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  • PHP/Javascript: Restart Form Button Isn't Working, Code Insde

    - by Josh K
    Basically I want a confirmation box to pop up when they click the button asking if they sure they want to restart, if yes, then it destroys session and takes them to the first page. Heres what i got... echo "<form id=\"form\" name=\"form\" method=\"post\" action=\"nextpage.php\">\n"; echo " <input type=\"button\" name='restart' value='Restart' id='restart' onclick='restartForm()' />"; and for the script... <script type=\"text/javascript\"> <!-- function restartForm() { var answer = confirm('Are you sure you want to start over?'); if (answer) { form.action=\"firstpage.php\"; session_destroy(); form.submit(); } else alert ('Restart Cancelled'); } // -- </script>"; EDIT: Note that pressing the button brings up the confirm box, but if you click okay nothing happens sometimes. Sometimes if u click cancel it still submits the form (To the original action)

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  • change php variable on click event

    - by Claudiu
    I want to change php variable ($start and $end called by var b1 and var b2) if the user clicks on button. Now I know that php is server side, but how can I do it? I read something about using $get but I don't know how to implement it: <?php if ( is_user_logged_in() ) { ?> <input type="submit" value="Start Chat" id="start_chat" style="position: absolute; top: 30px; left: 10px;" /> <?php } ?> <script> jQuery('#start_chat').click(function(){ $(this).data('clicked', true); }); var b1 = '<?php echo $start; ?>'; var b2 = '<?php echo $end; ?>'; if(jQuery('#start_chat').data('clicked')) { // change var b1 and b2 } else { // do not change anything } </script> <div id="eu_la"> <?php $start = strtotime('9:30'); $end = strtotime('12:30'); $timenow = date('U'); if((date('w') == 4) && ($timenow >= $start && $timenow <= $end)) { // day 2 = Tuesday include('facut_mine.php'); } else { do_shortcode('[upzslider usingphp=true]'); } ?>

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  • How to do this in a smart way

    - by Azzyh
    Hello SO. My websystem im coding with now is integrated from the originally phpBB system. With this i mean when you log on my websystem, you have actually logged into the forum from the login page. Now i came to the "log out" part, and i want to make it smart. Right now its a simply link with logout: <a href="<?php echo BASEDIR; ?>../../ucp.php?mode=logout&sid=<? echo $user->data['session_id']; ?>" style="margin-left: 14px; font-size: 10px;">- Log Out</a> As you see in the link it refers to the forum´s ucp.php?mode=logout, and you need to have the SID variable in it in order to log out right.. anyway.. I want to do this log out part in a smart way, i mean, not so you land on the forum´s "you have now been logged out", i want something maybe like run this page in background when you click, and then it refreshes the current page your on. Or should it be smarter just try to modify the phpbb file ucp.php? I think its hard coded, and not the way I am coding, so thats why i find it abit tricky. Thank you for your answers and examples on how this could be done, in a smart way..

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  • php search database for row

    - by Brenden Morley
    Okay I got code the code to pull data based on a users account number well here is what im using (And yes I know it isnt safe now that is the reason for my post) <?php include('config.php'); $user_info = fetch_user_info($_GET['AccountNumber']); ?> <html> <body> <div> <?php if ($user_info === false){ $Output = 'http://www.MyDomain.Com/'; echo '<META HTTP-EQUIV=Refresh CONTENT="0; URL='.$Output.'">'; }else{ ?> <center> <title><?php echo $user_info['FirstName'], ' ', $user_info['LastName'], ' - ', $user_info['City'], ', ', $user_info['State']; ?> - Name of site</title> So basically what this code is allowing me to do is have a file called Profile.php And when a user visits this this page it will return the data Like this http://MyDomain.com/Profile.php?AccountNumber=50b9c965b7c3b How can I do this securely cause right now its using a get method really unsafe to retive the account number from the url bar.

