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PHP - Userlist With Button To Nextpage With $_post Variable
Hi Everyone,
I'm trying to make an userlist where I have a button (or better a picture with a hyperlink) which if I click it I will be forwarded to a new page where the user name is send as a $_Post variable. The code below is what I have this far. In some way when I click the send button it always gives the last username in the table in the $_Post variable. Can anyone help me please. Thnx Ryflex <?php require_once('auth.php'); require_once('config.php'); $global_dbh = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD) or die("Could not connect to database"); mysql_select_db(DB_DATABASE, $global_dbh) or die("Could not select database"); $user_query = "SELECT * FROM members"; $user_result = mysql_query($user_query); ?> <form ID="gotoresource" NAME="gotoresource" METHOD="POST" ACTION="resource_compose.php"> <table width="500" border="1" align="center" cellpadding="2" cellspacing="0"> <tr> <td><b>User</b></td> <td><b>Resource</b></td> </tr> <?php while($row = mysql_fetch_assoc($user_result)) { echo "<tr>"; echo "<td>"; echo $row['login']; $user = $row['login']; ?> <input name="User" type="hidden" maxlength="15" id="user" value="<?php echo $user; ?>"/> </td> <td> <input type="submit" name="Submit" value="Send" /> </td> <?php echo "</tr>"; } ?> <html> <head> <title>Userlist</title> <link href="loginmodule.css" rel="stylesheet" type="text/css" /> </head> <body> </body> </html> Similar Tutorialswhat would be the $_POST variable name for a radio button option? I have a form that creates rows of data input textboxes depending on a user input number of things. I have a naming convention for all these textboxes that basically just keeps incrementing a number suffix for each row. All this is working fine. My problem is I need to get the data inserted into this table of textboxes into an array. Here's my code where I attempt to to this (it does not work): Code: [Select] $temp = $_SESSION['Num_Part']; $count = 1; while ($count <= $temp){ $temp2[$count] = "'Participant_P".$count."'"; //echo $temp2[$count]."<br/>"; $temp3[$count]=$_POST[$temp2[$count]]; //here's the problem $temp4[$count] = "'Result_P".$count."'"; $temp5[$count]=$_POST[$temp4[$count]]; //here's the problem //echo $temp4[$count]."<br/>"; $count++; } The problem is that the $_POST does not work with the variable in the argument position - even though the argument is formatted with single quotes. Can a variable be used in a POST argument and if so what is the correct syntax? If not, is there some other simple solution to harvest the data into an array. I understand I can harvest by explicitly accessing each key in the post assoc array. But this could be dozens of rows of input fields. Thanks in advance for your help here. I couldn't find anything online re this topic. Code: [Select] $amountBoxName = "amount".$id; // Constructs the name of the input field $amountToPay = strip_tags($_POST["$amountBoxName"]); $amountBoxName then goes to 'amount22' says, however $amountToPay is then showing also as 'amount22', not getting the value of the text field whose name/id is amount22 on the previous page. How do I do this? I tried having " around the $amountBoxName, ' and no quotes although still can't get it working. Hi, I need to pass value of variable to another php file. I thought it is possible to do it as following: <form action="products.php"> <INPUT TYPE=hidden NAME='id' . VALUE='$id'> </form> But the problem is like that. The php file inside which I want to write above html code has not using <form> tag and it has no buttons. So how to initiate the transfer of variable into another php file? Is the above idea is not the good idea? Are there any another ways? Why doesn't this code work... Code: [Select] // Initialize variables. $form_value = ''; $form_value = $_POST['form_value']; I get this error... Quote Notice: Undefined index: form_value Thanks, Debbie So, i need this. Whenever I try and post a form with radio buttons like this
<input type="radio" name"whatever" value="0">and I do a var_dump($_POST)it always shows
array(11) {["whatever"]=> string(2) "on" }
but this works and will display 1 in the var_dump
