I have the following table in a MYSQL database:
Messages
MessageId (PK) int(10) - auto_inc
Message varchar(100)
I have the following PHP that echos a particular message:
<?php
//..connect to database
$query = "SELECT Message FROM Messages WHERE MessageId = '1'";
$result = mysql_query($query);
$num = mysql_num_rows( $result );
if ($num == 1){
$row = mysql_fetch_assoc($result);
echo json_encode($row);
}else{
echo('Invalid');
}
?>
Can anyone advise me on how best to integrate jQuery in order to allow the document browser window to write the response...
If you use jQuery, you can easily use jQuery.getJSON()[DOCS] as follows:
$.getJSON('mypage.php', function(data) {
//Do somewith with JSON data
});
For normal Javascript, use
var xmlhttp;
if(window.XMLHttpRequest)
xmlhttp = new XMLHttpRequest();
else//IE5, IE6
xmlhttp = ActiveXObject("Microsoft.XMLHTTP");
xmlhttp.onreadystatechange = function()
{
if(xmlhttp.readyState == 4 && xmlhttp.Status == 200)
{
var response = JSON.parse(xmlhttp.responseText);
//Do something with JSON Data
}
};
Related
I have a form where the user has to enter their reservation id and last name. If these two values match in the database then I need to return the corresponding values from the database.
I have two files, one that is html where I use ajax and one php file. When clicking on the button, nothing is being returned, I am not seeing any specific errors and I am sure that the value I put in are correct.
<script>
var ajax = getHTTPObject();
function getHTTPObject()
{
var xmlhttp;
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
} else if (window.ActiveXObject) {
// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
} else {
//alert("Your browser does not support XMLHTTP!");
}
return xmlhttp;
}
function updateCityState()
{
if (ajax)
{
var reservation_id = document.getElementById("reservation_id").value;
var guest_last_name = document.getElementById("guest_last_name").value;
if(reservation_id)
{
var param = "?reservation_id=" + reservation_id + "&guest_last_name=" + guest_last_name;
var url = "test04.php";
ajax.open("GET", url + param, true);
ajax.onreadystatechange = handleAjax;
ajax.send(null);
}
}
}
function handleAjax()
{
if (ajax.readyState == 4)
{
var guest_full_name = document.getElementById('guest_full_name');
var unit_number = document.getElementById('unit_number');
var floor = document.getElementById('floor');
var key_sa = document.getElementById('key_sa');
if(!!ajax.responseText) {
var result = JSON.parse(ajax.responseText);
if(!!result){
guest_full_name.innerHTML = (!!result.guest_full_name) ? result.guest_full_name : '';
unit_number.innerHTML = (!!result.unit_number) ? result.unit_number : '';
floor.innerHTML = (!!result.floor) ? result.floor : '';
key_sa.innerHTML = (!!result.key_sa) ? result.key_sa : '';
}
}
}
}
</script>
<p id='employee_name'></p>
<p id='employee_age'></p>
<p id='safe_code'></p>
My test04.php
<?php
$conn = mysqli_connect("","","","");
$reservation_id = mysqli_real_escape_string($conn, $_GET['reservation_id']);
$guest_last_name = mysqli_real_escape_string($conn, $_GET['guest_last_name']);
$query = "SELECT reservation_id, guest_full_name, guest_last_name unit_number, floor, key_sa FROM reservations2 INNER JOIN guest ON (reservations2.reservation_id=guest.reservation_idg) INNER JOIN unit USING (unit_id) where reservation_id ='".$reservation_id."'AND guest_last_name ='".$guest_last_name."";
$result = mysqli_query($conn, $query) or die(mysql_error());
$response = array();
if(mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$response['guest_full_name'] = ($row['guest_full_name'] != '') ? $row['guest_full_name'] : '';
$response['unit_number'] = ($row['unit_number'] != '') ? $row['unit_number'] : '';
$response['floor'] = ($row['floor'] != '') ? $row['floor'] : '';
$response['key_sa'] = ($row['key_sa'] != '') ? $row['key_sa'] : '';
}
}
echo json_encode($response, true);
?>
I am not seeing any specific errors
Where are you looking?
Did you check the raw response from the PHP script or just look at what was rendered in your browser?
Did you verify that error logging is working and did you check your logs?
The logic of your PHP is unclear - your JSON data and the PHP array can't handle multiple records yet you process multiple records. It would be nice to implement REST properly. This should also apply authentication and use CSRF for security - but I'll assume you left those out for illustrative purposes.
Your code is not written to handle failures or missing data. Consider (noting all the differences with what you posted):
<?php
$conn = mysqli_connect("","","","");
$response = array();
$reservation_id = mysqli_real_escape_string($conn, $_GET['reservation_id']);
$guest_last_name = mysqli_real_escape_string($conn, $_GET['guest_last_name']);
$query = "SELECT reservation_id, guest_full_name
, guest_last_name unit_number, floor, key_sa
FROM reservations2
INNER JOIN guest
ON (reservations2.reservation_id=guest.reservation_idg)
INNER JOIN unit USING (unit_id)
WHERE reservation_id ='".$reservation_id."'
