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PHP - Send A Link Value And Request It Back Using Ajax?
Hi,
I would like to send a value in a link and request it back in ajax and set it as a variable in php in the same page, can anyone help me here? thanks 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=$status"); } -------- Problem: I'm getting all sorts of errors with the simplexml_load_file() function and can't figure out why it won't work. Any input appreciated as this the only way I know how to pass the lead onward and then inform/update the other party of receipt of information. thanks! hey guys im tring to look for a good way of finding out if a ajax request is being performed on my page...now ive had a look about and found these two snippets.
Both seems to work fine...any suggestions on what one is best to use or even if there is a better way to do this would be helpful...thank you
if (isset($_SERVER['HTTP_X_REQUESTED_WITH']) &&
strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) === "xmlhttprequest")
{
// is ajax
}
if (strpos($_SERVER['HTTP_ACCEPT'], 'application/json') !== false){
// is ajax
}
hi everyone, can you please kindly check my code for error Code: [Select] $xmltosend = "<?xml version='1.0'?>\r\n". "<soap:Envelope xmlns:soap='http://www.w3.org/2001/12/soap-envelope' soap:encodingStyle='http://www.w3.org/2001/12/soap-encoding'>\r\n". "<soap:Body xmlns:m='https://www.trustwave.com/'>\r\n". "<request>\n". "<authentication>\n". "<username>username</username>\n". "<password>password</password>\n". "</authentication>\n". "<operation>getResellerProducts</operation>\n". "<params>\n". "<int>12345</int>\n". "</params>\n". "</request>\n". "</soap:Envelope>\n"; $reqheader = "POST /index.php HTTP/1.1\r\n". "Host:www.trustwave.com\r\n". "Content-Type: application/soap+xml; charset=utf-8\r\n". "Content-Length: ".strlen($xmltosend) ."\r\n\r\n"; $url = 'https://testapi.ssl.trustwave.com/3.0/'; // fake - obviosly! $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_POSTFIELDS, $xmltosend); curl_setopt($ch, CURLOPT_POST, true); // what to post curl_setopt($ch, CURLOPT_BINARYTRANSFER, 1); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 1); curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0); curl_setopt($ch, CURLOPT_CUSTOMREQUEST, $reqheader); curl_setopt($ch, CURLOPT_USERPWD, "username:password"); $result = curl_exec($ch); curl_close($ch); print $result; im getting this error Code: [Select] Authorization Required This server could not verify that you are authorized to access the document requested. Either you supplied the wrong credentials (e.g., bad password), or your browser doesn't understand how to supply the credentials required. anyone pleeeaassseee...!!! tia I want to be able to send a GET request to another website for example i want to be able to run this http://www.mywebsite.com/blog.php?clear_cache= using curl but when i use curl the website is loaded into my own heres my code Code: [Select] $curl = curl_init(); $json_url = $_SESSION['blog_base'] . 'blog.php?empty_cache='; curl_setopt ($curl, CURLOPT_URL, "$json_url"); $ANSWER=curl_exec ($curl); curl_close ($curl); This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=347835.0 This part of the question is based off of php so I'll put it here of course. The idea behind this is jquery/ajax is going to get passed an id (eventid) from a dropdown select box inside a form and it's going to run a query to find all the events in my database with that same id. After running the query it's going to list them out and find the highest number of another field (label) and its going to add one to its value and then what I would like for it to is send it back to the form in the jquery/ajax and and place that label inside of the existing input text field in the form. So I'm asking is if my code is presented and is doing what I"m attempting it to do on the php side and if so if I can take out the print tag of the input text field. $eventid = $_GET['eventid']; $result = mysqli_query("SELECT * FROM `events` WHERE `event_id` = '$eventid'"); $list = mysqli_num_rows($result); $label = $list + 1; print '<input