_list[] .= "<strong>Products Options Name:</strong> $products_options_name <br />"; $arr_product_list[] .= "---------------------------------------------------------------"; //End Adding of New features } // end foreach ($productArray as $product) ?> Above this, there is another section that has attributes: <?php echo $product['attributeHiddenField']; if (isset($product['attributes']) && is_array($product['attributes'])) { echo '<div class="cartAttribsList">'; echo '<ul>'; reset($product['attributes']); foreach ($product['attributes'] as $option => $value) { ?> Can anyone help me figure out what is wrong? I'm not sure if the problem is on this page or if the attribute isn't being passed to this page. TIA

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  • Sort array by two specifics values in PHP

    - by Roger
    The folks have already showed me how to sort an array by a specific value using usort and a fallback function in PHP. What if this specifc value doesn't exist and we have to use two values? in the example bellow the values [4] and [5]... In other words, I want to do this: order all objects numericaly by the fith value of each object from the highest to the lowest, and addicionally, for those objects that have the fifht value is empty (in the examplem '-'), order them by the fourth value. Array( [0] => Array( [0] => links-patrocinados [1] => adwords [2] => 0,5 [3] => R$92,34 [4] => 823000 [5] => 49500 ) [1] => Array( [0] => adwords [1] => google adwords como funciona [2] => 0,38 [3] => R$0,20 [4] => 480 [5] => 480 ) [2] => Array( [0] => links-patrocinados [1] => adword [2] => 0,39 [3] => R$58,77 [4] => 49500 [5] => 2900 ) [3] => Array( [0] => agencia [1] => agencias viagens espanholas [2] => - [3] => R$0,20 [4] => 58 [5] => - ) [4] => Array( [0] => agencia [1] => era agencia imobiliaria [2] => - [3] => R$0,20 [4] => 73 [5] => - ) )

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  • PHP, MySQL: Display only required parts of my website in sister website

    - by Devner
    Hi all, Now I have my website built on PHP & Mysql. Consider this like a forum. Now when a user posts a reply in my website 1 (ex. www.website1.com), I want to be able to show the starting thread and it's related replies in a sister website of mine. I want to do this in a way that it does not show the rest of the page & other page contents (like logo etc.). I don't think iframe would be a solution because an iframe would embed the whole page and the users visiting my sister website (totally different domain i.e. www.website2.com) would be able to see all the page contents, like logo etc. I want to avoid that. I want to make them see only limited information from website 1 and only the info. that I intend. I hope that makes sense. In a way, you could say that I am trying to replicate my 1 website, and show only a limited part of it. Users browsing 2nd website can post a reply in the 2nd website and it should automatically be posted & visible to the visitors of the website 1. Users of website 1 should not know that a user of website 2 has posted it. They would feel that some user from website 1 has posted it. Do I have to use 2 separate mysql DB or just 1? I think it would be problematic if I am trying to use different DB. I also feel I might have to face DB connectivity issues as I can connect to only 1 DB at a time. It's basically like users of website1.com should feel that they are replying to users of website1.com & users of website2.com should feel that they are replying to users of website2.com. (I need it this way to bridge the gap between them). At the same time I want to make the front end of the websites different so that they don't feel that they are replying to some other users outside the domain. These websites would be under my control and I will have access to the source code at any time. If I need to change the source code, these changes are welcome. Is this really possible? Thank you in advance.

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  • php connecting to mysql server(localhost) very slow

    - by Ahmad
    actually its little complicated: summary: the connection to DB is very slow. the page rendering takes around 10 seconds but the last statement on the page is an echo and i can see its output while the page is loading in firefox (IE is same). in google chrome the output becomes visible only when the loading finishes. loading time is approximately the same across browsers. on debugging i found out that its the DB connectivity that is creating problem. the DB was on another machine. to debug further. i deployed the DB on my local machine .. so now the DB connection is at 127.0.0.1 but the connectivity still takes long time. this means that the issue is with APACHE/PHP and not with mysql. but then i deployed my code on another machine which connects to DB remotely.and everything seems fine. basically the application uses couple of mod_rewrite.. but i removed all the .htaccess files and the slow connectivity issue remains.. i installed another APACHE on my machine and used default settings. the connection was still very slow. i added following statements to measure the execution time $stime = microtime(); $stime = explode(" ",$stime); $stime = $stime[1] + $stime[0]; // my code -- it involves connection to DB $mtime = microtime(); $mtime = explode(" ",$mtime); $mtime = $mtime[1] + $mtime[0]; $totaltime = ($mtime - $stime); echo $totaltime; the output is 0.0631899833679 but firebug Net panel shows total loading time of 10-11 seconds. same is the case with google chrome i tried to turn off windows firewall.. connectivity is still slow and i just can't quite find the reason.. i've tried multiple DB servers.. multiple apaches.. nothing seems to be working.. any idea of what might be the problem?