<input type="radio" name"whatever" value="1">Why is this. I need the value set to 0 because that's the value going into the database. I don't really want to do a str_replace just to replace "on" with "0" User clicks on a url, ie: example.com/AEQ438J When I perform this in the code below: Code: [Select] $referrer = $_GET['_url']; // echo $referrer displays the referrer ID contents correctly as "AEQ438J" if ( ! empty($referrer)) { $mysqli->query("UPDATE coming_soon_emails SET clicks = clicks + 1 WHERE code='" . $referrer ."'"); } // this also updates the database correctly as it should if (!empty($_POST['email'])){ // echo $referrer displays the referrer ID contents as BLANK. It should display "AEQ438J"! ..... $referrer displays correctly BEFORE if($_POST['form']), however during the if($_POST['form']) $referrer is empty. How can I fix my code so that $referrer is not empty during the time the user posts their email address in the form? Thank you! Complete PHP and HTML Code: [Select] <?php require "includes/connect.php"; //var_dump($_GET);die; function gen_code($codeLen = 7) { $code = ''; for ($i=0; $i<$codeLen; $i++) { $d=rand(1,30)%2; $code .= $d ? chr(rand(65,90)) : chr(rand(48,57)); } return $code; } function add_code($email_id) { global $mysqli; $code = gen_code(7); $mysqli->query("UPDATE coming_soon_emails SET code='" . $code ."' WHERE email_id='" . $email_id . "'"); if($mysqli->affected_rows != 1) { add_code($email_id); } else return $code; } $msg = ''; $referrer = $_GET['_url']; // echo $referrer displays the referrer ID contents correctly if ( ! empty($referrer)) { $mysqli->query("UPDATE coming_soon_emails SET clicks = clicks + 1 WHERE code='" . $referrer ."'"); } if (!empty($_POST['email'])){ // echo $referrer displays the referrer ID contents as BLANK // Requested with AJAX: $ajax = ($_SERVER['HTTP_X_REQUESTED_WITH'] == 'XMLHttpRequest'); try{ if(!filter_input(INPUT_POST,'email',FILTER_VALIDATE_EMAIL)){ throw new Exception('Invalid Email!'); } $mysqli->query("INSERT INTO coming_soon_emails SET email='".$mysqli->real_escape_string($_POST['email'])."'"); if($mysqli->affected_rows != 1){ throw new Exception('This email already exists in the database.'); } else { $email_code = add_code($mysqli->insert_id); } $msg = "http://www.example.com/" . $email_code; //the following doesn't work as referrer is now empty :( if ( ! empty($referrer)) { $mysqli->query("UPDATE coming_soon_emails SET signup = signup + 1 WHERE code='" . $referrer ."'"); } if($ajax){ die(json_encode(array('msg' => $msg))); } } catch (Exception $e){ if($ajax){ die(json_encode(array('error'=>$e->getMessage()))); } $msg = $e->getMessage(); } } ?> <!DOCTYPE html> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title></title> <link rel="stylesheet" type="text/css" href="css/styles.css" /> </head> <body> <div id="launch"> <form id="form" method="post" action=""> <input type="text" id="email" name="email" value="<?php echo $msg;?>" /> <input type="submit" value="Submit" id="submitButton" /> </form> <div id="invite"> <p style="margin-top:20px;">The ID of who referred you: <?php echo $referrer; //this displays correctly?>)</p> <p style="margin-top:20px;"><span id="code" style="font-weight:bold;"> </span></p> </div> </div> <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.3/jquery.min.js"></script> <script src="js/script.js"></script> </body> </html> script.js Code: [Select] $(document).ready(function(){ // Binding event listeners for the form on document ready $('#email').defaultText('Your Email Address'); // 'working' prevents multiple submissions var working = false; $('#form').submit(function(){ if(working){ return false; } working = true; $.post("./index.php",{email:$('#email').val()},function(r){ if(r.error){ $('#email').val(r.error); } else { $('#email').val(r.msg); // not needed but gets hidden anyways... $('#launch form').hide(); $("#code").html(r.msg); $("#invite").fadeIn('slow'); } working = false; },'json'); return false; }); }); // A custom jQuery method for placeholder text: $.fn.defaultText = function(value){ var element = this.eq(0); element.data('defaultText',value); element.focus(function(){ if(element.val() == value){ element.val('').removeClass('defaultText'); } }).blur(function(){ if(element.val() == '' || element.val() == value){ element.addClass('defaultText').val(value); } }); return element.blur(); } htaccess Code: [Select] RewriteEngine on RewriteCond %{HTTP_HOST} ^my-url.com RewriteRule (.*) http://www.my-url.com/$1 [R=301,L] RewriteCond %{REQUEST_FILENAME} !-f RewriteCond %{REQUEST_FILENAME} !-d RewriteRule ^([a-z0-9]+)$ /index.php?_url=$1 [NC,L,QSA] table.sql Code: [Select] CREATE TABLE IF NOT EXISTS `coming_soon_emails` ( `email_id` int(11) NOT NULL auto_increment, `email` varchar(64) collate utf8_unicode_ci NOT NULL, `code` char(7) collate utf8_unicode_ci DEFAULT NULL, `clicks` int(64) collate utf8_unicode_ci DEFAULT 0, `signup` int(64) collate utf8_unicode_ci DEFAULT 0, `ts` timestamp NOT NULL default CURRENT_TIMESTAMP, PRIMARY KEY (`email_id`), UNIQUE KEY `email` (`email`), UNIQUE KEY `code` (`code`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci; I am trying to allow the user to update a variable he chooses by radio buttons, which they will then input text into a box, and submit, to change some attributes. I really need some help here. It works just fine until I add the second layer of variables on top of it, and I can't find the answer to this question anywhere. <?PHP require('connect.php'); ?