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  • What is wrong with this php loop?

    - by Mark R
    I made a loop but it doesn't work. here's what I did: <?php if(is_tree('4')) { ?> <?php $show_after_p = 2; $content = apply_filters('the_content', $post->post_content); if(substr_count($content, '<p>') > $show_after_p) { $contents = explode("</p>", $content); $p_count = 1; foreach($contents as $content) { echo $content; if($p_count == $show_after_p) { ?> YOUR AD CODE GOES HERE < ? } echo "</p>"; $p_count++; } } ?> <?php else : ?> <?php the_content('<p class="serif">Read the rest of this page &raquo;</p>'); ?> <?php endif; } ?> I need to make it work but don't know how. I'm guessing it's a simple syntax error I'm not seeing?

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  • Why does this sql statement keep saying it is a boolean and not a paramter? (php/Mysql)

    - by ggfan
    In this statement, I am trying to see if there if the latest posting in the database that has the exact same title, price, city, state, detail. If there is, then it would say to the user that the exact post has been already made; if not then insert the posting into the dbc. (This is one type of check so that users can't accidentally post twice. This may not be the best check, but this statement error is annoying me, so I want it to work :)) Why won't this sql work? I think it's not letting the title=$title and not getting the value in the $title... ERROR: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in postad.php on line 365 //there is a form that users fill out that has title, price, city, etc <form> blah blah </form> //if users click submit, then does all the checks and if all okay, insert to dbc if (isset($_POST['submit'])) { // Grab the pposting data from the POST and gets rid of any funny stuff $title = mysqli_real_escape_string($dbc, trim($_POST['title'])); $price = mysqli_real_escape_string($dbc, trim($_POST['price'])); $city = mysqli_real_escape_string($dbc, trim($_POST['city'])); $state = mysqli_real_escape_string($dbc, trim($_POST['state'])); $detail = mysqli_real_escape_string($dbc, trim($_POST['detail'])); if (!is_numeric($price) && !empty($price)) { echo "<p class='error'>The price can only be numbers. No special characters, etc</p>"; } //Error problem...won't let me set title=$title, detail=$detail, etc. //this statement after all the checks so that none of the variables are empty $query="Select * FROM posting WHERE user_id={$_SESSION['user_id']} AND title=$title AND price=$price AND city=$city AND state=$state AND detail=$detail"; $data = mysqli_query($dbc, $query); if(mysqli_num_rows($data)==1) { echo "You already posted this ad. Most likely caused by refreshing too many times."; } }

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  • Variable won't store in session

    - by Mittens
    So I'm trying to store the "rank" of a user when they log in to a control panel which displays different options depending on the given rank. I used the same method as I did for storing and displaying the username, which is displayed on the top of each page and works just fine. I can't for the life of me figure out why it won't work for the rank value, but I do know that it is not saving it in the session. Here is the bit that's not working; $username = ($_POST['username']); $password = hash('sha512', $_POST['password']); $dbhost = 'mysql:host=¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦;dbname=¦¦¦¦¦¦¦¦¦¦¦'; $dbuser = '¦¦¦¦¦¦¦¦¦¦¦'; $dbpassword = '¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦'; try { $db = new PDO($dbhost, $dbuser, $dbpassword); $statement = $db->prepare("select password from users where email = :name"); $statement->execute(array(':name' => $username)); $result = $statement->fetch(); $pass = $result[password]; $rank = $result[rank];} catch(PDOException $e) {echo $e->getMessage();} if ($password == $pass) { session_start(); $_SESSION['username'] = $username; $_SESSION['rank'] = $rank; header('Location: http://¦¦¦¦¦¦¦¦¦.ca/manage.php'); } else{ include'../../includes/head.inc'; echo '<h1>Incorrect username or password.</h1>'; include'../../includes/footer.inc'; } I'm also new to the whole PDO thing, hence why my method of authenticating the password is pretty sketchy.