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  • PHP Array problems....if anyone can assist!

    - by Homer_J
    First off, the code which brings back my data into an array: function fetch_questions($page) { global $link; $proc = mysqli_prepare($link, "SELECT * FROM tques WHERE page = $page"); mysqli_stmt_bind_param($proc, "i", $page); mysqli_stmt_execute($proc); $rowq = array(); stmt_bind_assoc($proc, $rowq); // loop through all result rows while ($proc->fetch()) { // print_r($rowq); } mysqli_stmt_close($proc); mysqli_clean_connection($link); return($rowq); } Now, when I `print_r($rowq);' I get the following, which is all good: Array ( [questions] => q1 [qnum] => 1 [qtext] => I find my job meaningful [page] => 1 ) Array ( [questions] => q2 [qnum] => 2 [qtext] => I find my job interesting [page] => 1 ) Array ( [questions] => q3 [qnum] => 3 [qtext] => My work supports ABC's objective [page] => 1 ) Array ( [questions] => q4 [qnum] => 4 [qtext] => I am able to balance my work and home life [page] => 1 ) Array ( [questions] => q5 [qnum] => 5 [qtext] => I am clear about what is expected of me in my job [page] => 1 ) Array ( [questions] => q6 [qnum] => 6 [qtext] => My induction helped me to settle into my job [page] => 1 ) Array ( [questions] => q7 [qnum] => 7 [qtext] => I understand the ABC vision [page] => 1 ) Array ( [questions] => q8 [qnum] => 8 [qtext] => I know how what I do fits into my team's objectives [page] => 1 ) Now, in my php page I have the following piece of script: $questions = fetch_questions($page); And when I print_r $questions, as below: print_r($questions); I only get the following back from the array, 1 row: Array ( [questions] => q8 [qnum] => 8 [qtext] => I know how what I do fits into my team's objectives [page] => 1 ) Any ideas why that might be? Thanks in advance, Homer.

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  • Remove specific definitions from a variable in PHP

    - by Amit
    Hi everyone, I have a PHP mail script that validates name, email address, and phone number before sending the mail. This then means that the Name, Email address, and Phone fields are Required Fields. I want to have it so that the Name and EITHER Email or Phone are required. Such that if a name and phone are inputted, it sends the mail, or if a name and an email are inputted, it also sends the email. The way the script works right now is it has several IF statements that check for (1) name, (2) email and (3) phone. Here's an example of an if statement of the code: if ( ($email == "") ) { $errors .= $emailError; // no email address entered $email_error = true; } if ( !(preg_match($match,$email)) ) { $errors .= $invalidEmailError; // checks validity of email $email_error = true; } And here's how it sends the mail: if ( !($errors) ) { mail ($to, $subject, $message, $headers); echo "<p id='correct'>"; echo "?????? ????? ??????!"; echo "</p>"; } else { if (($email_error == true)) { $errors != $phoneError; /*echo "<p id='errors'>"; echo $errors; echo "</p>";*/ } if (($phone_error == true)) { $errors != $emailError; $errors != $invalidEmailError; /*echo "<p id='errors'>"; echo $errors; echo "</p>";*/ } echo "<p id='errors'>"; echo $errors; echo "</p>"; } This doesn't work though. Basically this is what I want to do: If no email address was entered or if it was entered incorrectly, set a variable called $email_error to be true. Then check for that variable, and if it's true, then remove the $phoneError part of the $errors variable. Man I hope I'm making some sense here. Does anyone know why this doesn't work? It reports all errors if all fields are left empty :( Thanks! Amit

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  • Insert Registration Data in MySQL using PHP