type="text" name="label" class="text" readonly="readonly" value="' . $list . '" />'; hi there - first attempt at programming in PHP so please bear with my novice attempts (not much programming experience in any languege really) i am building a site which hosts mp3s. I have MySQL database with one of the fields contating descriptive keywords for the tracks (such as slow, chilled, intense etc) i have created an html page with search form that send data to php file which runs a query of the keywords field - and all works fine now i want that in addition to the search form (with an input text box for the search criteria and the submit button) i will have a 'cloud' of key words - clickable links - that will send the data to the PHP query file (clicking on slow will search for tracks described with 'slow' and so on) - any ideas how do send data by clicking on a link? my html form is: Code: [Select] <body bgcolor="white"> <form action="test3.php" target="main" method="post"> <input type="text" name="searchkeywords" size=100><br> <input type="submit" value="Search"> <input name="searchkeywords" value="slow" type="submit"> </form> <DIV align="center" style="left : 240px; position : absolute; top : 100px;"><iframe src ="test3.php" width="1000" height="800" name="main" frameborder="2" scrolling="no"> <p>Your browser does not support iframes.</p> </iframe></DIV> </body> my php query file is: Code: [Select] <?php $con = mysql_connect("localhost","root","password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("test", $con); $result = mysql_query("SELECT * FROM music WHERE keywords LIKE '%$_POST[searchkeywords]%'"); $txt1='<A HREF="/audioplayer-standalone/'; $txt2='" ><IMG src="images/misc_23.png" width="32" height="32" align="left" border="0"></A>'; $txt3='<script language="JavaScript" src="/audioplayer-standalone/audio-player.js"></script><object type="application/x-shockwave-flash" data="/audioplayer-standalone/player.swf" id="audioplayer'; $txt4='" height="24" width="290"><param name="movie" value="/audioplayer-standalone/player.swf"><param name="FlashVars" value="playerID=audioplayer'; $txt5='&soundFile=/audioplayer-standalone/'; $txt6='"><param name="quality" value="high"><param name="menu" value="false"><param name="wmode" value="transparent"> </object>'; echo "<table border='1'> <tr> <th width='220' align='left'>Track Name</th> <th width='220' align='left'>Artist</th> <th width='220' align='left'>Description</th> <th></th> <th></th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['track'] . "</td>"; echo "<td>" . $row['artist'] . "</td>"; echo "<td>" . $row['description'] . "</td>"; echo "<td>" . $txt1.$row['filename'].$txt2 . "</td>"; echo "<td>" . $txt3.$row['id'].$txt4.$row['id'].$txt5.$row['filename'].$txt6 ."</td>"; echo "</tr>"; } echo "</table>"; // some code mysql_close($con); ?> This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=322771.0 How do I send multiple headers with this code? Code: [Select] <?php $postdata = http_build_query( array( 'var1' => 'some content', 'var2' => 'doh' ) ); $opts = array('http' => array( 'method' => 'POST', 'header' => 'Content-type: application/x-www-form-urlencoded', 'content' => $postdata ) ); $context = stream_context_create($opts); $result = file_get_contents('http://example.com/submit.php', false, $context); ?> *this header: Code: [Select] User-Agent: Mozilla/5.0 (iPhone; U; CPU iPhone OS 3_0 like Mac OS X; en-us) AppleWebKit/528.18 (KHTML, like Gecko) Version/4.0 Mobile/7A341 Safari/528.16 Hello, If i am not posting this question at right location then i apologize for it. My problem is: I need to send HTTP get request to the webservice end point which must include the following http headers: GET / HTTP/1.1 Host: apis.live.net Accept: application/json Content-Type: application/json Authorization: WRAP access_token=AuthToken (here AuthToken is the Access Token String) Note: I am using OAuth for getting an access Token but for getting contents I need to send an HTTP get request. Any kind of help will be appreciated. Thanks Hello forum,
So I've been developing an app mostly in PHP, but am rather afraid of JS. Hope to fix that.