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  • Why won't the following PDO transaction won't work in PHP?

    - by jfizz
    I am using PHP version 5.4.4, and a MySQL database using InnoDB. I had been using PDO for awhile without utilizing transactions, and everything was working flawlessly. Then, I decided to try to implement transactions, and I keep getting Internal Server Error 500. The following code worked for me (doesn't contain transactions). try { $DB = new PDO('mysql:host=localhost;dbname=database', 'root', 'root'); $DB->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); $dbh = $DB->prepare("SELECT * FROM user WHERE username = :test"); $dbh->bindValue(':test', $test, PDO::PARAM_STR); $dbh->execute(); } catch(Exception $e){ $dbh->rollback(); echo "an error has occured"; } Then I attempted to utilize transactions with the following code (which doesn't work). try { $DB = new PDO('mysql:host=localhost;dbname=database', 'root', 'root'); $DB->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); $dbh = $DB->beginTransaction(); $dbh->prepare("SELECT * FROM user WHERE username = :test"); $dbh->bindValue(':test', $test, PDO::PARAM_STR); $dbh->execute(); $dbh->commit(); } catch(Exception $e){ $dbh->rollback(); echo "an error has occured"; } When I run the previous code, I get an Internal Server Error 500. Any help would be greatly appreciated! Thanks!

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  • Unable to parse variable length string separated by delimiter

    - by Technext
    Hi, I have a problem with parsing a string, which consists only of directory path. For ex. My input string is Abc\Program Files\sample\ My output should be Abc//Program Files//sample The script should work for input path of any length i.e., it can contain any no. of subdirectories. (For ex., abc\temp\sample\folder\joe) I have looked for help in many links but to no avail. Looks like FOR command extracts only one whole line or a string (when we use ‘token’ keyword in FOR syntax) but my problem is that I am not aware of the input path length and hence, the no. of tokens. My idea was to use \ as a delimiter and then extract each word before and after it (), and put the words to an output file along with // till we reach the end of the string. I tried implementing the following but it did not work: @echo off FOR /F "delims=\" %%x in (orig.txt) do ( IF NOT %%x == "" echo.%%x//output.txt ) The file orig.txt contains only one line i.e, Abc\Program Files\sample\ The output that I get contains only: Abc// The above output contains blank spaces as well after ‘Abc//’ My desired output should be: Abc//program Files//sample// Can anyone please help me with this? Regards, Technext

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  • Wordpress Admin Panel Code Input

    - by Wes
    I've got a wordpress admin panel for one of my themes and one of the boxes has an input for some code to drive google adsense. when I put the code into the box and call it with my php tags the code comes out like this: <script type="\&quot;text/javascript\&quot;"><!-- google_ad_client = \"pub-9295546347478163\"; /* Leaderboard 5/17/2010 */ google_ad_slot = \"7593465074\"; google_ad_width = 728; google_ad_height = 90; //--> </script> <script type="\&quot;text/javascript\&quot;" src="%5C%22http://pagead2.googlesyndication.com/pagead/show_ads.js%5C%22"> </script> Which I assume is a feature to stop SQL injections. How can I call pure code form a box? This is how I currently have that textbox setup. array( "name" => "Code for Top ad", "desc" => "Enter the HTML that will drive the banner ad for the page header", "id" => $shortname."_headerAd", "type" => "textarea"), and then echo it out with this: <?php echo get_option('lifestyle_headerAd'); ?>

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  • Why doesn't this short php script send email?

