I am using the well known ajaxForm Jquery Form Plugin (http://jquery.malsup.com/form/). I 'll present to you my code:
HTML code:
<script type="text/javascript">
$(document).ready(function() {
$('#users_form1').ajaxForm({
dataType: 'json',
success: processJson
});
});
function processJson(data) {
$("#first").val(data[1].elem1);
$("#second").val(data[1].elem2);
}
</script>
PHP code:
...
$result=$db->query($query);
if ($result->num_rows>=1)
{
$counter=0;
while ($row = $result->fetch_assoc()) {
$counter++;
$data1=$row["req_created"];
$data2=$row["subject"];
$temp[$counter] = array(
'elem1' => $data1,
'elem2' => $data2,
);
}
echo json_encode($temp);
}
As you may see from the above code, $temp is passed to var data inside function processJson. I'd like to know if array $temp is accessible outside processJson? For example, I want to choose $temp[3]["elem2"] upon a button click, however is it possible to get this data without searching again the database? If yes, how?
Thank you very much
You can have the data in variable, this will be like temporary storage.
<script type="text/javascript">
$(document).ready(function() {
$('#users_form1').ajaxForm({
dataType: 'json',
success: processJson
});
});
var tem_data;
function processJson(data) {
$("#first").val(data[1].elem1);
$("#second").val(data[1].elem2);
tem_data = data;
}
// Use tem_data anywhere;
</script>
But only last requested data will be the tem_data.
If you want all data then do it in array with array push method
Related
I want to access a javascript variable inside a php query
function myFunction2(){
Total=parseInt(point2)
<?php
$con->query("UPDATE eventlist SET P2 = $this.Total WHERE Eid=$PID");
?>
}
I want the query to set p2=value of total
I understand that php is a serverside script and I cant do this like this. What is an alternative to this.?
EDIT
ok i got this on the JS side
function myFunction2(){
var Total = parseInt(point1)+parseInt(point2);
$.ajax({ url: 'ajax.php',
data: {'total' : Total},
type: 'post',
dataType:'json',
success: function(output) {
alert(output);
},
error: function(request, status, error){
alert("Error");
}
and if i put
echo $_POST['total']
in ajax.php i get an alert with the value passed. So i think the value is being passed properly.
But what i need to do is a Mysql Query.
$con->query("UPDATE eventlist SET P2 = $_POST['total']; WHERE Eid=1");
Something like this. How do i do this
Try send javascript value to another php page which contain your query
function myFunction () {
var param = "value";
$.post('query.php', { postvalue: param}, function(data) {
//do what you want with returned data
//postvalue should be the name of post parameter in your query page
})
}
Change your PHP in this way:
$total = $_POST['total'];
$con->query("UPDATE eventlist SET P2 = $total WHERE Eid=1");
I have a button Next that changes the page -
<a class="disabled" href="step">
<button class="btn btn-primary launch-btn disabled next">NEXT</button>
</a>
Also, when this button is clicked, I have an Ajax function that sends data to a controller function -
<script type="text/javascript">
$(function(){
$('.next').click(function() {
var a = $('#box').data('val');
$.ajax({
type: "POST",
url: "<?php echo base_url(); ?>study/CreateData",
data: a,
success: function(data)
{
console.log(data);
},
error: function()
{
console.log("fail");
}
});
});
});
</script>
And I'm using this to receive the value from Ajax -
public function CreateData() {
$data = $this->input->post('data');
return $data;
}
When the Next button hits its controller where it changes to a new page, I'm trying to retrieve the value posted by Ajax to the CreateData() function -
public function step()
{
$data = $this->CreateData();
if(!empty($data)){
print_r($data); exit;
}
else {
echo "blank"; exit;
}
$this->load->view('step2');
}
However, this keeps echoing blank. This obviously means that the $data being returned is empty but I'm not sure why. The success function in my Ajax script is logging data, so it's posting the value to CreateData(). What am I doing wrong here?
Your javascript should be something like this
<script type="text/javascript">
$(function(){
$('.next').click(function() {
var a = $('#box').data('val');
$.ajax({
type: "POST",
url: "<?php echo base_url(); ?>study/CreateData",
data: {my_data: a},
success: function(data)
{
console.log(data);
},
error: function()
{
console.log("fail");
}
});
});
});
</script>
Notice the data property. I have changed it to an object containing the variable a.