I have an AJAX dropdown using JQuery to search locations. It works great. However, I want to make it similar to what is seen on this site:
http://placefinder.com/
As you can see, the dropdown, when clicked populates a box. Then the user submits the form and the data is used in the application.
I have no clue how to make the form populate with data from the DB (I'm using mySQL) when clicked. So far, I've only been able to make it clickable as a URL (not what I want, obvioiusly!)
Is there a way to do this on a really small, simple script for starters? I'm certain their is, but don't even know where to begin.
Any help appreciated
/*I'm trying to use dropzone js plugin for drag/drop single phote but it require me to create another form for file upload, but i want to use single form for both image and name input, i have no idea on how to combine this field in sinle request, the form to submit both image and name look like*/
<form method="POST" enctype="multipart/form-data">
<input type="text" name="name" id="name">
<!--how to replace this field with dropzone but in this form in order to use the same ajax as below-->
<input type="file" name="photo" id="photo">
<button type="submit">send</button>
</form>
//ajax, how to add dropzone data in
$("form").on('submit', function(e) {
$.ajax({
url: 'add.php',
type: 'POST',
data: new FormData(this),
dataType: 'JSON',
contentType: false,
cache: false,
processData:false,
}).done( function (data) {
if(data.success == false) { //for error message response
if(data.errors.name) {
$('#name').append('<span class="text-danger">' + data.errors.name + '</span>');
}
if(data.errors.photo) {
$('#photo').append('<span class="text-danger">' + data.errors.photo + '</span>');
}
}
});
e.preventDefault();
});
OK, I think this should belong on this board. I am trying to write something that does the following (simplified for ease of understanding): User has signed into my website and has a session variable with their user id in it. There is a table in the mySQL databse that contains details about the user. (this part is already working) User wants to pay for a trip, so clicks a customised Paypal button that takes him to the Paypal site. The item has an trip id associated with it. Once the transaction has been completed, I want to feed information back to my website, and insert the 'item id' and 'user id' into a field in a table in the database. Any idea how I might do this!? Whilst I have a basic script that will insert a row into my table, it relies on variables being sent from the page it is on, thus will not work when the user is redirected to Paypal, as I cannot find a way to send variables to paypal about the item, and then have them send it back to me once the transaction is complete. My PHP is self-taught so sorry if this way of doing things seems a bit scrappy. Hi All, I have a table (with an ID of example) that can be seen he http://tinyurl.com/cfla7wp The following script calls jeditable.js and makes the Telephone & Mobile columns editable. Once edited, the table shows the new value plus a * symbol. Code: [Select] <script type="text/javascript"> $(document).ready(function() { /* Init DataTables */ var oTable = $('#example').dataTable(); /* Apply the jEditable handlers to the table */ $('td:eq(4), td:eq(5)', oTable.fnGetNodes()).editable( 'update_table_tenants.php', { "callback": function( sValue, y ) { var aPos = oTable.fnGetPosition( this ); oTable.fnUpdate( sValue, aPos[0], aPos[1] ); }, "submitdata": function ( value, settings ) { return { "row_id": this.parentNode.getAttribute("1"), "column": oTable.fnGetPosition( this )[2] }; }, "height": "14px" } ); } ); </script> What I need to do however is add the value entered into my database, but I do not know how to do this. I believe I need to do something like: UPDATE my_table VALUE whatever column has been edited in the table WHERE tenID is the same as the row selected in the table i.e. if I update the MOBILE NUMBER of the person called BILL GATES it would find his tenID, update numMobile in the table called my_table All help appreciated Dave I'm working on learning AJAX, all is fine with that script as it's from a book. The issue I'm having is the PHP server side code to get a response from the server. I'm still rather new to all of this so try to