    - by J M 4
    I may not be asking this in the best way possible but i will try my hardest. Thank you ahead of time for your help: I am creating an enrollment website which allows an individual OR manager to enroll for medical testing services for professional athletes. I will NOT be using the site as a query DB which anybody can view information stored within the database. The information is instead simply stored, and passed along in a CSV format to our network provider so they can use as needed after the fact. There are two possible scenarios: Scenario 1 - Individual Enrollment If an individual athlete chooses to enroll him/herself, they enter their personal information, submit their payment information (credit/bank account) for processing, and their information is stored in an online database as Athlete1. Scenario 2 - Manager Enrollment If a manager chooses to enroll several athletes he manages/ promotes for, he enters his personal information, then enters the personal information for each athlete he wishes to pay for (name, address, ssn, dob, etc), then submits payment information for ALL athletes he is enrolling. This number can range from 1 single athlete, up to 20 athletes per single enrollment (he can return and complete a follow up enrollment for additional athletes). Initially, I was building the database to house ALL information regardless of enrollment type in a single table which housed over 400 columns (think 20 athletes with over 10 fields per athlete such as name, dob, ssn, etc). Now that I think about it more, I believe create multiple tables (manager(s), athlete(s)) may be a better idea here but still not quite sure how to go about it for the following very important reasons: Issue 1 If I list the manager as the parent table, I am afraid the individual enrolling athlete will not show up in the primary table and will not be included in the overall registration file which needs to be sent on to the network providers. Issue 2 All athletes being enrolled by a manager are being stored in SESSION as F1FirstName, F2FirstName where F1 and F2 relate to the id of the fighter. I am not sure technically speaking how to store multiple pieces of information within the same table under separate rows using PHP. For example, all athleteswill have a first name. The very basic theory of what i am trying to do is: If number_of_athletes 1, store F1FirstName in row 1, column 1 of Table "Athletes"; store F1LastName in row 1, column 2 of Table "Athletes"; store F2FirstName in row 2, column 1 of Table "Athletes"; store F2LastName in row 2, column 2 of table "Athletes"; Does this make sense? I know this question is very long and probably difficult so i appreciate the guidance.

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  • Converting this code from ASP to PHP

    - by jethomas
    I'll admit I'm a novice programmer and really the only experience I have is in classic ASP. I'm looking for a way to convert this asp code to PHP. For a customer who only has access to a linux box but also as a learning tool for me. Thanks in advance for the help: Recordset and Function: Function pd(n, totalDigits) if totalDigits > len(n) then pd = String(totalDigits-len(n),"0") & n else pd = n end if End Function 'declare the variables Dim Connection Dim Recordset Dim SQL Dim SQLDate SQLDate = Year(Date)& "-" & pd(Month(Date()),2)& "-" & pd(Day(Date()),2) 'declare the SQL statement that will query the database SQL = "SELECT * FROM tblXYZ WHERE element_8 = 2 AND element_9 > '" & SQLDate &"'" 'create an instance of the ADO connection and recordset objects Set Connection = Server.CreateObject("ADODB.Connection") Set Recordset = Server.CreateObject("ADODB.Recordset") 'open the connection to the database Connection.Open "PROVIDER=MSDASQL;DRIVER={MySQL ODBC 5.1 Driver};SERVER=localhost;UID=xxxxx;PWD=xxxxx;database=xxxxx;Option=3;" 'Open the recordset object executing the SQL statement and return records Recordset.Open SQL,Connection Display page/loop: Dim counter counter = 0 While Not Recordset.EOF counter = counter + 1 response.write("<div><td width='200' valign='top' align='center'><a href='" & Recordset("element_6") & "' style='text-decoration: none;'><div id='ad_header'>" & Recordset("element_3") & "</div><div id='store_name' valign='bottom'>" & Recordset("element_5") & "</div><img id='photo-small-img' src='http://xyz.com/files/" & Recordset("element_7") & "' /><br /><div id='ad_details'>"& Recordset("element_4") & "</div></a></td></div>") Recordset.MoveNext If counter = 3 Then Response.Write("</tr><tr>") counter = 0 End If Wend

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  • posting php code using jquery .html()