    - by RoryG
    I can't seem to get my php script to send email. <?php echo "Does this page work?"; mail('my email address', 'test subject', 'test message'); ?> First, I have set the mail function settings in the php.ini file as follows: I checked my email account settings on outlook. It does not require authentication, its port is 25, and its type of encrypted connection is 'Auto'. Given this I configured my php.ini file accordingly: SMTP = ssl://smtp1.iis.com smtp_port = 25 Then I set: sendmail_from: my email address The echo statement prints out in the browser, so I know the php file is recognized and processed. But the browser also shows the following error: Warning: mail() [function.mail]: "sendmail_from" not set in php.ini or custom "From:" header missing in C:\xampp\htdocs\mailtest.php on line 3 I have clearly set the sendmail_from so I don't know what else to do. I have also tried removing the 'ssl://' part from the SMTP setting in the php.ini file, and configuring the php5.ini file. Which of these .ini files should I be configuring anyways?

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  • Error after redirection using CakePHP

    - by Praveen kalal
    I have created some code called LoginController. Whenever Admin gets successfully logged in I redirect the page to index. However, I got an error like "problem on loading page". This is my code: <?php class LoginController extends AdminAppController { var $name = 'Login'; var $uses = array('Admin.Login'); var $sessionkey= ''; /*function beforeFilter() { if($this->Session->read('user')=='Admin' || $this->params['action']=='login') { echo "in"; exit; } else { echo "else"; exit; $this->Session->setFlash('Login first','flash_failure'); $this->redirect(array('action'=>'login')); } }*/ function index() { } function login() { //pr($this->data); exit; if(!empty($this->data)) { $results = $this->Login->findByEmail($this->data['Login']['email']); if(!empty($results) && $results['Login']['password']== md5($this->data['Login']['password'])) { $this->Session->write('user', 'Admin'); $results['Login']['last_login']=date("Y-m-d H:i:s"); $this->Login->save($results); $this->Session->setFlash('Login successfully.', 'flash_success'); $this->redirect(array('controller'=>'login','action' => 'index')); } } } } ?> Can anyone help me? Thanks.

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  • Looping Through Database Query

    - by DrakeNET
    I am creating a very simple script. The purpose of the script is to pull a question from the database and display all answers associated with that particular question. We are dealing with two tables here and there is a foreign key from the question database to the answer database so answers are associated with questions. Hope that is enough explanation. Here is my code. I was wondering if this is the most efficient way to complete this or is there an easier way? <html> <head> <title>Advise Me</title> <head> <body> <h1>Today's Question</h1> <?php //Establish connection to database require_once('config.php'); require_once('open_connection.php'); //Pull the "active" question from the database $todays_question = mysql_query("SELECT name, question FROM approvedQuestions WHERE status = active") or die(mysql_error()); //Variable to hold $todays_question aQID $questionID = mysql_query("SELECT commentID FROM approvedQuestions WHERE status = active") or die(mysql_error()); //Print today's question echo $todays_question; //Print comments associated with today's question $sql = "SELECT commentID FROM approvedQuestions WHERE status = active"; $result_set = mysql_query($sql); $result_num = mysql_numrows($result_set); for ($a = 0; $a < $result_num; $a++) { echo $sql; } ?> </body> </html>

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  • Reference problem when returning an object from array in PHP

    - by avastreg
    I've a reference problem; the example should be more descriptive than me :P I have a class that has an array of objects and retrieve them through a key (string), like an associative array: class Collection { public $elements; function __construct() { $this->elements = array(); } public function get_element($key) { foreach($this->elements as $element) { if ($element->key == $key) { return $element; break; } } return null; } public function add_element ($element) { $this->elements[] = $element; } } Then i have an object (generic), with a key and some variables: class Element { public $key; public $another_var; public function __construct($key) { $this->key = $key; $this->another_var = "default"; } } Now, i create my collection: $collection = new Collection(); $collection->add_element(new Element("test1")); $collection->add_element(new Element("test2")); And then i try to change variable of an element contained in my "array": $element = $collection->get_element("test1"); $element->another_var = "random_string"; echo $collection->get_element("test1")->another_var; Ok, the output is random_string so i know that my object is passed to $element in reference mode. But if i do, instead: $element = $collection-get_element("test1"); $element = null; //or $element = new GenericObject(); $element-another_var = "bla"; echo $collection-get_element("test1")-another_var; the output is default like if it lost the reference. So, what's wrong? I have got the references to the variables of the element and not to the element itself? Any ideas?