> <form action ='' method='post'> <select name="id"> <?php $extract = mysql_query("SELECT * FROM cars"); while($row=mysql_fetch_assoc($extract)){ $id = $row['id']; $make= $row['make']; $model= $row['model']; $year= $row['year']; $color= $row['color']; echo "<option value=$id>$color $year $make $model</option> ";}?> </select> Which attribute would you like to change?<br /> <input type="radio" name="getchanged" value="make"/>Make<br /> <input type="radio" name="getchanged" value="model"/>Model<br /> <input type="radio" name="getchanged" value="year" />Year<br /> <input type="radio" name="getchanged" value="color" />Color<br /><br /> <br /><input type='text' value='' name='tochange'> <input type='submit' value='Change' name='submit'> </form> //This is where I need help... <?PHP if(isset($_POST['submit'])&&($_POST['tochange'])){ mysql_query(" UPDATE cars SET '$_POST[getchanged]'='$_POST[tochange]' where id = '$_POST[id]' ");}?> Hi, I have set up 2 php pages page 1 - add_entry2.php In this page I have a invoice table created where I can dynamically add/delete rows. This has a View Bill button which takes me to page 2- add_entry3.php In this page it shows up the rows added in page 1 in read only format, so if the user wants to modify the data that he/she entered then he must Click on <back> that i have provided in the page 2 which will direct him to page 1 Now the problem starts here on click of Back the dynamically added rows dissappear..which is frustrating..I know its something to do with my code..but can anyone help me fix it. One more thing is that i dont want to store the data into DB till the finalise button is clicked on page 2 so that means till page 2 is submitted nothing goes to DB from Page 1. I am able to retain values if I use the code Code: [Select] <form action="add_entry2.php" name="eval_edit" method="post" format="html"> i,e if I submit back to the same page and retrieve values form $_POST but If I use the code Code: [Select] <input type="button" value="Back" onClick="history.go(-1);return true;">to get back to add_entry2.ph it looses all the values. Is there any other way to code the BACK link retaining my $_POST values(Do you think $_SESSION would work in this case?) preg_replace() asks that "Delimiter must not be alphanumeric or backslash" in the pattern. So I changed $new_text = preg_replace($_POST['withthis'] ,$_POST['withthis'],$_POST['text']); to this $replacethis = $_POST['replacethis']; $new_text = preg_replace("/$replacethis/",$_POST['withthis'],$_POST['text']); It works fine, but out of curiosity, is there any way to have the POST variable as a parameter directly, and why does it not work? Just to try it, I attempted: "/$_POST['withthis']/" and $_POST["/'withthis'/"] and both do not work. str_replace is a better option I think, but I am just trying to get a better understanding of this delimiter rule. Thanks for your time! ok so I just want to get the username into the url after the button is clicked Code: [Select] <?php $val = 123; ?> <a href="index.php>Button</a> How do I get the href to be Code: [Select] index.php?user=123 all help greatly appreciated. thanks I know this is 'old stuff', but I am new to this and very frustrated- I am trying to pass a variable to a new url via an html onclick event, and failing. I have found several posts about it but the code posted in those posts does not work when I try it. I have struggled mightly with this. Here is my code: first an echo of two versions of the onclick statement, and two versions of the actual button code. The echos from the two versions appear identical, but the first button with no variable works while the second using a variable results in no action. Can anyone point me in the direction of a solution? I suspect there is some simple principal I am missing relating to parsing strings. <?php echo "window.location.href='ManageTitles.php?dog=2'"; echo "\n" ; echo "window.location.href='ManageTitles.php?dog={$row_rsDogsowned['idDogInfo']}'"; ?> <form methd="POST" Name="form3" action="<?php echo $editFormAction; ?