explain any response given in a noob friendly manner. So I have the following so far: // Connection Values $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); // Get the variable sent by AJAX script if (isset($_GET['username'])) { // Secure the information $passname = mysqli_real_escape_string($dbc, trim($_GET['username'])); // Lookup the username in the database $query = "SELECT username FROM information WHERE username = '$passname'"; $data = mysqli_query($dbc, $query); // Allow some time to get the response sleep(2); // If 1 is returned that name exsists, if 0 then we can move forward if (mysqli_num_rows($data) == 0) { // Send okay back to the ajax script so it knows to stop bothering the user echo 'okay'; } else { // Send denied back to the ajax script so the user knows that name is taken echo 'denied'; } } Hi, For about a month, I have been trying to figure out why my code will not return anything after posting a wwwForm (I have also tried the newer equivalent of this function but I had no luck with that either.) The nameField and passwordField are taken from text boxes within the game and the code used in my login script is copied and pasted from a Register script but I have changed the file location to the login.php file. The register script works fine and I can add new users to my database but the login script only outputs "Form Sent." and not the "present" that should return when the form is returned and it never gets any further than that point meaning that it lets the user through with no consequence if they use an invalid name because the script never returns an answer. What should I do to fix this? Thanks, Unity Code:
using System.Collections;
using UnityEngine;
using UnityEngine.UI;
using UnityEngine.Networking;
public class Login : MonoBehaviour
{
public InputField nameField;
public InputField passwordField;
public Button acceptSubmissionButton;
public void CallLogInCoroutine()
{
StartCoroutine(LogIn());
}
IEnumerator LogIn()
{
WWWForm form = new WWWForm();
form.AddField("username", nameField.text);
form.AddField("password", passwordField.text);
WWW www = new WWW("http://localhost/sqlconnect/login.php", form);
Debug.Log("Form Sent.");
yield return www;
Debug.Log("Present"); if (www.text[0] == '0')
{
Debug.Log("Present2");
DatabaseManager.username = nameField.text;
DatabaseManager.score = int.Parse(www.text.Split('\t')[1]);
Debug.Log("Log In Success.");
}
else
{
Debug.Log("User Login Failed. Error #" + www.text);
}
}
public void Validation()
{
acceptSubmissionButton.interactable = nameField.text.Length >= 7 && passwordField.text.Length >= 8;
}
}
login.php:
<?php
echo "Test String2";
$con = mysqli_connect('localhost', 'root', 'root', 'computer science coursework');
// check for successful connection.
if (mysqli_connect_errno())
{
echo "1: Connection failed"; // Error code #1 - connection failed.
exit();
}
$username = mysqli_escape_string($con, $_POST["username"]);
$usernameClean = filter_var($username, FILTER_SANITIZE_STRING, FILTER_FLAG_STRIP_LOW | FILTER_FLAG_STRIP_HIGH);
$password = $_POST["password"];
if($username != $usernameClean)
{
echo "7: Illegal Username, Potential SQL Injection Query. Access Denied.";
exit();
}
// check for if the name already exists.
$namecheckquery = "SELECT username, salt, hash, score FROM players WHERE username='" . $usernameClean . "';";
$namecheck = mysqli_query($con, $namecheckquery) or die("2: Name check query failed"); // Error code # 2 - name check query failed.
if (mysqli_num_rows($namecheck) != 1)
{
echo "5: No User With Your Log In Details Were Found Or More Than One User With Your Log In Details Were Found"; // Error code #5 - other than 1 user found with login details
exit();
}
// get login info from query
$existinginfo = mysqli_fetch_assoc($namecheck);
$salt = $existinginfo["salt"];
$hash = $existinginfo["hash"];
$loginhash = crypt($password, $salt);
if ($hash != $loginhash)
{
echo "6: Incorrect Password"; // error code #6 - password does not hash to match table
exit;
}
echo "Test String2";
echo"0\t" . $existinginfo["score"];
?>
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>
I am retrieving Google Books info in JSON format and displaying it inside a div. I would like to send the contents of this div (name, title, description) to my database using Ajax.