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  • Show Color Picker Value Code Within Input Text Field in Wordpress

    - by Shwan Namiq
    Hi i have options page for my theme i used color picker jquery plugin in my options page as show in image below. i want when i change the color from color picker automatically show the color value code within the text field.how can do this? this is the code within my options page related to appearing the color picker and text field function to register the options setting function YPE_register_settings_sections_fields() { register_setting ( 'YPE_header_option_group', 'YPE_header_option_name', 'YPE_sanitize_validate_callback' ); add_settings_section ( 'YPE_header_section', 'Header Section', 'YPE_header_section_callback', 'YPE_menu_page_options' ); add_settings_field ( 'YPE_header_bg', 'Header Background', 'YPE_header_bg_callback', 'YPE_menu_page_options', 'YPE_header_section' ); } add_action('admin_init', 'YPE_register_settings_sections_fields'); function to appear the text field and color picker function YPE_header_bg_callback() { $YPE_options = get_option('YPE_header_option_name'); $YPE_header_bg = isset($YPE_options['YPE_header_bg']) ? $YPE_options['YPE_header_bg'] : ''; ?> <div class="input-group color-picker"> <input class="form-control" style="width:80px;" name="YPE_header_option_name[YPE_header_bg]" id="<?php echo 'YPE_header_bg'; ?>" type="text" value="<?php echo $YPE_header_bg; ?>" /> <span class="input-group-btn"> <div id="colorSelector"> <div nam style="background-color: #0000ff"> </div> </div> </span> </div> <script> $("#colorSelector").ColorPicker({ color: '#0000ff', onShow: function (colpkr) { $(colpkr).fadeIn(500); return false; }, onHide: function (colpkr) { $(colpkr).fadeOut(500); return false; }, onChange: function (hsb, hex, rgb) { $('#colorSelector div').css('backgroundColor', '#' + hex); }); </script> <?php }

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  • $_FILES is null, $_POST is not null

    - by Cory Dee
    When I am going to upload a file, my $_POST variable knows the file name, but the $_FILES variable is null. I've used this code before, so I'm really stumped. Here's what I'm using for input: <label for="importFile">Attach Resume:</label> <input type="hidden" name="MAX_FILE_SIZE" value="10000000"> <input type="file" name="importFile" id="importFile" class="validate['required']"> And for processing: $uploaddir = "E:/Sites/OPL/2008/assets/apps/newjobs/resumes/"; $uploadfile = $uploaddir . time() . '-' . urlencode(basename($_FILES['importFile']['name'])); if (!move_uploaded_file($_FILES['importFile']['tmp_name'], $uploadfile)) { echo 'Error uploading file. Error number: ' . $_FILES['importFile']['error']; var_dump($_FILES['importFile']); echo $_POST['importFile']; die(); } Which is giving me this result: Error uploading file. Error number: NULL Maintaining The OPL Website.doc Any help would be greatly appreciated.

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  • php while loop throwing a error over a $var < 10 statement

    - by William
    Okay, so for some reason this is giving me a error as seen here: http://prime.programming-designs.com/test_forum/viewboard.php?board=0 However if I take away the '&& $cur_row < 10' it works fine. Why is the '&& $cur_row < 10' causing me a problem? $sql_result = mysql_query("SELECT post, name, trip, Thread FROM (SELECT MIN(ID) AS min_id, MAX(ID) AS max_id, MAX(Date) AS max_date FROM test_posts GROUP BY Thread ) t_min_max INNER JOIN test_posts ON test_posts.ID = t_min_max.min_id WHERE Board=".$board." ORDER BY max_date DESC", $db); $num_rows = mysql_num_rows($sql_result); $cur_row = 0; while($row = mysql_fetch_row($sql_result) && $cur_row < 10) { $sql_max_post_query = mysql_query("SELECT ID FROM test_posts WHERE Thread=".$row[3].""); $post_num = mysql_num_rows($sql_max_post_query); $post_num--; $cur_row++; echo''.$cur_row.'<br/>'; echo'<div class="postbox"><h4>'.$row[1].'['.$row[2].']</h4><hr />' .$row[0]. '<br /><hr />[<a href="http://prime.programming-designs.com/test_forum/viewthread.php?thread='.$row[3].'">Reply</a>] '.$post_num.' posts omitted.</div>'; }