>"> <input type="button" value="works" onclick="window.location.href='ManageTitles.php?dog=2'" /> <input type=button onclick="window.location.href='ManageTitles.php?dog={$row_rsDogsowned['idDogInfo']}'" value="fails" /> </form Hey all, I'm almost too embarrased to post this as I'm sure it is very simple - but for the life of me I can't get this to work. I am a complete newbie to PHP so please bear with me.. What I'm trying to do is to write a program that will go on to calculate a user's Body Mass Index (BMI) based on user inputted data of height and weight. However, I want to be able to accept heights and weights in a number of different units for maximum ease of use. I am trying to write some PHP code that will handle this , the main goal is to convert everything into 'cm' and 'kg' before going on and doing the simple BMI calculation later on. However I am stuck at this point: I have a text field for users to input their weight into. This is immediately followed by two radio buttons for the user to select which units they are inputting their weight in (kilograms or pounds). A third option to input their weight is given after this for those wishing to input their weight in 'stones and pounds'. My problem is as follows: I can't get my code to recognise which radio button (either 'kg' or 'lbs') has been pressed. What should happen is that my code can tell something has been inputted in the text box AND which of the radio buttons has been selected. From this it does one of the two; for the kilogram option it leaves the value as it is, but if the weight has been inputted in pounds (and the 'lbs' radio button selected) then I want the code to convert this into kilos by multiplying by 0.4535. Here is what I have so far (sorry its a bit messy - like i say I am a newbie): <?php //convert.php $ft = $cm = $inches = $weight = $stones =$pounds = $kilos = $units = ""; if(isset($_POST['cm'])) $cm = sanitizeString($_POST['cm']); if(isset($_POST['ft'])) $ft = sanitizeString($_POST['ft']); if(isset($_POST['inches'])) $inches = sanitizeString($_POST['inches']); if(isset($_POST['weight'])) $weight = sanitizeString($_POST['weight']); if(isset($_POST['stones'])) $stones = sanitizeString($_POST['stones']); if(isset($_POST['pounds'])) $pounds = sanitizeString($_POST['pounds']); if(isset($_POST['kilos'])) $kilos = sanitizeString($_POST['kilos']); if(isset($_POST['units'])) $units = sanitizeString($_POST['units']); if ($ft != '') { $height = intval(($ft * 30.48) + ($inches * 2.54)); $out = "you are $height cm tall"; } elseif($cm != '') { $height = intval($cm); $out = "you are $height cm tall"; } else $out = ""; if ($stones != '') { $kilos = intval((($stones * 14) + $pounds) * 0.45359237); $out2 = "you weigh $kilos Kg"; } elseif($weight != '' AND $units = "kg") { $kilos = $weight; $out2 = "you weigh $kilos Kg"; } elseif($weight != '' AND $units = "lbs") { $kilos = ($weight * 0.45359237); $out2 = "you weigh $kilos Kg"; } else $out2 = ""; echo <<<_END <html><head><title>Height & Weight Converter</title> </head><body><pre> Please enter your details below <b>$out$out2</b> <form method="post" action="convert.php"> Height: <input type="text" name="cm" size="3"> cm OR <input type="text" name="ft" size="1">ft <input type="text" name="inches" size="2">inches Weight: <input type="text" name="weight" size="4" /> Kg<input type="radio" name="units" value="kg" /> lbs<input type="radio" name="units" value="lbs" /> OR <input type="text" name="stones" size="2">stone <input type="text" name="pounds" size="2">pounds <input type="submit" value="Submit" /> </form></pre></body></html> _END; function sanitizeString($var) { $var = stripslashes($var); $var = htmlentities($var); $var = strip_tags($var); return $var; } ?> Hello World, I need some tips on how to make a button that does something like the code below. .......echo '<tr><td align="left"><a href="add_cart.php?pid=' . $row1['item_id'] . '">Add to Cart</a></td>........ I've been messing with html form and button type with no success. This code does the job but I prefer a button. Thanks This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=350095.0 I am new at php. I am trying to loop a number in each click of a button. See attached picture for reference. 1 should iterate after the click of the button and will stop to iterate when it gets to 5.
Here is my code:
<!DOCTYPE html>
<html>
<body>
<div>Question <?php $num = 5; $n = 1; $n <= $num; echo $n; ?> of <?php echo $num;?></div>
<form method="post"> <button id="button" class="button" value="add" name="add">Click</button> </form>
</body>
</html>
If I try to loop it this way,
<!DOCTYPE html>
<html>
<body>
<div>Question <?php $num = 5; for($i = 1; $i <= $num; $i++){ echo $i; } ?> of <?php echo $num;?></div>
<form method="post"> <button id="button" class="button" value="add" name="add">Click</button> </form>
</body>
</html>
the result is Question 12345 of 5.
I am trying to create a page for customers to enter their details. I am using a html form.
When the submit button is pressed the form posts the inputs to the same page, which then checks if the inputs are empty. If they are not then each post variable is allocated a session variable so this info can be accessed late on in the system.