    - by Emmanuel Imwene
    simple query,but it's giving me a headache, i need a division to be updated with a changed session variable each time a user clicks on a name,i figured i'd use .html() using jquery to update the division, i don't know if you can do this, but here goes: $("#inner").html('<?php session_start(); if(file_exists($_SESSION['full'])||file_exists($_SESSION['str'])){ if(file_exists($_SESSION['full'])) { $full=$_SESSION['full']; $handlle = fopen($full, "r"); $contents = fread($handlle, filesize($full)); fclose($handlle); echo $contents; echo '<script type="text/javascript" src="jquery-1.8.0.min (1).js">'; echo '</script>'; echo '<script type="text/javascript">'; echo 'function loadLog(){ var oldscrollHeight = $("#inner").attr("scrollHeight") - 20; $.ajax({ url: \''.$_SESSION['full'].'\', cache: false, success: function(html){ $("#inner").html(html); //Insert chat log into the #chatbox div var newscrollHeight = $("#inner").attr("scrollHeight") - 20; if(newscrollHeight > oldscrollHeight){ $("#inner").animate({ scrollTop: newscrollHeight }, \'normal\'); //Autoscroll to bottom of div } }, }); } setInterval (loadLog, 2500);'; echo '</script>'; } else { $str=$_SESSION['str']; if(file_exists($str)) { $handle = fopen($str, 'r'); $contents = fread($handle, filesize($str)); fclose($handle); echo $contents; $full=$_SESSION['full']; $handlle = fopen($full, "r"); $contents = fread($handlle, filesize($full)); fclose($handlle); echo $contents; echo '<script type="text/javascript" src="jquery-1.8.0.min (1).js">'; echo '</script>'; echo '<script type="text/javascript">'; echo 'function loadLog(){ var oldscrollHeight = $("#inner").attr("scrollHeight") - 20; $.ajax({ url: \''.$_SESSION['str'].'\', cache: false, success: function(html){ $("#inner").html(html); //Insert chat log into the #chatbox div var newscrollHeight = $("#inner").attr("scrollHeight") - 20; if(newscrollHeight > oldscrollHeight){ $("#inner").animate({ scrollTop: newscrollHeight }, \'normal\'); //Autoscroll to bottom of div } }, }); } setInterval (loadLog, 2500);'; echo '</script>'; } } } ?>'); is that legal, if not, how would i accomplish this?

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  • Attach file to mail using php

    - by ktsixit
    Hi all, I've created a form which contains an upload field file and some other text fields. I'm using php to send the form's data via email and attach the file. This is the code I'm using but it's not working properly. The file is normally attached to the message but the rest of the data is not sent. $body="bla bla bla"; $attachment = $_FILES['cv']['tmp_name']; $attachment_name = $_FILES['cv']['name']; if (is_uploaded_file($attachment)) { $fp = fopen($attachment, "rb"); $data = fread($fp, filesize($attachment)); $data = chunk_split(base64_encode($data)); fclose($fp); } $headers = "From: $email<$email>\n"; $headers .= "Reply-To: <$email>\n"; $headers .= "MIME-Version: 1.0\n"; $headers .= "Content-Type: multipart/related; type=\"multipart/alternative\"; boundary=\"----=MIME_BOUNDRY_main_message\"\n"; $headers .= "X-Sender: $first_name $family_name<$email>\n"; $headers .= "X-Mailer: PHP4\n"; $headers .= "X-Priority: 3\n"; $headers .= "Return-Path: <$email>\n"; $headers .= "This is a multi-part message in MIME format.\n"; $headers .= "------=MIME_BOUNDRY_main_message \n"; $headers .= "Content-Type: multipart/alternative; boundary=\"----=MIME_BOUNDRY_message_parts\"\n"; $message = "------=MIME_BOUNDRY_message_parts\n"; $message .= "Content-Type: text/html; charset=\"utf-8\"\n"; $message .= "Content-Transfer-Encoding: quoted-printable\n"; $message .= "\n"; $message .= "$body\n"; $message .= "\n"; $message .= "------=MIME_BOUNDRY_message_parts--\n"; $message .= "\n"; $message .= "------=MIME_BOUNDRY_main_message\n"; $message .= "Content-Type: application/octet-stream;\n\tname=\"" . $attachment_name . "\"\n"; $message .= "Content-Transfer-Encoding: base64\n"; $message .= "Content-Disposition: attachment;\n\tfilename=\"" . $attachment_name . "\"\n\n"; $message .= $data; //The base64 encoded message $message .= "\n"; $message .= "------=MIME_BOUNDRY_main_message--\n"; $subject = 'bla bla bla'; $to="[email protected]"; mail($to,$subject,$message,$headers); Why isn't the $body data not sent? Can you help me fix it?

