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  • Problem carrying Session over to other pages

    - by AAA
    I am able to login a user, but while processing to the next page (memebers area) I can't display any user info let alone print the $_SESSION[email]. I am not sure what's up. Below is the login code and the testing members are page. Login page: session_start(); //also in a real app you would get the id dynamically $sql = "select `email`, `password` from `accounts` where `email` = '$_POST[email]'"; $query = mysql_query($sql) or die ("Error: ".mysql_error()); while ($row = mysql_fetch_array($query)){ $email = $row['email']; $secret = $row['password']; //we will echo these into the proper fields } mysql_free_result($query); // Process the POST variables $email = $_POST["email"]; //Variables $_SESSION["email"] = $_POST["email"]; $secret = $info['password']; //Checks if there is a login cookie if(isset($_COOKIE['ID_my_site'])) //if there is, it logs you in and directes you to the members page { $email = $_COOKIE['ID_my_site']; $pass = $_COOKIE['Key_my_site']; $check = mysql_query("SELECT email, password FROM accounts WHERE email = '$email'")or die(mysql_error()); while($info = mysql_fetch_array( $check )) { if (@ $info['password'] != $pass) { } else { header("Location: home.php"); } } } //if the login form is submitted if (isset($_POST['submit'])) { // if form has been submitted // makes sure they filled it in if(!$_POST['email'] | !$_POST['password']) { die('You did not fill in a required field.'); } // checks it against the database if (!get_magic_quotes_gpc()) { $_POST['email'] = addslashes($_POST['email']); } $check = mysql_query("SELECT email,password FROM accounts WHERE email = '".$_POST['email']."'")or die(mysql_error()); //Gives error if user dosen't exist $check2 = mysql_num_rows($check); if ($check2 == 0) { die('That user does not exist in our database. <a href=add.php>Click Here to Register</a>'); } while($info = mysql_fetch_array( $check )) //gives error if the password is wrong if (@ $_POST['password'] != $info['password']) { die('Incorrect password, please try again'); } else { // if login is ok then we add a cookie $_POST['email'] = stripslashes($_POST['email']); $hour = time() + 3600; setcookie(ID_my_site, $_POST['email'], $hour); setcookie(Key_my_site, $_POST['password'], $hour); //then redirect them to the members area header("Location: home.php"); } } } else { // if they are not logged in ?> <?php } ?> home.php session_start(); if(!isset($_SESSION['email'])) { header('Location: login_test3.php'); die('<a href="login_test3.php">Login first!</a>'); } //Variables $_SESSION["email"] = $email; print $_SESSION['name']; UPDATE Just realized the existing code gets in to the home.php file but will not echo anything. But as soon as you hit refresh the session is gone.

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  • Image upload not functioning correctly

    - by PurpleSmurf
    I'm still trying to get to grips with PHP, and I'm trying to make a form that uploads a picture to a database. I don't have permissions for move_uploaded_file so I'm using the copy() as an alternative. Everywhere I've seen experiencing similar problems have all been with move_uploaded_file so I'm rather stuck. Trying to copy the image from the desktop doesn't seem to be working, it's not throwing up any more PHP errors but is displaying the error message for if something goes wrong. The form sends data to two tables in the database but I'm mainly concerned with the upload not working. There's over 200 lines so I'll post a snippet of the upload code, thank you in advance: function getExtension($str) { $i = strrpos($str,"."); if (!$i) { return ""; } $l = strlen($str) - $i; $ext = substr($str,$i+1,$l); return $ext; } //This variable is used as a flag. The value is initialized with 0 (meaning no error found) and it will be changed to 1 if an errro occures. If the error occures the file will not be uploaded. $errors=0; //checks if the form has been submitted if(isset($_POST['submitted'])) { //reads the name of the file the user submitted for uploading $image=$_FILES['image']['name']; //if it is not empty if ($image) { //get the original name of the file from the clients machine $filename = stripslashes($_FILES['image']['name']); //get the extension of the file in a lower case format $extension = getExtension($filename); $extension = strtolower($extension); //if it is unknown extension, class as an error and do not upload the file, otherwise continue if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { //print error message echo '<h1>Unknown extension!</h1>'; $errors=1; } else { //get the size of the image in bytes //$_FILES['image']['tmp_name'] is the temporary filename of the file in which the uploaded file was stored on the server $size=filesize($_FILES['image']['tmp_name']); //give an unique name, for example the time in unix time format $image_name=time().'.'.$extension; //the new name will be containing the full path where will be stored (images folder) $newname="/home/k0929907/www/uploads/".$image_name; //verify if the image has been uploaded, and print error instead $copied = copy($_FILES['image']['tmp_name'], $newname); if (!$copied) { echo '<h1>Copy unsuccessfull!</h1>'; $errors=1; } } } }

