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PHP - Send Data To External Site, Then Retrieve The Results.
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.
Similar TutorialsHi, I'm trying to setup a quick PHP script that will grab the email from the url (see below) and after inserting into MySQL db - which is working fine - the script will complete two additional tasks: 1. send that same captured email out to a external db as in shown via http://domain1.com/insert.php?email=$lead (example), but then send to a DIFFERENT source - the originator of the lead - a portback acknowledgement using Header (sending the status and email to http://domain2.com/check.php?e=$lead&s=$status for their records). See the code below: ------------------------- Code: [Select] $lead = $_REQUEST['e_mail']; // will grab email from posted url string and assign to local variable $result = mysql_query($command); // this is just to execute the MySQL insert which works just fine but included here to explain validation below // Create API Call string to insert lead into iContact folder $requestURL = "http://domain1.com/insert.php?email=$lead"; // Execute API Call to CAKE $xml = simplexml_load_file($requestURL) or die("feed not loading"); if ($result) { $status = 1; // mark lead as sucess // send postback on lead status header("Location: http://domain2.com/check.php?e=$lead&s=$status"); } -------- Problem: I'm getting all sorts of errors with the simplexml_load_file() function and can't figure out why it won't work. Any input appreciated as this the only way I know how to pass the lead onward and then inform/update the other party of receipt of information. thanks! Hello, im new here, and i have little experience to php and mysql as i started for 2 weeks ago. I started out with some tutorials and feeling im getting the hang of it. Enough of me, lets get to the point: <?php $con = mysql_connect('localhost',$user,$pass)or die(mysql_error()); $selectdb = mysql_select_db($selectdb)or die(mysql_error()); $sql = "SELECT * From table"; $result = mysql_query($sql); $num = mysql_num_rows($result); $myarray = array($result); $i =0; while ($i < $num){ echo $myarray[$i]; $i++; } ?> Here i have written a dummyscript that does what the original script does, it tries to fetch the keys from the table and then trying to loop it and echo out the results. The output in the browser is this: Resource id #3 I know this probably is a simple fix but i cant seem to get it sorted out. Hope some of you could help me get this baby work, or maybe have another way of doing it more "simple". Thanks in advance! Dan-Levi hi guys, im new to this forum I'm new also to php, I need help from you guys: I want to display personal information from a certain person (the data is on the mysql database) using his name as a link: example: (index.php) names 1. Bill Gates 2. Mr. nice Guy i want to click Bill Gates (output.php) Name: Bill Gates Country:xxxx Age: xx etc. How can i make this or how to learn this? I'm currently running a classified ads site and planning to display my own content from database combined with and external site rss. So here is what i got right now after the db query for the jobs ads (procedural php),
while ($row = mysqli_fetch_array($results, MYSQLI_ASSOC)){
echo '<div class="media margin-none">
<a class="pull-left bg-inverse innerAll text-center" href="#"><img src="'.$foto.'" share_alt="" width="100" height="100"></a>
<div class="media-body innerAll">
<h4 class="media-heading innerT">
<a href="' . $row['title'] .'-da' . $row['id_ad'] . '" class="text-inverse">'. $remuneracion .' ' . substr(ucfirst(strtolower($row['title'])), 0, 53) . '</a> <small class="pull-right label label-default"><i class="fa fa-fw fa-calendar-o"></i> ' . $row['date_created'] . '</small></h4>
<p>' . substr(ucfirst(strtolower($row['description'])), 0, 80) . ' ...</p>';
echo '</div>
</div>
<div class="col-separator-h"></div>';
}
echo pagination($statement,$per_page,$page, $url_filtros, $filtros);
?>
it is the while loop that i use to display ads from my database, what could be the best way to display (in this same loop?) other site's rss feed so i can show my content combined with the external rss?
Thanks
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 guys, I'm not so good with PHP (hence my post here). Other forums have suggested using javascript, but if not enabled by the user, then it's not much good). I'm trying to set up a search box with submit button. For any keywords searched for, I would like the search terms sent to an external site in the format: http://www.domain.com/default.aspx?st=FT&ss=XXXXX (where XXXXX is the keyword). So far, I have: <?php $keyword = htmlspecialchars($_POST['keyword']); ?> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <form method="post" id="searchform" action="http://www.domain.com/default.aspx?st=FT&ss=<?php echo 'keyword'; ?>" /> <input type="text" name="keyword" id="se" size="35" onblur="if (this.value == '') {this.value = 'search...';}" onfocus="if (this.value == 'search...') {this.value = '';}" value="search..." class="text" /> <input type="submit" id="searchsubmit" class="submit" value="Send" /> </form> </body> </html> .and although it directs to the url, the keyword(s) is not present. Any help would be appreciated. I am creating a library app (personal dev) and have ran into some trouble. I'm very new to Ajax and Php
I have a page named addEntryISBN which shows the results of a user search in a div named #results. I want to post the contents of #results to my database. The contents of #results comes from a page named searchISBN.