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  • how to pass value to controller??

    - by rajesh
    When I try to pass url value to controller action, action is not getting the required value. I'm sending the value like this: function value(url,id) { alert(url); document.getElementById('rating').innerHTML=id; var params = 'artist='+id; alert(params); // var newurl='http://localhost/songs_full/public/eslresult/ratesong/userid/1/id/27'; var myAjax = new Ajax.Request(newurl,{method: 'post',parameters:params,onComplete: loadResponse}); //var myAjax = new Ajax.Request(url,{method:'POST',parameters:params,onComplete: load}); //alert(myAjax); } function load(http) { alert('success'); } and in the controller I have: public function ratesongAction() { $user=$_POST['rating']; echo $user; $post= $this->getRequest()->getPost(); //echo $post; $ratesongid= $this->_getParam('id'); } But still not getting the result. I am using zend framework.

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  • PHP & MySQL username validation and storage problem.

    - by php
    For some reason when a user enters a brand new username the error message <p>Username unavailable</p> is displayed and the name is not stored. I was wondering if some can help find the flaw in my code so I can fix this error? Thanks Here is the PHP code. if($_POST['username'] && trim($_POST['username'])!=='') { $u = "SELECT * FROM users WHERE username = '$username' AND user_id <> '$user_id'"; $r = mysqli_query ($mysqli, $u) or trigger_error("Query: $u\n<br />MySQL Error: " . mysqli_error($mysqli)); if (mysqli_num_rows($r) == TRUE) { echo '<p>Username unavailable</p>'; $_POST['username'] = NULL; } else if(isset($_POST['username']) && mysqli_num_rows($r) == 0 && strlen($_POST['username']) <= 255) { $username = mysqli_real_escape_string($mysqli, $_POST['username']); } else if($_POST['username'] && strlen($_POST['username']) >= 256) { echo '<p>Username can not exceed 255 characters</p>'; } }

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  • Why doesn't sed's automatic printing deliver the expected results?

    - by CodeGnome
    What Works This sed script works as intended: $ echo -e "2\n1\n4\n3" | sed -n 'h; n; G; p' 1 2 3 4 It takes pair of input lines at a time, and swaps the lines. So far, so good. What Doesn't Work What I don't understand is why I can't use sed's automatic printing. Since sed automatically prints the pattern space at the end of each execution cycle (except when it's suppressed), why is this not equivalent? $ echo -e "2\n1\n4\n3" | sed 'h; n; G' 2 1 2 4 3 4 What I think the code says is: The input line is copied to the hold space. The next line is read into the pattern space. The hold space is appended to the pattern space. The pattern space (line1 + newline + line2) is printed automatically because we've reached the end of the execution cycle. Obviously, I'm wrong...but I don't understand why. Can anyone explain why this second example breaks, and why print suppression is needed to yield the correct results?

