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PHP - Problem To Retrieve Form Data From Xml Page
Similar TutorialsHi all, I have the following code that generates a table of results from a MySQL query: $i=1; while($arr = mysql_fetch_array($result, MYSQL_NUM)) { $table .= "<tr>" ."<td width='5px';><input type='checkbox' name='transcheck".$i."'></td>" ."<td id='parent".$i."A'>".$arr[0]." ".$arr[1]."</td>" ."<td id='parent".$i."B'>".$arr[2]."</td></tr>" ."<tr style='display:none'><input type='hidden'; name='transemail".$i."'; value='".$arr[9]."'></tr>"; $i++; } $table .= "</table>"; echo $table; As you can see it generates a table containing (amongst other things): checkboxes: transcheck1,2........9...etc. hidden inputs: transemail1,2........9...etc. This table is inside a form, so that the above gets posted to another php file. What I now want to do in this 2nd php file, is to retrieve all the checkboxes and hidden inputs and then to display the values of the hidden inputs, where the corresponding checkbox has been checked. So e.g. if transcheck1, transcheck3 and transcheck12 have been checked, then I want to display transemail1, transemail3 and transemail12. I can see that this should be relatively straightforward, but I'm fairly new to this stuff, could someone pls help me out? Thanks! Hi, I have an HTML form with a "submit" button that sends form data to a PHP script. Firstly, the form doesn't seem to send the data to the PHP script, unless I include a line in the PHP script as follows: $var1 = $_REQUEST['var1']; Is this always the case or is there a way to automatically have the form send through data? The reason I ask is because the form includes elements that can be disabled, in which case (if element 'var1' is disabled) the above PHP line returns an error "Notice: Undefined index: var1" i.e. I want to only send through data from enabled elements and not from disabled elements. Alternatively, is there a line I can include in the PHP script something along these lines: if (defined($var1)==FALSE) {$var=" ";} so that when the PHP script tries to retrieve var1, which is undefined because the corresponding form element has been disabled, it sets it to a particular value. my HTML form is along these lines: Code: [Select] <form action=something.php method=POST> <select id="var1" name="var1"> <option> </option> <option selected="selected"> option1</option> <option> option2</option> <option> option3</option> </select> <input type="submit" value="next"> </form> Thanks! I'm creating a report page and can't figure out how to retrieve multiple rows from a table and display them in html. There are 5 data elements in each row and I'm thinking that using a form in the html might be the best way as I'm totally ignorant about tables in html. I'm a newbie to php and can't figure out how to accomplish this. Here's the code that I have so far:
<?php
$filename = NULL;
session_start();
// start of script every time.
// setup a path for all of your canned php scripts
$php_scripts = '../php/'; // a folder above the web accessible tree
// load the pdo connection module
require $php_scripts . 'PDO_Connection_Select.php';
//*******************************
// Begin the script here
// Connect to the database
if (!$con = PDOConnect("foxclone")):
{
echo "Failed to connect to database" ;
exit;
}
else:
{
$sql = 'SELECT COUNT(IP_ADDRESS) FROM download WHERE FILENAME IS NOT NULL';
$sql1 = 'Update download t2, ip_lookup t1 set t2.country = t1.country, t2.area = t1.area, t2.city = t1.city where ((t2.IP_ADDRESS) = (t1.start_ip) OR (t2.IP_ADDRESS) > (t1.start_ip)) AND ((t2.IP_ADDRESS) = (t1.end_ip) OR (t2.IP_ADDRESS) < (t1.end_ip)) AND (t2.FILENAME is not null and t2.country is null)';
$sql2 = 'SELECT (IP_ADDRESS, FILENAME, country, area, city) from download where FILENAME is not null';
// Update the table
$stmt = $con->prepare($sql1);
$stmt->execute();
// Get count of rows to be displayed in table
$stmt = $con->prepare($sql);
$stmt->execute() ;
$cnt = $stmt->fetch(PDO::FETCH_NUM);
// retrieve one row at a time
$i = 1;
while($i <= $cnt){
$stmt = $con->prepare($sql2);
$row->execute(array('')); // Do I need an array here?