What is the most effective way of doing this? My code so far is aas follows;
addEntryISBN.php
$(document).ready(function() {
$('form').on('submit', function (e) {
e.preventDefault();
$.ajax({
type: 'post',
url: 'searchIsbn.php',
data: $('form').serialize(),
success: function (result) {
$('.result').html(result);
}
});
return true;
});
});
<div class="result"</div>
Here's the code that deals with the client side:
<?php
session_start();
if(!isset($_SESSION['Logged_in'])){
header("Location: /page.php?page=login");
}
?>
<!DOCTYPE Html>
<html>
<head>
<!--Connections made and head included-->
<?php require_once("../INC/head.php"); ?>
<?php require_once("../Scripts/DB/connect.php"); ?>
<!--Asynchronously Return User Names-->
<script>
$(document).ready(function(){
function search(){
var textboxvalue = $('input[name=search]').val();
$.ajax(
{
type: "GET",
url: 'search.php',
data: {Search: textboxvalue},
success: function(result)
{
$("#results").html(result);
}
});
};
</script>
</head>
<body>
<div id="header-wrapper">
<?php include_once("../INC/nav2.php"); ?>
</div>
<div id="content">
<h1 style="color: red; text-align: center;">Member Directory</h1>
<form onsubmit="search()">
<label for="search">Search for User:</label>
<input type="text" size="70px" id="search" name="search">
</form>
<a href="index.php?do=">Show All Users</a>|<a href="index.php?do=ONLINE">Show All Online Users</a>
<div id="results">
<!--Results will be returned HERE!-->
</div>
search.php
<?php //testing if data is sent ok echo "<h1>Hello</h1><br>" . $_GET['search']; ?>This is the link I get after sending foo. http://www.family-li...php?&search=foo Is that mean it was sent, but I'm not processing it correctly? I'm new to the whole AJAX thing. This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=322859.0 I insert multiple id from my checkbox to mysql database using php post form. in e.x i insert id (checkbox value table test) to mysql. no i need to any function for retrieve data from mysql and print to my page with my e.x output.(print horizontal list name of table test where data = userid) my checkbox value ( name table is test ) : Code: [Select] 01 ---id----- name ---- 02 ---1 ----- test1 ---- 03 ---2 ----- test2 ---- 04 ---3 ----- test3 ---- 05 ---4 ----- test4 ---- 06 ---5 ----- test5 ---- 07 ---6 ----- test6 ---- 08 ---7 ----- test7 ---- 09 ---8 ----- test8 ---- 10 ---9 ----- test9 ---- mysql data Insert ( name of table usertest ): Code: [Select] 1 ---id----- data ---- userid ----- 2 ---1 ----- 1:4:6:9 ---- 2 ----- 3 ---2 ----- 1:2:3:4 ---- 5 ----- 4 ---3 ----- 1:2 ---- 7 ----- example outout : ( print horizontal list name of table test where data = userid ) print? Code: [Select] 1 user id 2 choise : test1 - test4 - test6 - test9 Thanks I have code for displaying result from database:
foreach ($_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..
Hi 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 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 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? 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 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
}
}
?>
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... 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) {$var=" ";} so that when the PHP script tries to retrieve var1, which is undefined because the corresponding form element has been disabled, it sets it to a particular value. my HTML form is along these lines: Code: [Select] <form action=something.php method=POST> <select id="var1" name="var1"> <option> </option> <option selected="selected"> option1</option> <option> option2</option> <option> option3</option> </select> <input type="submit" value="next"> </form> Thanks! I am trying to create a code where the user enters the zip code and if found in the database, a location will be retrieved. I am getting two undefined variables as errors. I am still learning the POST and GET method and I fear that is where I am getting something mixed up. I have also attached an image of my DB. Thank you so much any input is greatly appreciated. Notice: Undefined variable: address in C:\xampp\htdocs\index.php on line 53 Notice: Undefined variable: zip in C:\xampp\htdocs\index.php on line 54
<?php
include ('header.php');
include ('function.php');
?>
<div class="wrapper">
<div>
<?php
echo'<form method="POST" action"'.getLocations($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);
}
}
}
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