AND guest_last_name ='".$guest_last_name."";
$result = mysqli_query($conn, $query);
if (!$result) {
$response['status']=503
$response['msg']="Error";
trigger_error(mysql_error());
finish($response);
exit;
}
$response['status']=200;
$response['msg']='OK';
$response['guest_full_name'] = htmlentities($_GET['guest_last_name']);
$response['reservations']=array();
while($row = mysqli_fetch_assoc($result)) {
$response['reservations'][]=array(
'unit_number'=>$row['unit_number'],
'floor'=>$row['floor'],
'key_sa'=>$row['floor_sa']);
}
}
finish($response);
exit;
function finish($response)
{
header("HTTP/1.1 $response[status] $response[msg]");
header("Content-type: application/json");
echo json_encode($response, true);
}
I have a database called opera_house with a table called room that has a field room_name and a field capacity. I want to show the room_name 's that have a capacity larger than the one entered by the user.
The Available Room text disappears, but my code only shows the MySQL query if I echo it, but I'm not sure if it is reaching to search the database.
This is my script code:
<script src="https://code.jquery.com/jquery-3.2.1.min.js"></script>
<script type="text/javascript">
function showRoom(str) {
if (str === "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (this.readyState === 4 && this.status === 200) {
document.getElementById("txtHint").innerHTML = this.responseText;
}
};
xmlhttp.open("GET","ajax_events.php?q="+str,true);
xmlhttp.send();
}
}
This is my html:
<body>
<form>
<input type="text" name="room" onkeyup="showRoom(this.value)">
</form>
<br>
<div id="txtHint"><b>Available Room...</b></div>
</body>
This is my php:
<?php
include('dbconnect.php');
$q = intval($_GET['q']);
mysqli_select_db($connection,"opera_house");
$sql="SELECT room_name FROM room WHERE capacity >= '".$q."'";
echo $sql;
$result = mysqli_query($connection,$sql);
while($row = mysqli_fetch_array($result)) {
echo "<td>" . $row['room_name'] . "</td>";
}
?>
My php file is called ajax_events.php
And my dbconnect.php is one that I constantly use to connect to this database.
Would really appreciate some help!!
I propose an answer using jquery. You've embedded it in your question but you're not using it ...
Explanations : You call the following url ajax_events.php with the parameter "q" only if str is defined, otherwise it fills the selector txtHint with nothing.
AJAX
if (str != "") {
$.ajax({
type: 'GET',
url: 'ajax_events.php',
dataType: 'JSON',
data : {
q: str
}
}).done(function (data) {
$('#txtHint').text = data;
}).fail(function() {
alert('Fatal error');
})
} else {
$('#txtHint').text = '';
}
With this configuration, it is important to return result with echo json_encode in your server side code.
PHP
<?php
include('dbconnect.php');
$q = intval($_GET['q']);
mysqli_select_db($connection, "opera_house");
$sql = 'SELECT room_name FROM room WHERE capacity >= '.$q; // Some corrections
$result = mysqli_query($connection, $sql);
$return = '';
while($row = mysqli_fetch_array($result)) {
$return .= '<td>' . $row["room_name"] . '</td>';
}
echo json_encode($return); // Return Json to ajax
?>
While thinking of JS it's fine. I think the problems are in the php code. Try this one.
<?php
include('dbconnect.php');
$q = intval($_GET['q']);
mysqli_select_db($connection,"opera_house");
$sql="SELECT room_name FROM room WHERE capacity >= " . $q;
$result = mysqli_query($connection,$sql);
if (!$result) {
echo mysqli_error();
exit();
} // this is to do debugging. remove when you get it fixed
$ret = ""; //variable to hold return string
while($row = mysqli_fetch_array($result)) {
$ret .= "<td>" . $row['room_name'] . "</td>";
}
echo $ret;
I created a text filed for customer name. Already registered customer will show while clicking that text field. Only that registered customer can entry the data. I used ajax for prevent the new user addition.
this is the ajax:
function validateForm()
{
var customerName = document.getElementById('customerName').value;
var customerId = document.getElementById('customerId').value;
var getResponse=0;
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
// document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
getResponse= xmlhttp.responseText;alert(getResponse);
if(getResponse=="0")
{
return false;
}
}
};
xmlhttp.open("GET", "customer_check_ajax.php?name="+customerName+"&id="+customerId, true);
xmlhttp.send();
}
and here is the ajax page customer_check_ajax.php :
<?php
require("../../config/config.inc.php");
require("../../config/Database.class.php");
require("../../config/Application.class.php");
if($_SESSION['travelType']=='Admin')
{
$check = 1;
}
else
{
$check = '';
$logId = $_SESSION['travelId'];
$proId = $_SESSION['proId'];
$check = "proId='$proId'";
}
$cusName = $_REQUEST['name'];
$cusId = $_REQUEST['id'];
$connection = mysqli_connect(DB_SERVER, DB_USER, DB_PASS, DB_DATABASE);
$qry="select * FROM ".TABLE_ACCOUNTS." WHERE ID='$cusId' and accountName='$cusName' and proId='$proId'";
$res = mysqli_query($connection, $qry);//echo $qry;
$num = mysqli_num_rows($res);
if($num>0)
echo '1';
else
echo '0';
?>
here i would get the responses as 0 for new user and 1 for already registered user. But i want to stop saving data when new user addition. Please help me.