If some of the inputs are empty then the value of the input forms become equal to the session variables that they were just allocated to so that the customer doesn’t have to retype their information.
This is where the problem occurs. When I load the page each input box has a slash inside it and when the submit button is pressed a mother slash is added.
My code is below:
<?php
session_start();
if(isset($_POST['NextPage'])){
if (!empty($_POST['CName'])){
$_SESSION["CName"] = $_POST['CName'];
if (!empty($_POST['CStreet'])){
$_SESSION["CStreet"] = $_POST['CStreet'];
if (!empty($_POST['CTown'])){
$_SESSION["CTown"] = $_POST['CTown'];
if ($_POST['Counties'] != "-"){
$_SESSION["CCounty"] = $_POST['Counties'];
if (!empty($_POST['CPostcode'])){
$_SESSION["CPostcode"] = $_POST['CPostcode'];
if (!empty($_POST['CEmail'])){
$_SESSION["CEmail"] = $_POST['CEmail'];
if (!empty($_POST['CNumb'])){
$_SESSION["CNumb"] = $_POST['CNumb'];
$NotEmpty = true;
}else{
$ErrorMsg = "Number is empty. </br>";
}
}else{
$ErrorMsg = "Email is empty. </br>";
}
}else{
$ErrorMsg = "Postcode is empty. </br>";
}
}else{
$ErrorMsg = "County is empty. </br>";
}
}else{
$ErrorMsg = "Town is empty. </br>";
}
}else{
$ErrorMsg = "Street is empty. </br>";
}
}else{
$ErrorMsg = "Name is empty. </br>";
}
}
$content = ' <h3 id="CTitle"> Customer Details </h3>
<p><i>'.$ErrorMsg.'</i></p>
<form action=" " method="POST" name="CDetails" id="CDetails">
Name: * <input type="text" name="CName" size="30" value='.$_SESSION["CName"].'/></br>
First line of your address: * <input type="text" name="CStreet" size="40" value='.$_SESSION["CStreet”];.’/></br>
Town: * <input type="text" name="CTown" size="25" value='.$_SESSION["CTown"].'/></br>
Postcode: * <input type="text" name="CPostcode" size="11" value=‘.$_SESSION["CPostcode"].'/></br>
Email address: * <input type="text" name="CEmail" size ="35" value='.$_SESSION["CEmail”];.’/></br>
Phone Number: * <input type="text" name="CNumb" value='.$_SESSION["CNumb"].'/></br>
<input type="submit" name="NextPage" value="Next" id="Next”/>
</form>
?>
I use a form to sent date and use php to display it . However , the code only can working on one page. I can not turn the page. I do not why . Please tell me reason . Thank you very much. <select name="kind"> <option >kind</option> <option value="Copier Toner">Copier Toner</option> <option value="Laser Toner">Laser Toner</option> <option value="MICR Toner">MICR Toner</option> <option value="Inkjet">Inkjet</option> php code if (isset($_GET["page"])) { $page = $_GET["page"]; } else { $page=1; }; $start_from = ($page-1) * 18; $select="select * from $chun where brand = '$_POST[brand]' or sort ='$_POST[kind]' or type='$_POST[type]' LIMIT $start_from, 18"; $result2=mysql_query($select, $connection) or die (mysql_error()); <?php $sql = "select count(*) from $chun where brand = '$_POST[brand]' or sort='$_POST[kind]' or type='$_POST[type]' "; $rs_result = mysql_query($sql,$connection); $row = mysql_fetch_row($rs_result); $total_records = $row[0]; $total_pages = ceil($total_records / 18); for ($i=1; $i<=$total_pages; $i++) { ?> <div class="trunpage"><a href='table2.php?page=<?php echo "$i" ; ?>&id=<?php echo "$_POST[brand]";?>&cd=<?php echo "$_POST[kind]";?>&td=<?php echo "$_POST[type]";?>' ><?php echo "$i" ; ?></a> </div> The first page is working fine. The second page I get error message. Undefined index: brand in C:\wamp\www\php1000\table2.php on line 234 Hi guys, I have a form using radio buttons. For the radio button, lets just say i have the id name as 'rim' + number eg; rim0, rim1, rim2.... When i post the data to another file to execute the data collected, naturally i would use the: Code: [Select] $rim0=$_POST['rim0']; $rim1=$_POST['rim1']; .... $rim10=$_POST['rim10']; i tried to shorten this process using this method: Code: [Select] //$q is part of the post variable. for($x = 0; $x < count($q); $x++){ $rim[]=$_POST['"rim"."$x"']; } But i get this error Code: [Select] Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Is there something wrong with my POST syntax? |
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