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  • Accessing a numeric property in a SimpleXMLElement

    - by Webnet
    I'm trying to access the number in the below element, but I'm having trouble getting the value out of it. echo $object->0; //returns Parse error: syntax error, unexpected T_LNUMBER, expecting T_STRING or T_VARIABLE or '{' or '$' SimpleXMLElement Object ( [0:public] => 15810 ) Any ideas on how I can get that value?

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  • Not allow a href tags in form textarea

    - by saquib
    Hello friends, How can i prevent user to enter any url or link in contact form text area, i have tried it with this but its not working - if (!isset($_POST['submit']) && preg_match_all('/<a.*>.*<\/a>/', $_POST['query'])) { echo "<h1 style='color:red;'>HTML Tag Not allowed </h1>"; } else { //sendmail } Please help me

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  • Different data for different dates

    - by nsw1475
    I am making a web page which will allow users to input and view data for different dates, and to change the date, there are two buttons, one of which will display the previous day, and one which will show the next day. I know that I can do this by submitting forms and reloading the page every time they press one of these buttons, but I would rather use javascript and not have to submit a form, but I am having troubles getting it to work. Currently, I have the two buttons, and the date stored in a PHP variable, as shown in my code below: <script> function init() { <? $nutrDate = $this->parseDate(date('m/d/Y')); ?> } function nutrPrevDay() { <? $nutrDate = mktime(0, 0, 0, date('m',$nutrDate), date('d',$nutrDate)-1, date('Y',$nutrDate)); ?> alert("<? echo(date("m/d/y", $nutrDate)) ?>"); } function nutrNextDay() { <? $nutrDate = mktime(0, 0, 0, date('m',$nutrDate), date('d',$nutrDate)+1, date('Y',$nutrDate)); ?> } window.onload = init; </script> <p style="text-align:center; font-size:14px; color:#03F"> <button onclick="nutrPrevDay()" style="width:200px" >< Show Previous Day</button> <? echo(date('m/d/Y', $nutrDate)) ?> <button onclick="nutrNextDay()" style="width:200px">Show Next Day ></button> </p> I have the alert in the nutrPrevDay() only as debugging. What happens is when I click on the button, the alert shows that the day is correctly decreased (for example from May 17 to May 16), but only decreases one day, and not one day for every click. Also, I do not know how to make the text on the page (created by the line ) change to display the new date after a button is clicked. So here are my questions: 1) Is it possible to dynamically change data (such as text, and in the future, SQL queries) on a page using javascript without having to reload a page when clicking on a button? 2) If possible, how can I make those changes? 3) How can I fix this so that it will increment and decrement through dates every time a button is clicked?

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  • how to disable on click event