// from here on, I'm lost
$i++;
I'd appreciate any guidance you can provide or understandable tutorials you can point me to. Larry As a complete newbie to php and webdesigning i have a following problem.I would like to retrieve the data from database and display it in a drop down menu.Then i should allow the user to select the values from drop down list along with other details,in other words i have to embed the drop down output as the form input for the user and store the form data in another table.I am running a xampp server and i am using php 5.4 version.Please help.My code is as follows.In this case project_name is displayed as the drop down output.but how do i use the same drop down output as a input in the form. <html> <head></head> <body> <?php error_reporting(E_ALL ^ E_DEPRECATED); include 'connect.php' ; $tbl_name="projects"; $sql="SELECT project_name FROM $tbl_name "; $result=mysql_query($sql); if($result === FALSE) { die(mysql_error()); } ?> <form name="resources" action="hourssubmit.php" method="post" > <?php echo "<select name='project_name'>"; while ($row = mysql_fetch_array($result)) { echo "<option value='" . $row['project_name'] ."'>" . $row['project_name'] ."</option>"; } echo "</select>"; ?> </form> </body> </html> hi guys, im new to this forum I'm new also to php, I need help from you guys: I want to display personal information from a certain person (the data is on the mysql database) using his name as a link: example: (index.php) names 1. Bill Gates 2. Mr. nice Guy i want to click Bill Gates (output.php) Name: Bill Gates Country:xxxx Age: xx etc. How can i make this or how to learn this? Hey everyone. I am creating a website so my family can select from the list of present my daugther has asked for. I have them logging in, and that works. I have the table data set and the search and table display works. But I would like to display the list (as in the code below) but with a checkbox in the first column (add a column). From that the authenticated user can click on the item they have purchased and that selection will be moved to another table (called purchased). I have that code. The only thing I cannot figure out is to present the data in list form with a checkbox (like you would see on an order form). Here is the display code (there is no error checking at this point): <head><LINK REL="SHORTCUT ICON" HREF="cmwschl.ico"></head> <body bgcolor="#C0C0C0"> <font face="Arial" color="#000080">Books from the Selected Series</font> </h1> <hr font color="Navy" font size="3"> <center> <table border="1" cellpadding="5" cellspacing="0" bordercolor="#000000"> <tr> <td width="170" bgcolor="#FFCC66"><center><b><font color="navy" size="2" face="Trebuchet MS"><b><center>Book Number</center></font></b></center></td> <td width="140" bgcolor="#FFCC66"><center><b><font color="navy" size="2" face="Trebuchet MS"><b><center>Book Group</center></font></b></center></td> <td width="100" bgcolor="#FFCC66"><center><b><font color="navy" size="2" face="Trebuchet MS"><b><center>Book Title</center></font></b></center></td> </tr> <?php $con = mysql_connect("localhost","twilson","R00tb33r!") or die('Connection: ' . mysql_error());; mysql_select_db("CMWWeb", $con) or die('Database: ' . mysql_error()); ?> <? $group = $_POST['bookgrp']; //if ($Prod <> 0) { $sql = "SELECT * FROM `OpenBooks` WHERE `bookgrp`= '$group'"; $results = mysql_query($sql); if ($results) { //this will check if the query failed or not if (mysql_num_rows($results) > 0) { //this will check if results were returned while($copy = mysql_fetch_array($results)) { $variable1=$copy['booknum']; $variable2=$copy['bookgrp']; $variable3=$copy['booktitle']; //table layout for results print ("<tr>"); print ("<td><center>$variable1</center></td>"); print ("<td><center>$variable2</center></td>"); print ("<td><left>$variable3</left></td>"); print ("</tr>"); } } else { echo "No results returned"; } } else { echo "Query error: ".mysql_error(); } mysql_close($con); ?> </table> </center> Hi folks, Complete No0b here when it comes to PHP and MySQL. I am in the middle of creating a PHP website. What I want to do is have the contents of a page in a MySQL table and have PHP gather the page content and display it on the page. Here is what I have in my home.tpl file: Code: [Select] <?php $query = "SELECT home, FROM $database_name"; $result = mysql_query($query); ?> And, in my index.php I simply call that home.tpl to display the data by using: Code: [Select] <?php include 'templates/default/home.tpl'; ?> Now, all I get is a blank page on index.php. Is there something else I should be doing? Remember, I am a complete No0b! Thanks folks. I have code for displaying result from database:
foreach ($_SESSION["products"] as $cart_itm)
{
$product_code = $cart_itm["code"];
$results = $mysqli->query("SELECT * FROM products WHERE product_code='$product_code' LIMIT 1");
$obj = $results->fetch_object();
Now, I display data with: $obj->product_name
When I get the list, how to put inline numbers (1 2 3 4 ...)?
1 products1
2 products1
3 products1
4 products1..