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  • PHP Shared Sessions across Domain

    - by bigstylee
    Hi, I have seen a few answers to this on SOO but most of these are concerned with the use of subdomains, of which none have worked for me. The common one being that the use of session.cookie_domain, which from my understanding will only work with subdomains. I am interested in a solution that deals with deals with entirely different domains (and includes the possibility of subdomains). Unfortunately project deadlines being what they are, time is not on my side, so I turn to SOO's expertise and experience. The current project brief is to be able to log into one site which currently only stores the user_id in the session and then be able to retrieve this value while on a different domain within the same server enviroment. Session data is being stored/retrieved from a database where the session id is the primary key. I am hoping to find a "light wieght" and "easy" to implement solution. The system is utlising an in-house Model View Controller design pattern, so all requests (including different domains) are run through a single bootstrap script. Using the domain name as a variable, this determines what context to display to the user. One option that did look like to have potential is the use of a hidden image and using the alt tag to set the user id. My first impressions suggest this immediately seems "too easy" (if possible) and riddled with security flaws. Disscuss? Another option which I considered is using the IP and User Agent for authentication but again I feel this not going to be a reliable option due to shared networks and changing IP addresses. My third option (and preferred) which I considered and as yet not seen discussed is using htaccess to fool the user into thinking that they are on a different domain when infact apache is redirecting; something like www.foo.com/index.php?domain=bar.com&controller=news/categoires/1 but displays to the user as www.bar.com/news/categories/1 foo.com represents the "main site domain" which all requests are run through and bar.com is what the user thinks they are accessing. The controller request dictates the page and view being requested. Is this possible? Are there other options? Pros/Cons? Thanks in advanced!!!

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  • PHP - not sure how to ask - regarding variables and $_POST

    - by Phil
    I have a PHP form. The form works but I'm trying to test to see if a value other than the first item has been selected. I can't figure out how to write the If statement. $products = array( '' => 1, 'Item 2' => 2, 'Item 3' => 3, 'Item 4' => 4, 'Item 5' => 5, 'Item 6' => 6 ); $html = generateSelect('products', $products); function generateSelect($name = '', $options = array()) { $html = '<select name="'.$name.'">'; foreach ($options as $option => $value) { $html .= '<option value='.$value.'>'.$option.'</option>'; } $html .= '</select>'; return $html; } In my table, the drop down box is displayed: <tr> <td style="width:{$left_col_width}; text-align:left; vertical-align:center; padding:{$cell_padding}; font-weight:bold; {$product[3]}">{$product[0]}</td> <td style="text-align:left; vertical-align:top; padding:{$cell_padding};"><select name="{$product[1]}"> <option value="1"></option> <option value="2">Item 2</option> <option value="3">Item 3</option> <option value="4">Item 4</option> <option value="5">Item 5</option> <option value="6">Item 6</option> </select></td> </tr> I use the following if statement to check to see if someone has entered a phone number. if they have not entered a phone number, then the "Phone:" text turns red. How do I do an if statement similar to this to verify that someone has selected a product option from the drop down box? if(!empty($_POST['phone'])) { $phone[2] = clean_var($_POST['phone']); if (function_exists('htmlspecialchars')) $phone[2] = htmlspecialchars($phone[2], ENT_QUOTES); } else { $error = 1; $phone[3] = 'color:#d20128;'; } it seems simple but I can't figure it out.

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  • Securing paths in PHP

    - by tjm
    I'm writing some PHP which takes some paths to different content directories, and uses these to include various parts of pages later. I'm trying to ensure that the paths are as they seem, and none of them break the rules of the application. I have PRIVATEDIR which must lie above DOCUMENT_ROOT (aka) PUBLICDIR. CONTENTDIR which must lie within PRIVATEDIR and not go back below PUBLICDIR and some other *DIR's which must remain within CONTENTDIR. Currently I set up some defaults, and then override the ones the user specifies and then sanity check them with the following. private function __construct($options) { error_reporting(0); if(is_array($options)) { $this->opts = array_merge($this->opts, $options); } if($this->opts['STATUS']==='debug') { error_reporting(E_ALL | E_NOTICE | E_STRICT); } $this->opts['PUBLICDIR'] = realpath($_SERVER['DOCUMENT_ROOT']) .DIRECTORY_SEPARATOR; $this->opts['PRIVATEDIR'] = realpath($this->opts['PUBLICDIR'] .$this->opts['PRIVATEDIR']) .DIRECTORY_SEPARATOR; $this->opts['CONTENTDIR'] = realpath($this->opts['PRIVATEDIR'] .$this->opts['CONTENTDIR']) .DIRECTORY_SEPARATOR; $this->opts['CACHEDIR'] = realpath($this->opts['PRIVATEDIR'] .$this->opts['CACHEDIR']) .DIRECTORY_SEPARATOR; $this->opts['ERRORDIR'] = realpath($this->opts['CONTENTDIR'] .$this->opts['ERRORDIR']) .DIRECTORY_SEPARATOR; $this->opts['TEMPLATEDIR' = realpath($this->opts['CONTENTDIR'] .$this->opts['TEMPLATEDIR']) .DIRECTORY_SEPARATOR; // then here I have to check that PRIVATEDIR is above PUBLICDIR // and that all the rest remain within private dir and don't drop // down into (or below) PUBLICDIR again. And die with an error if // they don't conform. } The thing is this seems like a lot of work to do, especially as it must be run, every time a page is accessed, before I can do anything else, e.g check for a cached version of the page I'm serving. Part of me is thinking, since all of these paths are predefined by the maintainer of the site, they SHOULD be aware of what paths they are allowing access to and ensuring they are secure. But, I think I'm thinking that because currently I am said maintainer, and I KNOW my paths conform to the rules. That said, I do want to secure this thing from any accidental errors by future maintainers (and I bet, now I've said above "I KNOW...", probably from myself somewhere down the line). This just feels like a suboptimal solution. I wonder how fast this would really be and what you would suggest to improve it or as an alternative? Thanks.