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  • Session in php are not enough clear to me

    - by Lulzim
    I find sessions in php kind of confusing, can anybody of you explain those to me. I have an example which is not working in my case: I register sessions this way, would you please tell me is this the right way of registering sessions //this is the page from where i register myusername in sessions if($count==1){ session_start(); $_SESSION['myusername'] = $_POST['myusername']; include("enterpincover.php"); } else { echo "Wrong Pin"; } here i check first whether the username is registered in sessions in oder to open his account , otherwise open again login. It works, if user is not loged in, it will show login page which is right, if user is loged it shows welcome message but not the Welcome the name of the user as I want. for ex: Welcome David <?php session_start(); if(isset($_SESSION['myusername'])) { echo 'Welcome '.$_SESSION['myusername']; } else { include("leftmodules.php"); include("rightmodules.php"); include("login.php"); } ?>

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  • PHP the SELECT FROM WHERE col IN no working using array

    - by Sam Ram San
    I'm trying to pull some data from MySQL using an Array that was fetch from a first query. Everything's fine all the way to the implode after that, it's been a headache for me. Can someone help me? <?php $var = 'somedata'; include("config/conect.php"); $zip="SELECT * FROM table WHERE firstcol LIKE '%$var%' ORDER BY seconcol"; $results = $connetction->query($zip); while ($row = $results->fetch_array()){ $mycode[]=$row['zip']; } array_pop($mycode); $mycode = implode(', ',$mycode); //print_r ($mycode); echo '<br /><br /><br />'; $usr="SELECT * FROM reg_temp WHERE zip IN('".join("','", $mycode)."')"; $results = $asies->query($usr); while ($row = $results-> fetch_arra()) { $name = $row['name']; echo $name; } ?>

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  • Textbox auto generate by php and html

    - by user2892997
    i have few field of data such as product , amount and barcode.The system just show 1 row of data insert form that contain the 3 field.when i completed insert the first row, then second row of textbox will auto generate, i can do it by microsoft access, can i do so for php ? <?php $i=0; ?> <form method="post" action=""> <input type="text" name="<?php echo $i; ?>" /> </form> <?php if(isset($_POST[$i])){ $i++; ?> <form method="post" action=""> <input type="text" name="<?php echo $i; ?>" /> </form> <?php }?> it work for the first and second textbox, but how can i continue to create more textbox accordingly?

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  • Using curl to submit/retrieve a forms results

    - by Jason
    I need help in trying to use curl to post data to a page and retrieve the results after the form has been submitted. I created a simple form: <form name="test" method="post" action="form.php"> <input type="text" name="name" size="40" />e <input type="text" name="comment" size="140" /> <input type="submit" name="submit" value="submit" /> </form> In addition, I have php code to handle this form in the same page. All it does is echo back the form values. The curl that I have been using is this: $h = curl_init(); curl_setopt($h, CURLOPT_URL, "path/to/form.php"); curl_setopt($h, CURLOPT_POST, true); curl_setopt($h, CURLOPT_POSTFIELDS, array( 'name' = 'yes', 'comment' = 'no' )); curl_setopt($h, CURLOPT_HEADER, false); curl_setopt($h, CURLOPT_RETURNTRANSFER, 1); $result = curl_exec($h); When I launch the page with the curl code in it, I get the form.php page contents but it doesn't not show the variables that PHP should have echo'd when the form is submitted. would appreciate any help with this. Thanks.

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  • PHP: Need a double check on an error in this small code

    - by Josh K
    I have this simple Select box that is suppose to store the selected value in a hidden input, which can then be used for POST (I am doing it this way to use data from disabled drop down menus) <body> <?php $Z = $_POST[hdn]; ?> <form id="form1" name="form1" method="post" action="test.php"> <select name="whatever" id="whatever" onchange="document.getElementById('hdn').value = this.value"> <option value="1">1Value</option> <option value="2">2Value</option> <option value="3">3Value</option> <option value="4">4Value</option> </select> <input type="hidden" name ='hdn' id="hdn" /> <input type="submit" id='submit' /> <?php echo "<p>".$Z."</p>"; ?> </form> </body> The echo call works for the last 3 options (2,3,4) but if I select the first one it doesnt output anything, and even if i change first one it still doesnt output anything. Can someone explain to me whats going on, I think it might be a syntax issue.

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