This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=322859.0 I have some php code on view invoice and it's getting all the data I need to from three database tables but for some reason it's not getting the data for one specific column and unsure why I have done a echo sql and it is echoing the sql query ok and have ran the sql query in phpmyadmin and is displaying the column in phpmyadmin that is not displaying on the php page so am bit confused, below is the coding I have. I took out a lot of it and tried to keep to the relevant parts <?php
$sql = "SELECT t1.id, t1.invoice_number, DATE_FORMAT(t1.invoice_date,'%d/%m/%Y') AS invoice_date, t1.user_id, t2.invoice_number, t2.service_name, t2.service_price, t3.user_id, t3.customer_name, t3.customer_phone, t3.customer_email, t2.service_name, t2.service_price, t1.invoice_total, t1.balance, t1.notes, DATE_FORMAT(t1.payment_date,'%d/%m/%Y') AS payment_date, t1.payment_method
FROM invoices as t1
LEFT JOIN invoice_products as t2
ON t1.id = t2.invoice_id
LEFT JOIN users as t3
ON t1.user_id = t3.user_id
WHERE t1.user_id = '".$_SESSION['user_id']."' and t1.id = '".$_GET['id']."'";
$result = mysqli_query($connect, $sql);
if(mysqli_num_rows($result) > 0)
{
if($row = mysqli_fetch_array($result))
{
?>
Notes: <?php echo $row['notes'];?>
<?php
}
}
?>
Hi I have coded a drop down menu with php and i am trying to retrieve the data when a user select a option from the menu and the data is retrieved from the database. So far i have tried and nothing is displaying when i tried to process the php form. Sales.php Page <form action="saleprocess.php" method="GET"> <?php echo 'Product Model:'; $query="SELECT * FROM products"; /* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */ $result = mysql_query ($query); echo "<select name=product_model value=Select>Product Model</option>"; // printing the list box select command while($rows=mysql_fetch_array($result)){//Array or records stored in $nt echo "<option name=product_model value='.$rows[product_id].'>$rows[product_model]</option>"; /* Option values are added by looping through the array */ } echo "</select><br>"; ?> <input type='submit' name='submit' value='Create'></input> <br> </form> ******************************************************************************** Salesprocess.php page <?php include("connect.php"); if(isset($_GET['product_id'])){ $product_id = $_GET['product_id']; $query = mysql_query("SELECT * FROM products WHERE product_id= $product_id"); while($rows = mysql_fetch_assoc($query)) { echo 'Product Model<br>'; echo $rows['product_id']; echo $rows['product_model']; } } ?> Muchly appreciated if someone can help me I insert multiple id from my checkbox to mysql database using php post form. in e.x i insert id (checkbox value table test) to mysql. no i need to any function for retrieve data from mysql and print to my page with my e.x output.(print horizontal list name of table test where data = userid) my checkbox value ( name table is test ) : Code: [Select] 01 ---id----- name ---- 02 ---1 ----- test1 ---- 03 ---2 ----- test2 ---- 04 ---3 ----- test3 ---- 05 ---4 ----- test4 ---- 06 ---5 ----- test5 ---- 07 ---6 ----- test6 ---- 08 ---7 ----- test7 ---- 09 ---8 ----- test8 ---- 10 ---9 ----- test9 ---- mysql data Insert ( name of table usertest ): Code: [Select] 1 ---id----- data ---- userid ----- 2 ---1 ----- 1:4:6:9 ---- 2 ----- 3 ---2 ----- 1:2:3:4 ---- 5 ----- 4 ---3 ----- 1:2 ---- 7 ----- example outout : ( print horizontal list name of table test where data = userid ) print? Code: [Select] 1 user id 2 choise : test1 - test4 - test6 - test9 Thanks I want to use session to do a query and will I be able to do this? I have a session that was gathered from login and now i was to use this session to do a query If Yes, How? I m trying to fetch a image from mysql (blob) with header..here is my coding..."<?php include("db.php"); $query=mysql_query("select * from table where id='3' "); $row=mysql_fetch_array($query); $r=$row['image']; header("content-type:image"); echo $r; ?>" i want to fetch another fields from the database....but when i try to echo another fields...the page shows error or it does not echo other fields of database....please help me...how can i resolve it...i want to fetch other fields from database,like'username'password'firstname'lastname and image...thanks in advance... I am trying to create a code where the user enters the zip code and if found in the database, a location will be retrieved. I am getting two undefined variables as errors. I am still learning the POST and GET method and I fear that is where I am getting something mixed up. I have also attached an image of my DB. Thank you so much any input is greatly appreciated. Notice: Undefined variable: address in C:\xampp\htdocs\index.php on line 53 Notice: Undefined variable: zip in C:\xampp\htdocs\index.php on line 54
<?php
include ('header.php');
include ('function.php');
?>
<div class="wrapper">
<div>
<?php
echo'<form method="POST" action"'.getLocations($conn).'">
<input type="text" name="zip" class="search" placeholder="Zip code"><br>
<button type="submit" value="submit" id="submit">Submit</button>
</form>';
getLocations($conn);
?>
</div>
</div>
<div>
<!--foreach($zip as $zip) : -->
<?php
echo $address['address'];
echo $zip['zip'];
?>
</div>
function.php file code:
<?php
$dBServername = "localhost";
$dBUsername = "root";
$dBPassword = "";
$dBName = "addresses";
// Create connection
$conn = mysqli_connect($dBServername, $dBUsername, $dBPassword, $dBName);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
function getLocations($conn) {
if (isset($_POST['submit'])){
// Validate Zip code field
if (!empty ($_POST['zip']) && is_numeric ($_POST['zip'])) {
$zip = (int)$_POST['zip'];
$query = "SELECT * FROM locations WHERE zip = '$zip'";
//get the results
$result = mysqli_query($conn, $query);
//fetch the data
$zip = mysqli_fetch_all($result, MYSQLI_ASSOC);
var_dump($zip);
mysqli_free_result($result);
//close connection
mysqli_close($conn);
}
}
}
I have been given a task, and I gotta say it is kicking my butt.