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  • php code is not fetching data from mysql database using wamp server

    - by john
    I want to display a table from database in phpMyAdmin by putting the following conditions that in every different options in drop down menu it displays different table from database by pressing the button of search. But it is not doing so. <p class="h2">Quick Search</p> <div class="sb2_opts"> <p></p> <form method="post" action="" > <p>Enter your source and destination.</p> <p>From:</p> <select name="from"> <option value="Islamabad">Islamabad</option> <option value="Lahore">Lahore</option> <option value="murree">Murree</option> <option value="Muzaffarabad">Muzaffarabad</option> </select> <p>To:</p> <select name="To"> <option value="Islamabad">Islamabad</option> <option value="Lahore">Lahore</option> <option value="murree">Murree</option> <option value="Muzaffarabad">Muzaffarabad</option> </select> <input type="submit" value="search" /> </form> </form> </table> <?php $con=mysqli_connect("localhost","root","","test"); if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } if(isset($_POST['from']) and isset($_POST['To'])) { $from = $_POST['from'] ; $to = $_POST['To'] ; $table = array($from, $to); switch ($table) { case array ("Islamabad", "Lahore") : $result = mysqli_query($con,"SELECT * FROM flights"); echo "</flights>"; //table name is flights break; case array ("Islamabad", "Murree") : $result = mysqli_query($con,"SELECT * FROM isb to murree"); echo "</isb to murree>"; //table name isb to murree ; break; case array ("Islamabad", "Muzaffarabad") : $result = mysqli_query($con,"SELECT * FROM isb to muzz"); echo "</isb to muzz>"; break; //..... //...... default: echo "Your choice is nor valid !!"; } } mysqli_close($con); ?>

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  • PHP timeslot booking*

    - by boyee007
    regarding of this question.. PHP Booking timeslot I tried 'GROUP BY' id_timeslot still didnt work, as its only showing the booked timeslot not available i tried that solution, but give me an error and not quite understand how to use 'coelence' table timeslot (id_timeslot integer); table doctor (id_doctor integer); table bookslot (id_bookslot, id_doctor, id_timeslot integer); insert into doctor (id_doctor) values (1 = doc_A), (2 = doc_B), (3 = doc_C); insert into TimeSlot (id_timeslot) values (1 = 10:00:00), (2 = 10:15:00), (3 = 10:30:00), (4 = 10:45:00); insert into bookslot (id_doctor,id_timeslot) values (1,1), (1,5), (2,1), (2,4), (3,1); Join mysql table $q = $mysqli->query("SELECT * FROM bookslot RIGHT JOIN timeslot ON bookslot.id_timeslot = timeslot.id_timeslot LEFT JOIN doctor ON bookslot.id_doctor = doctor.id_doctor "); echoing result and checking if it matches todays date or else set available while($r = $q->fetch_array(MYSQLI_ASSOC)) : echo '<tr>'; echo '<td align="center">' . $r['times'] . '</td>'; if($r['booked_date'] == date('Y-m-d') && $r['id_doctor'] == 1): echo '<td><a href="#available" class="booked">booked</a></td>'; else : echo '<td><a href="#" class="available">available</a></td>'; endif; if($r['booked_date'] == date('Y-m-d') && $r['id_doctor'] == 2): echo '<td><a href="#available" class="booked">booked</a></td>'; else : echo '<td><a href="#" class="available">available</a></td>'; endif; if($r['booked_date'] == date('Y-m-d') && $r['id_doctor'] == 3): echo '<td><a href="#available" class="booked">booked</a></td>'; else : echo '<td><a href="#" class="available">available</a></td>'; endif; echo '</tr>'; endwhile; result from webpage and i want the result look like: id_timeslot doc_A doc_B doc_C ---------------------------------------------- 1 booked booked booked 2 available available available 3 available available available 4 available booked available 5 booked available available Any other solution please!