Now you should be able to retrieve this in your php function using
public function CreateData() {
$data = $this->input->post('my_data');
return $data;
}
Adr's answer solved part of the problem. In order to actually store and access the data, storing it in the session instead of a variable did the trick -
public function CreateData() {
$data = $this->input->post();
$this->session->unset_userdata('createData');
$this->session->set_userdata('createData', $data);
}
This prevented the $data variable from being overwritten when called the second time.
I'm trying post data to PHP file but i can't receive any data from PHP file. Let me add codes.
This is my jQuery function:
$(document).ready(function () {
$(function () {
$('a[class="some-class"]').click(function(){
var somedata = $(this).attr("id");
$.ajax({
url: "foo.php",
type: "POST",
data: "id=" + somedata,
success: function(){
$("#someid").html();
},
error:function(){
alert("AJAX request was a failure");
}
});
});
});
});
This is my PHP file:
<?php
$data = $_POST['id'];
$con = mysqli_connect('localhost','root','','somedatabase');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"database");
$sql="SELECT * FROM sometable WHERE id = '".$data."'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result)) {
echo $row['info'];
}
mysqli_close($con);
?>
This what i have in HTML file:
<p id="someid"></p>
Data1
Data2
Note: This website is horizontal scrolling and shouldn't be refreshed. When i'm clicking links (like Data1) it's going to another page without getting data from PHP file
You have a few problems:
You are not using the data as mentioned in the other answers:success: function(data){
$("#someid").html(data);
},
You are not cancelling the default click action so your link will be followed:$('a[class="some-class"]').click(function(e){
e.preventDefault();
...;
As the id's are integers, you can use data: "id=" + somedata, although sending an object is safer in case somedata contains characters that need to be escaped:data: {"id": somedata},;
You have an sql injection problem. You should cast the variable to an integer or use a prepared statement:$data = (int) $_POST['id'];;
As also mentioned in another answer, you have two $(document).ready() functions, one wrapping the other. You only need one.
success: function(){
$("#someid").html();
},
should be:
success: function(data){
$("#someid").html(data);
},
You should add parameter in success
success: function(data){ //Added data parameter
console.log(data);
$("#someid").html(data);
},
The data get the values what you echo in PHP end.
This:
success: function(data){
$("#someid").html(data);
},
and you have two document ready, so get rid of:
$(document).ready(function () { ...
});
data: "id=" + somedata,
Change it to:
data: { id : somedata }
I am trying to get the following code to work:
//In Javascript
function updateContentEditable(){
var span = $(this);
var data = new Object();
data.pid = '1';
data.content = 'this is a test';
data.action = 'update_content'; //This should run update_content php function
$.post(ajaxPath, data, onContentSaved); //ajaxPath returns: /wp-admin/admin-ajax.php
}
//In PHP
function update_content(){
echo "<script>alert (\"php was reached\")</script>";
}
NOTE:
//onContentSaved is this:
function onContentSaved(data){
console.log(data);
}
My problem is the the php function is not being run.
What I'm I doing wrong?
Assuming you are using jquery, try this:
function SendRequestCallBack(webMethod, parameters, callBack) {
$.ajax({
type: "POST",
url: webMethod,
data: parameters,
contentType: "application/json; charset=utf-8",
success: function(results) {
$(".ajaxImage").hide();
eval(callBack(results.d));
}
});
}
function formatData(obj){
alert(obj);
}
$(document).ready(function()
{
SendRequestCallBack("http://url-to-php","{'any':'parameters'}",formatData);
});
Then all the PHP needs to do is echo valid JSON. Or if you like you can change the contentType: parameter to text/xml or text/plain etc depending on what you want it to return. Keep in mind that your PHP needs to output the matching content type as well. for example
<?php
header("Content-type: text/json");
echo {"d":[{"firstname":"John","lastname":"Doe"},{"firstname":"Jane","lastname":"Doe"}]}
?>
If you want to pass data in to the PHP function, set it in the parameters, and make sure that your PHP will accommodate it using the $_POST array.
I am exploring the ajax synchronous json import into my javascript code.