You are making Asynchronous request and the answer of the ajax will come after the function is over ..
To do what you want you must do synchronous ajax request
xmlhttp.open("GET", "customer_check_ajax.php?name="+customerName+"&id="+customerId, false); // `false` makes the request synchronous
I am new to AJAX and am trying to insert form data into a database using ajax and php. It is simply not working, I had errors to start off but now there are no more errors and the database doesn't update and there is no response text. Please Help!
AJAX
function submit(){
var vid = <?php echo $user['vid'] ?>;
var type = document.getElementById("type").value;
var rules = document.getElementById("rules").value;
var comments = document.getElementById("comments").value;
var xhttp;
if (window.XMLHttpRequest) {
xhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xhttp.onreadystatechange = function() {
if (xhttp.readyState == 4 && xhttp.status == 200) {
document.getElementById("page").innerHTML = xhttp.responseText;
}
}
xhttp.open("POST", "/content/training/request/submit.php", false);
xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xhttp.send("vid="+vid+"&type="+type+"&rules="+rules+"&comments="+comments);
}
</script>
PHP FILE AJAX IS CALLING:
<?php
require("http://".$_SERVER["HTTP_HOST"]."/config/db.php");
$stmt = $db->prepare("INSERT INTO trainingRequests (vid, type, rules, comments, timeSubmitted) VALUES (:vid,:type,:rules,:comments,:timeSubmitted)");
$stmt->bindParam(':vid', $vid);
$stmt->bindParam(':type', $type);
$stmt->bindParam(':rules', $rules);
$stmt->bindParam(':comments', $comments);
$stmt->bindParam(':timeSubmitted', $time);
$vid = $_POST["vid"];
$type = $_POST["type"];
$rules = $_POST["rules"];
$comments = $_POST["comments"];
$time = time();
$stmt->execute();
echo "Submitted!";
Don't use an HTTP URL in require. When you access the PHP script through the webserver, it runs the script (in another server instance, so it doesn't affect the current script), it doesn't return the script contents. Change it to:
require("../../config/db.php");
I have 2 pages
The first page for calling a second page to get all records from database, like this:
<script language="JavaScript">
var HttPRequest = false;
function doCallAjax(Mode,users_ID) {
HttPRequest = false;
if (window.XMLHttpRequest) { // Mozilla, Safari,...
HttPRequest = new XMLHttpRequest();
if (HttPRequest.overrideMimeType) {
HttPRequest.overrideMimeType('text/html');
}
} else if (window.ActiveXObject) { // IE
try {
HttPRequest = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try {
HttPRequest = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e) {}
}
}
if (!HttPRequest) {
alert('Cannot create XMLHTTP instance');
return false;
}
var url = 'AjaxDeleteRecord.php';
var pmeters = "tMode=" + Mode +
"&tusers_ID=" + users_ID;
HttPRequest.open('POST',url,true);
HttPRequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
HttPRequest.setRequestHeader("Content-length", pmeters.length);
HttPRequest.setRequestHeader("Connection", "close");
HttPRequest.send(pmeters);
HttPRequest.onreadystatechange = function()
{
if(HttPRequest.readyState == 3) // Loading Request
{
document.getElementById("mySpan").innerHTML = "Now is Loading...";
}
if(HttPRequest.readyState == 4) // Return Request
{
document.getElementById("mySpan").innerHTML = HttPRequest.responseText;
}
}
}
</script>
<body Onload="JavaScript:doCallAjax('LIST','');">
<form name="frmMain">
<div style="margin-right:10px">
<span id="mySpan"></span>
</div>
and the second page is php that fetches data:
$strMode = $_POST["tMode"];
$users_ID = $_POST["tusers_ID"];
if($strMode == "DELETE")
{
//$strSQL = "DELETE FROM users , stores WHERE users.users_StoreId=stores.stores_ID AND users.users_ID = '".$users_ID."'";
$strSQL = "update users set users_delete='1' where users.users_ID = '".$users_ID."'";
$objQuery = mysql_query($strSQL) or die (mysql_error());
}
$strSQL = "SELECT * FROM users , stores WHERE users.users_StoreId=stores.stores_ID and users.users_delete='0' ORDER BY stores.stores_Name , users.users_Type ASC ";
$objQuery = mysql_query($strSQL) or die ("Error Query [".$strSQL."]");
The problem that I cannot make paging work for this.
Can any one help me?
Use Google's Visualization API's
table chart
Visualization: Table.
It will automatically enable paging as per your count like 10 or 20 and sorting your content ascending or descending on the basis of column.