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  • PHP MSSQL : How to display output when query return no row

    - by vamps
    i have a problem with my PHP-MSSQL query. i have a join table that need to give a result something be like this: Department Group A Group B Total A+B WORKHOUR A OTHOUR A WORKHOUR B OTHOUR B WORKHOUR OTHOUR HR 10 15 25 0 35 15 IT 5 5 5 5 Admin 12 12 12 12 the query will count how many employee as per given date (admin will enter data and once submitted, the query will give the above result). The problem is, the final output is a mess when there's no row to be displayed. the column is shifted to the right. i.e: only Group A in IT only Group B in Admin Department Group A Group B Total A+B WORKHOUR A OTHOUR A WORKHOUR B OTHOUR B WORKHOUR OTHOUR HR 10 15 25 0 35 15 IT 5 5 5 5 Admin 12 12 12 12 my question is, how to prevent this to happen? i've tried everything with While.... if else.. but the result is still the same. how to display output "0" if no rows to return? echo "0"; this is my QUERY: select DD.DPT_ID,DPT.DEPARTMENT_NAME,TU.EMP_GROUP, sum(DD.WORK_HOUR) AS WORK_HOUR, sum(DD.OT_HOUR) AS OT_HOUR FROM DEPARTMENT_DETAIL DD left join DEPARTMENT DPT ON (DD.DEPT_ID=DPT.DEPT_ID) LEFT JOIN TBL_USERS TU ON (TU.EMP_ID=DD.EMP_ID) WHERE DD_DATE>='2012-01-01' AND DD_DATE<='2012-01-31' AND TU.EMP_GROUP!=2 GROUP BY DD.DEPT_ID, DPT.DEPARTMENT_NAME,TU.EMP_GROUP ORDER BY DPT.DEPARTMENT_NAME this is one of the logic that i've used, but doesn't return the result that i want:: while($row = mssql_fetch_array($displayResult)) { if ((!$row["WORK_HOUR"])&&(!$row["OT_HOUR"])) { echo "<td >"; echo "empty"; echo "&nbsp;</td>"; echo "<td >"; echo "empty"; echo "&nbsp;</td>"; } else { echo "<td>"; echo $row["WORK_HOUR"]; echo "&nbsp;</td>"; echo "<td>"; echo $row["OT_HOUR"]; echo "&nbsp;</td>"; } } please help. i've been doing this for 2 days. @__@

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  • PHP Array Not Working in Function

    - by lemonpole
    Hello all. I'm currently experimenting with arrays in PHP, and I created a fake environment where a team's information will be displayed. $t1 = array ( "basicInfo" => array ( "The Sineps", "December 25, 2010", "lemonpole" ), "overallRecord" => array (0, 0, 0, 0), "overallSeasons" => array ( 1 => array (14, 0, 0), 2 => array (9, 5, 2), 3 => array (12, 4, 0), 4 => array (3, 11, 2) ), "games" => array ( "<img src=\"images/cs.gif\" alt=\"Counter-Strike\" />", "<img src=\"images/cs.gif\" alt=\"Counter-Strike\" />", "<img src=\"images/cs.gif\" alt=\"Counter-Strike\" />", "<img src=\"images/cs.gif\" alt=\"Counter-Strike\" />" ), "seasonHistory" => array ( "Season I", "Season II", "Season III", "Season IV" ), "divisions" => array ( "Open", "Main", "Main", "Invite" ) ); // Displays the seasons the team has been in along // with the record of each season. function seasonHistory() { // Make array variable local-scope. global $t1; // Count the number of seasons. $numrows = count($t1["seasonHistory"]); // Loop through all the variables until // it reaches the last entry made and display // each item seperately. for($v = 0; $v <= $numrows; $v++) { // Echo each season. echo "<tr><td>{$t1["games"][$v]}</td>"; echo "<td>{$t1["seasonHistory"][$v]}</td>"; echo "<td>{$t1["divisions"][$v]}</td></tr>"; } } I have tested several possible problems out and after narrowing them down I have come down to one conclusion and that is my function is not connecting to the array for some reason. I don't know what else to do because I thought making the array global would fix that problem. What works: I can echo $t1["games"][0] on the page I need it to display and it gives me the content. I tried echo $t1["games"][0] INSIDE the function and then calling the function and it doesn't display anything.