    - by user1819709
    function insertQuestion(form) { var x = "<img src='Images/plussigndisabled.jpg' width='30' height='30' alt='Look Up Previous Question' class='plusimage' name='plusbuttonrow'/><span id='plussignmsg'>(Click Plus Sign to look <br/> up Previous Questions)</span>" ; if (qnum == <?php echo (int)$_SESSION['textQuestion']; ?>) { $('#mainPlusbutton').replaceWith(x); } //append rows into a table code, not needed for this question } .... $('.plusimage').live('click', function() { plusbutton($(this)); }); function plusbutton(plus_id) { // Set global info plusbutton_clicked = plus_id; // Display an external page using an iframe var src = "previousquestions.php"; $.modal('<iframe src="' + src + '" style="border:0;width:100%;height:100%;">'); return false; } <form id="QandA" action="<?php echo htmlentities($action); ?>" method="post"> <table id="question"> <tr> <td colspan="2"> <a onclick="return plusbutton();"> <img src="Images/plussign.jpg" width="30" height="30" alt="Look Up Previous Question" class="plusimage" id="mainPlusbutton" name="plusbuttonrow"/> </a> <span id="plussignmsg">(Click Plus Sign to look up Previous Questions)</span> </td> </tr> </table> <table id="questionBtn" align="center"> <tr> <th> <input id="addQuestionBtn" name="addQuestion" type="button" value="Add Question" onClick="insertQuestion(this.form)" /> </th> </tr> </table> </form> In the code above I am able to replace an image with another image when the if statement is met. But my problem is that when the image is replaced, it does not disable the on click event. My question is that when the image is replaced, how do I disable the onclick event onclick="return plusbutton();? Could unbind click work in this situation. I don't want to use href=# because I don't want to include # at the end of the url

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  • reading from file

    - by lego69
    can somebody help me, how can I filter my file, inside the file I have rows with 3, 4, 5 elements, I want print using echo only these which have 5 elements, thanks in advance (I'm talkin about scripts)

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  • how to get category value from xml params?

    - by C-link Nepal
    I have the following field in xml: <field name="cat1" type="category" extension="COM_CONTENT" label="MOD_ITEM_CAT1" description="MOD_ITEM_CAT_DESC" /> Now, I wanted to get selected value: echo $params->get('cat1'); Which shows me 8 (the value of selected option) not it's title. I've category title people and it should show people instead of 8. So, how to extract the title of the category?

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  • Why save output until the end?

    - by user509006
    Very quick question about programming practices here: I've always used echo() to output HTML code to the user as soon as it was generated, and used ob_start() at the same time to be able to output headers later in the code. Recently, I was made aware that this is bad programming practice and I should be saving HTML output until the end. Is there a reason for this? What is it, and why isn't output buffering a good alternative? Thanks!

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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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  • login not working when changing from mysql to mysqli

    - by user1438647
    I have a code below where it logs a teacher in by matching it's username and password in the database, if correct, then log in, if incorrect, then display a message. <?php session_start(); $username="xxx"; $password="xxx"; $database="mobile_app"; $link = mysqli_connect('localhost',$username,$password); mysqli_select_db($link, $database) or die( "Unable to select database"); foreach (array('teacherusername','teacherpassword') as $varname) { $$varname = (isset($_POST[$varname])) ? $_POST[$varname] : ''; } ?> <form action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post" id="teachLoginForm"> <p>Username</p><p><input type="text" name="teacherusername" /></p> <!-- Enter Teacher Username--> <p>Password</p><p><input type="password" name="teacherpassword" /></p> <!-- Enter Teacher Password--> <p><input id="loginSubmit" type="submit" value="Login" name="submit" /></p> </form> <?php if (isset($_POST['submit'])) { $query = " SELECT * FROM Teacher t WHERE (t.TeacherUsername = '".mysqli_real_escape_string($teacherusername)."') AND (t.TeacherPassword = '".mysqli_real_escape_string($teacherpassword)."') "; $result = mysqli_query($link, $query); $num = mysqli_num_rows($result); $loged = false; while($row = mysqli_fetch_array($result)) { if ($_POST['teacherusername'] == ($row['TeacherUsername']) && $_POST['teacherpassword'] == ($row['TeacherPassword'])) { $loged = true; } $_SESSION['teacherforename'] = $row['TeacherForename']; $_SESSION['teachersurname'] = $row['TeacherSurname']; $_SESSION['teacherusername'] = $row['TeacherUsername']; } if ($loged == true){ header( 'Location: menu.php' ) ; }else{ echo "The Username or Password that you Entered is not Valid. Try Entering it Again."; } mysqli_close($link); } ?> Now the problem is that even if the teacher has entered in the correct username and password, it still doesn't let the teacher log in. When the code above was the old mysql() code, it worked fine as teacher was able to login when username and password match, but when trying to change the code into mysqli then it causes login to not work even though username and password match. What am I doing wrong?