Here is what I have to do.
1. Have user fill out and submit a form.
2. Data gets sent to: http://www.ffiec.gov...de/Default.aspx
3. Data is set as values for input fields in the sites form.
4. Form executes.
5. Retrieve result data.
6. Display data back to my site.
I have no idea how to do this.
Usually when I have done something like this I use an API.
Hope my question is clear.
Thanks for the help.
I am not sure if the title is correct; I tried my best.
I'm a PHP/MySQL beginner and I really need some help.
I have a small script that I am using for sending SMS. I recently added a phonebook. The problem with the phonebook right now is that it's available to all users, i.e. they can all update and delete all rows. What I would like to do is make it so that each user can update and delete only their own contacts.
I have a table call contacts. Inside that table there is first name, last name, company and phonenumber.
How can I accomplish this with PHP & MySQL?
CREATE TABLE IF NOT EXISTS `contacts` ( `contact_id` int(10) NOT NULL AUTO_INCREMENT, `firstname` varchar(255) NOT NULL, `lastname` varchar(255) NOT NULL, `company` varchar(255) NOT NULL, `cell_no` text NOT NULL, PRIMARY KEY (`contact_id`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=8 ; CREATE TABLE IF NOT EXISTS `users` ( `user_id` int(11) NOT NULL AUTO_INCREMENT, `users_name` varchar(30) NOT NULL, `uname` varchar(30) NOT NULL, `u_pass` varchar(60) NOT NULL, `utype` varchar(30) NOT NULL, `timezone` varchar(30) NOT NULL, `uapi_user` varchar(30) NOT NULL, `uapi_pass` varchar(60) NOT NULL, PRIMARY KEY (`user_id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=20 ;
<?php
if (isset($_POST['submit'])){
//form has been submitted1
$firstname = trim($_POST['firstname']);
$lastname = trim($_POST['lastname']);
$company = trim($_POST['company']);
$cellno = trim($_POST['cell_no']);
if($firstname == ''){
echo '<div class="alert alert-danger">First Name is not Valid!</div>';
exit;
}elseif($lastname == ''){
echo '<div class="alert alert-danger">Last Name is not Valid!</div>';
exit;
}elseif($company == ''){
echo '<div class="alert alert-danger">Company is not Valid!</div>';
exit;
}elseif($cellno == ''){
echo '<div class="alert alert-danger">Cellphone Number is not Valid!</div>';
exit;
}else{
$query = "Select cell_no from contacts where cell_no = '".$cellno."' ";
$result = mysql_query($query);
if (!mysql_num_rows($result)) {
$sql = "INSERT INTO contacts(firstname, lastname, company, cell_no)
values('{$firstname}','{$lastname}', '{$company}', '{$cellno}')";
$result = mysql_query($sql);
confirm_query($result);
//echo '<div class="alert alert-success">Successfully added.</div>';
//exit;
?>
<script type="text/javascript">
window.location = "contact_list.php";
</script>
<?php
}
else{
echo '<div class="alert alert-danger">Username. already exist!.</div>';
echo '<p><a href="new_contact.php" class="btn btn-success"> Back </a></p>';
exit;
}}
}else{
$firstname = "";
$lastname = "";
$company = "";
$cellno = "";
}
?>
Hi, I am newbie to PHP. If any one explain me with a simple example, it would be really great. I have a form page "queue_info.php". After user hits SUBMIT, this page will redirect to an action page "queue_action.php". from form page data are passed using "$quests=$_POST['quests'];" to this action page. Now i need to resend this same data back to the form page "queue_info.php" from action page. How to do this? Appreciate your inputs on this. thanks, prakash Hi, I am sending data from a form to a URL using $_GET but I would also like the user to be redirected to a thankyou page which is a different URL. Does anyone know if this is possible? Thanks, I ve been having trouble processing form data to the same page as the do on facebook comments. Can php scripts do that. Help me guys!! |
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