Currently only the ISBN field sends because I have it declared as a variable. However my question is, how do I send the other fields (name, title, author). How do I declare these also, I'm not sure what format they need to be in etc.
My JS
$(document).ready(function() {
$('#submit').click(function(ev) {
ev.preventDefault();
var isbn = $('#isbn_search').val(); //get isbn direct from input
var url='https://www.googleapis.com/books/v1/volumes?q='+isbn;
$.getJSON(url,function(data){
$.each(data.items, function(entryIndex, entry){
var html = '<div class="results well">';
html += '<h3>' + entry.volumeInfo.title + '</h3>';
html += '<div class="author">' + entry.volumeInfo.authors + '</div>';
html += '<div class="description">' + entry.volumeInfo.description + '</div>';
});
});
});
});
My Ajax;
$.ajax({
type: 'POST',
url: 'addIsbnScript.php',
data: {
'isbn' : isbn,
'title' : title
'subtitle' : subtitle,
'authors' : authors,
'description' : description
},
success: function () {
$.growl({
message: " Record added"
});
}
});
Note, if i manually set the vars like below they all do successfully send to my database, so I know my query is working ok
var title = "some text", var author = "some text", Var description = "some text"Thanks in advance for any help, newbie here (incase it wasn't obvious!). J I am having trouble figuring this out myself, so hopefully someone can help me out. What I am trying to do is write a code to link an image to a logged in users "edit profile" page. Right now, the page address the image needs to be linked to would be: "MyCollegeFleaMarket.com/users/Username" Currently, the code I have in place is this: "users/<?php global $user; if ($user->uid) { print $user->name; } ?>"> but this is creating a link to, "MyCollegeFleaMarket.com/content/users/Username. Thanks for your help! Hi guys, I'm getting a bit tongue-tied trying to implement some next and previous link logic for my pagination (see lines under /* next and prev links */ comment). I've tried several alternatives but I can't seem to simply go forwards by one page through the pagination or back one. If you could give me some pointers I'd be really grateful. Thanks. MK Code: [Select] $eventdate = mktime(0,0,0,date("m"),date("d")+1,date("Y")); // to get total pages / last page $total_results = mysql_num_rows($result); $total_pages = ceil($total_results / $per_page); // check if the 'page' variable is set in the URL (ex: view-paginated.php?page=1) if (isset($_GET['page']) && is_numeric($_GET['page'])) { $show_page = $_GET['page']; // make sure the $show_page value is valid if ($show_page > 0 && $show_page <= $total_pages) { $start = ($show_page -1) * $per_page; $end = $start + $per_page; } else { // error - show first set of results $start = 0; $end = $per_page; } } else { // if page isn't set, show first set of results $start = 0; $end = $per_page; } $mysqldate = date( 'd-m-Y H:i:s', $eventdate ); $eventdate = strtotime( $mysqldate ); // loop through results of database query, displaying them in the table for ($i = $start; $i < $end; $i++) { // make sure that PHP doesn't try to show results that don't exist if ($i == $total_results) { break; } // echo out the contents of each row into a table echo "<br />"; /*echo mysql_result($result, $i, 'eventid');*/ echo mysql_result($result, $i, 'eventvenue'); /*echo " "; echo mysql_result($result, $i, 'eventdate');*/ echo ", "; echo mysql_result($result, $i, 'eventdate'); //echo date("Y/m/d"); echo "<br />"; } echo "</div><p class='pagination'>"; echo "<a href='events3.php?page=$i=1'>« first</a> "; for ($i = 1; $i <= $total_pages; $i++) { echo "<a href='events3.php?page=$i'>$i</a> "; } echo "<a href='events3.php?page=$total_pages'>last »</a></p>"; /* next and prev links */ $k = $page-1; if ($i > 1) { echo "<a href='{$_SERVER['PHP_SELF']}?page=$k'>Previous</a>"; } if ($i < $total_pages) { echo "NEXT"; } |
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