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  • $_SERVER['DOCUMENT_ROOT'] path not working

    - by aeonsleo
    I am using document root to provide absolute path which is not working. if i echo this path it turns out to be C:wamp/www/proman/header.php. I i give relative path it works fine what is the problem here? $path = $_SERVER['DOCUMENT_ROOT']."proman/header.php";

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  • What is the best way to include a php file as a template?

    - by Jon
    I have simple template that's html mostly and then pulls some stuff out of SQL via PHP and I want to include this template in three different spots of another php file. What is the best way to do this? Can I include it and then print the contents? Example of template: Price: <?php echo $price ?> and, for example, I have another php file that will show the template file only if the date is more than two days after a date in SQL.

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  • display php foreach values in div

    - by San82moon
    Hi, I have foreach ($a as $key = $value) { echo $value; } which displays result one below other, like 1234 5678 2010-05-20 5678 1590 2010-05-19 but i want it in a table like structure like 1234 5678 2010-05-20 5678 1590 2010-05-19 how can i do that?

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  • MySQL, how to use returned data?

    - by aejo
    Well, I know there is a funciton mysql_fetch_array() and we can use it like this: while ($row = mysql_fetch_array($result)) { echo $row['name'] . "<br />"; } But is there any other way? For example, if there is only one element that can be returned, and not an array. Thanks)

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  • Pass variables between separate instances of ruby (without writing to a text file or database)

    - by boulder_ruby
    Lets say I'm running a long worker-script in one of several open interactive rails consoles. The script is updating columns in a very, very, very large table of records. I've muted the ActiveRecord logger to speed up the process, and instruct the script to output some record of progress so I know how roughly how long the process is going to take. That is what I am currently doing and it would look something like this: ModelName.all.each_with_index do |r, i| puts i if i % 250 ...runs some process... r.save end Sometimes its two nested arrays running, such that there would be multiple iterators and other things running all at once. Is there a way that I could do something like this and access that variable from a separate rails console? (such that the variable would be overwritten every time the process is run without much slowdown) records = ModelName.all $total = records.count records.each_with_index do |r, i| $i = i ...runs some process... r.save end meanwhile mid-process in other console puts "#{($i/$total * 100).round(2)}% complete" #=> 67.43% complete I know passing global variables from one separate instance of ruby to the next doesn't work. I also just tried this to no effect as well unix console 1 $X=5 echo {$X} #=> 5 unix console 2 echo {$X} #=> "" Lastly, I also know using global variables like this is a major software design pattern no-no. I think that's reasonable, but I'd still like to know how to break that rule if I'd like. Writing to a text file obviously would work. So would writing to a separate database table or something. That's not a bad idea. But the really cool trick would be sharing a variable between two instances without writing to a text file or database column. What would this be called anyway? Tunneling? I don't quite know how to tag this question. Maybe bad-idea is one of them. But honestly design-patterns isn't what this question is about.

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  • What does `?` stand for in SQL?

    - by user295189
    I have this SQL by a programmer: $sql = " INSERT INTO `{$database}`.`table` ( `my_id`, `xType`, `subType`, `recordID`, `textarea` ) VALUES ( {$my_id}, ?xType, ?subType, {$recordID}, ?areaText ) "; My question is why is he using ? before values? How do I see what values are coming in? I did echo and it shows ?xType as ?xType. No values. What does ? stand for in SQL?

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  • SELECT DISTINCT multiple field search?

    - by Patrick
    I'm trying to search multiple fields zc_city and zc_zip for when user input and then return row results for zc_city zc_state and zc_zip $q = strtolower($_GET["q"]); if (!$q) return; $sql = "SELECT DISTINCT zc_city AS zcity FROM search_zipcodes WHERE zc_city LIKE '$q%'"; $rsd = mysql_query($sql); while($rs = mysql_fetch_array($rsd)) { $zcity = $rs['zcity']; echo "$zcity\n"; }

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