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  • PHP-OOP extending two classes?

    - by user1292810
    I am very beginner to OOP and now I am trying to write some PHP class to connect with FTP server. class ftpConnect { private $server; private $user; private $password; private $connection_id; private $connection_correct = false; public function __construct($server, $user = "anonymous", $password = "[email protected]") { $this->server = $server; $this->user = $user; $this->password = $password; $this->connection_id = ftp_connect($this->server); $this->connection_correct = ftp_login($this->connection_id, $this->user, $this->password); if ( (!$this->connection_id) || (!$this->connection_correct) ){ echo "Error! Couldn't connect to $this->server"; var_dump($this->connection_id); var_dump($this->connection_correct); return false; } else { echo "Successfully connected to $this->server, user: $this->user"; $this->connection_correct = true; return true; } } } I reckon that body of the class is insignificant at the moment. Main issue is that I have some problems with understanding OOP idea. I wanted to add sending emails every time, when the code is run. I have downloaded PHPMailer Class and extended my class with it: class ftpConnect extends PHPMailer {...} I have added some variables and methods and everything works as expected to that point. I thought: why not to add storing everything in database. Everytime user runs above code, proper information should be stored in database. I could edit my ftpConnect class and add database connecting to the constructor, and some other methods to updating tables. But database connecting and all that stuff could be used by other classes in the future, so it definitely should be implemented in seperate class. But my "main" ftpConnect class already extends one class and could not extend not a single one more. I have no idea how can I resolve this problem. Maybe my ftpConnect class is to complex and I should somehow divide it into couple smaller classes? Any help is much appreciated.

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  • Manipulating values from database table with php

    - by charliecodex23
    I currently have 5 tables in MySQL database. Some of them share foreign keys and are interdependent of each other. I am displaying classes accordingly to their majors. Each class is taught during the fall, spring or all_year. In my database I have a table named semester which has an id, year, and semester fields. The semester field in particular is a tinyint that has three values 0, 1, 2. This signifies the fall, spring or all_year. When I display the query instead of having it show 0 or 1 or 2 can I have it show fall, spring etc? Extra: How can I add space to the end of each loop so the data doesn't look clustered? Key 0 Fall 1 Spring 2 All-year PHP <? try { $pdo = new PDO ("mysql:host=$hostname;dbname=$dbname","$username","$pw"); } catch (PDOException $e) { echo "Failed to get DB handle: " . $e->getMessage() . "\n"; exit; } $query = $pdo->prepare("SELECT course.name, course.code, course.description, course.hours, semester.semester, semester.year FROM course LEFT JOIN major_course_xref ON course.id = major_course_xref.course_id LEFT JOIN major ON major.id = major_course_xref.major_id LEFT JOIN course_semester_xref ON course.id = course_semester_xref.course_id LEFT JOIN semester ON course_semester_xref.semester_id = semester.id"); $query->execute(); if ($query->execute()){ while ($row = $query->fetch(PDO::FETCH_ASSOC)){ print $row['name'] . "<br>"; print $row['code'] . "<br>"; print $row['description'] . "<br>"; print $row['hours'] . " hrs.<br>"; print $row['semester'] . "<br>"; print $row['year'] . "<br>"; } } else echo 'Could not fetch results.'; unset($pdo); unset($query); ?> Current Display Computer Programming I CPSC1400 Introduction to disciplined, object-oriented program development. 4 hrs. 0 2013 Desire Display Computer Programming I CPSC1400 Introduction to disciplined, object-oriented program development. 4 hrs. Fall 2013

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  • PHP-CGI slowly gains memory despite lack of requests.

    - by Kyle
    I know by now that PHP-CGI will sit there and hog my memory but I'm not looking to reduce it, just a way to reset the processes every so often. I'm using PHP-CGI with FastCGI/Lighttpd. So the best way of saying this is how could I reset these processes to prevent these annoying memory leaks?

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  • Trying to setup a PHP daemon using System_Daemon and I'm having issues getting it to run.

    - by yummm
    I get the following error when trying to start a daemon using Ubuntu 10.04 and the PHP5: PHP Warning: PHP Startup: Unable to load dynamic library 'usr/lib/php5/20060613/pcntl.so' - /usr/lib/php5/20060613/pcntl.so: cannot open shared object file: No such file or directory in Unknown on line 0 Does System_Daemon try to call pcntl? If so, why is it looking for the file where it does not exist?

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