The JSON source link I want to use is
http://www.nusantech.com/hendak/default.php?m=galaksi&galaksi=1&viewID=1&t=json
But to keep server loads down, a week ago or so I created a static page showing the same data at
http://www.nusantech.com/hendak/noobjson.php
My javascript import is as below:
<head>
<title>Nusantech</title>
<script src="\OpenLayers213\OpenLayers.js"></script>
<script type="text/javascript" src="http://code.jquery.com/jquery.min.js"></script>
<script type="text/javascript">
var jsonData = {};
$.ajax({
url: "http://hendak.seribudaya.com/noobjson.php",
async: false,
dataType: 'json',
success: function(data) {
jsonData = data;
}
});
alert("Galaksi value retrieved from JSON (expected: 1) : "+jsonData.galaksi);
</script>
<script type="text/javascript">
function kemasMaklumat(id,content) {
var container = document.getElementById(id);
container.innerHTML = content;
}
</script>
</head>
From there I retrieve the values I want on jsonData, eg, (x,y) coordinates as
(jsonData.planets[7].coordinates[0].x,jsonData.planets[7].coordinates[0].y)
It works fine with the noobjson.php link, but when I point it back to default.php, nothing appears. The page took a while to load which make it seem like its loading the json values, but the alert("Galaksi value retrieved") returns undefined.
I copy & pasted the output from the default.php page on a JSON verifier on the web and it showed OK. I don't know why the static link works but the $_GET based link doesn't.
Can someone suggest me what is happening?
EDIT
I have tried:
<script type="text/javascript">
var jsonData = {};
$.ajax({
// url: "http://hendak.seribudaya.com/noobjson.php",
url: "http://hendak.seribudaya.com/default.php?"+encodeURIComponent("galaksi=1&viewID=1&m=galaksi&t=json"),
// url: "http://hendak.seribudaya.com/default.php?galaksi=1&viewID=1&m=galaksi&t=json",
async: false,
dataType: 'json',
type: 'GET',
contentType: "application/json",
success: function(data) {
jsonData = JSON.parse(JSON.stringify(eval("("+data+")")));
alert("Success");
},
error: function(data) {
alert("Failed to download info." + data);
}
});
</SCRIPT>
enter code here
I always get the Failed to download info unless I use the noobjson URL.
It is as if that URL with the GET doesn't exist.
You have to encode the URL component before sending the request. Try:
$.ajax({
url: "http://www.nusantech.com/hendak/default.php?" + encodeURIComponent('m=galaksi&galaksi=1&viewID=1&t=json'),
async: false,
dataType: 'json',
success: function(data) {
jsonData = data;
}
});
Reference: encodeURIComponent()
I have solved it.
In the default.php, what I have done was:
if ($_GET["t"]=="json") {
$viewID=$_GET["viewID"];
$galaksi=$_GET["galaksi"];
$con=mysqli_connect($server, $user, $password, $database);
$sql="SELECT Hari FROM berita WHERE Galaksi=".$galaksi;
$hari=1;
$result = mysqli_query($con,$sql); while(($row = mysqli_fetch_array($result)) ){$hari=$row['Hari']; }
$lb="";
if ($_GET["t"]!="json") { echo "<PRE>\n"; $lb="\n"; }
echo "{\"galaksi\": ".$galaksi.",";
echo $lb."\"hari\": ".$hari.",";
echo $lb."\"planets\": [";
//etc
//etc
}
So I replaced all the individual echoes with $JSONstr like below.
if ($_GET["t"]=="json") {
$viewID=$_GET["viewID"];
$galaksi=$_GET["galaksi"];
$con=mysqli_connect($server, $user, $password, $database);
$sql="SELECT Hari FROM berita WHERE Galaksi=".$galaksi;
$hari=1;
$result = mysqli_query($con,$sql); while(($row = mysqli_fetch_array($result)) ){$hari=$row['Hari']; }
$lb="";
$JSONstr="";
// if ($_GET["t"]!="json") { $JSONstr="<PRE>\n"; $lb="\n"; }
$JSONstr=$JSONstr."{\"galaksi\": ".$galaksi.",";
$JSONstr=$JSONstr.$lb."\"hari\": ".$hari.",";
$JSONstr=$JSONstr.$lb."\"planets\": [";
//etc
//etc
//and at the end:
echo $JSONstr;
}
Then I added the echo $JSONstr; at the end. Originally I did that so that I can do :
echo json_encode($JSONstr);
but this creates {\"Galaksi\" : 1} at the JSON output instead of the intended { "Galaksi": 1 }
So I removed the json_encode and just output the string.
Also I had to remove the
if ($_GET["t"]!="json"){ $JSONstr="<PRE>\n"; $lb="\n"; }
I also used a different JSON tester this time.
Originally I used http://www.freeformatter.com/json-validator.html which says JSON Valid for my initial JSON output. Then I used this one, which said that my JSON output url was invalid, although if I copy+paste the output string it returned valid. http://jsonformatter.curiousconcept.com/
So after making those changes and removing the "<PRE>", the curiousconcept validator gave me a valid status.
Then I used this in the javascript, and I am now able to retrieve expected values.
Thank you all, hope this helps someone else too.