Im using the jquery .load function to query a php file that will output some data. Now sometimes the script will return nothing. In this case, can I have the load function not put any data into my specified div? (right now it clears out the div and just puts a blank white area.
Thanks!
try using $.get;
$.get('<url>',{param1:true},function(result){
if(result) {
$('selector').html(result);
}
else {
//code to handle if no results
}
});
Use $.get
http://api.jquery.com/jQuery.get/
in addition to #jerjer's post, you can also use this:
var paramData= 'param=' + param1 + '&user=<?echo $user;?>';
$.ajax({
type: "GET",
data:paramData,
url: "myUrl.php",
dataType: "json", // this line is optional
success: function(result) {
// do you code here
alert(result); // this can be an any value returned from myUrl.php
},
statusCode: {
404: function() {
alert('page not found');
}
}
});
Related
How can I load data from my PHP response via ajax into a panel?
My PHP outputs correctly and I can see a table in the response, but I can;t get it to build the data on my webpage.
Here is my jquery/ajax so far. It passed the value to PHP correctly and PHP builds the table via its echo, but what am I missing for AJAX to display the table?
PHP:
<?php
foreach ($lines as $value) {
echo "<input name='data[]' value='$value'><br/>";
}
?>
JQUERY:
$(function () {
$('#rotator').change(function (e) {
var rotator = $("#rotator").val();
$.ajax({
type: "POST",
url: "tmp/JFM/National/national.php",
data: {
rotator: rotator
},
success: function (result) {
$('#panel').load(result);
}
})
return false;
});
});
The answer to this was two fold.
I was attempting to append to my main div, which apparently can't happen. I created a new empty div and was able to load the results there.
Beyond that, the comments to change .load(results) to .html(results) were needed.
The correct jquery code is below.
$(function () {
$('#rotator').change(function (e) {
var rotator = $("#rotator").val();
$.ajax({
type: "POST",
url: "tmp/JFM/National/national.php",
data: {
rotator: rotator
},
success: function (result) {
console.log(result);
$('#test').html(result);
}
})
return false;
});
});
move your function from:
$.ajax({...,success: function(){...}});
to
$.ajax({..}).done(function(){...});
if it doesn't work, try to add async:false into the ajax object...
$.ajax({...,async:false}).done(function(){...});
Hope it helps... =}
Sorry for my terms incase they are incorrect, but I have a php function that I am going to interop with ajax so I can use a php function to get a value for a jquery variable.
So now I am just at the point where I have received the ajax callback and would like to set the jquery variable to the response value I got back. Here is my code as an example:
$(document).ready(function() {
$('body').videoBG({
position:"fixed",
zIndex:0,
mp4:'', //This is where I want to use the ajax response value
opacity:1,
});
})
jQuery.ajax({
type : "POST",
url : "index.php",
data : {
request : "getvideo_Action"
},
success : function(response) {
alert(response);
//Do my response stuff
}
});
So basically what I want to do is set the 'Mp4' var(or is it property?) with the value that I get from the response. Can anyone help me out with this? Thanks.
You can put the entire thing inside the success function, like this:
jQuery.ajax({
type: "POST",
url: "index.php",
data: {
request: "getvideo_Action"
},
success: function (response) {
$('body').videoBG({
position: "fixed",
zIndex: 0,
mp4: response,
opacity: 1
});
}
});
I have checked around, but can't seem to figure out how this is done.
I would like to send form data to PHP to have it processed and inserted into a database (this is working).
Then I would like to send a variable ($selected_moid) back from PHP to a JavaScript function (the same one if possible) so that it can be used again.
function submit_data() {
"use strict";
$.post('insert.php', $('#formName').formSerialize());
$.get('add_host.cgi?moid='.$selected_moid.');
}
Here is my latest attempt, but still getting errors:
PHP:
$get_moid = "
SELECT ID FROM nagios.view_all_monitored_objects
WHERE CoID='$company'
AND MoTypeID='$type'
AND MoName='$name'
AND DNS='$name.$selected_shortname.mon'
AND IP='$ip'
";
while($MonitoredObjectID = mysql_fetch_row($get_moid)){
//Sets MonitoredObjectID for added/edited device.
$Response = $MonitoredObjectID;
if ($logon_choice = '1') {
$Response = $Response'&'$logon_id;
$Response = $Response'&'$logon_pwd;
}
}
echo json_encode($response);
JS:
function submit_data(action, formName) {
"use strict";
$.ajax({
cache: false,
type: 'POST',
url: 'library/plugins/' + action + '.php',
data: $('#' + formName).serialize(),
success: function (response) {
// PROCESS DATA HERE
var resp = $.parseJSON(response);
$.get('/nagios/cgi-bin/add_host.cgi', {moid: resp });
alert('success!');
},
error: function (response) {
//PROCESS HERE FOR FAILURE
alert('failure 'response);
}
});
}
I am going out on a limb on this since your question is not 100% clear. First of all, Javascript AJAX calls are asynchronous, meaning both the $.get and $.post will be call almost simultaneously.
If you are trying to get the response from one and using it in a second call, then you need to nest them in the success function. Since you are using jQuery, take a look at their API to see the arguments your AJAX call can handle (http://api.jquery.com/jQuery.post/)
$.post('insert.php', $('#formName').formSerialize(),function(data){
$.get('add_host.cgi?moid='+data);
});
In your PHP script, after you have updated the database and everything, just echo the data want. Javascript will take the text and put it in the data variable in the success function.
You need to use a callback function to get the returned value.
function submit_data(action, formName) {
"use strict";
$.post('insert.php', $('#' + formName).formSerialize(), function (selected_moid) {
$.get('add_host.cgi', {moid: selected_moid });
});
}
$("ID OF THE SUBMIT BUTTON").click(function() {
$.ajax({
cache: false,
type: 'POST',
url: 'FILE IN HERE FOR PROCESSING',
data: $("ID HERE OF THE FORM").serialize(),
success: function(data) {
// PROCESS DATA HERE
},
error: function(data) {
//PROCESS HERE FOR FAILURE
}
});
return false; //This stops the Button from Actually Preforming
});
Now for the Php
<?php
start_session(); <-- This will make it share the same Session Princables
//error check and soforth use $_POST[] to get everything
$Response = array('success'=>true, 'VAR'=>'DATA'); <--- Success
$Response = array('success'=>false, 'VAR'=>'DATA'); <--- fails
echo json_encode($Response);
?>
I forgot to Mention, this is using JavaScript/jQuery, and ajax to do this.
Example of this as a Function
Var Form_Data = THIS IS THE DATA OF THE FORM;
function YOUR FUNCTION HERE(VARS HERE) {
$.ajax({
cache: false,
type: 'POST',
url: 'FILE IN HERE FOR PROCESSING',
data:Form_Data.serialize(),
success: function(data) {
// PROCESS DATA HERE
},
error: function(data) {
//PROCESS HERE FOR FAILURE
}
});
}
Now you could use this as the Button Click which would also function :3
I have a PHP populated table from Mysql and I am using JQuery to listen if a button is clicked and if clicked it will grab notes on the associated name that they clicked. It all works wonderful, there is just one problem. Sometimes when you click it and the dialog(JQuery UI) window opens, there in the text area there is nothing. If you are to click it again it will pop back up. So it seems sometimes, maybe the value is getting thrown out? I am not to sure and could use a hand.
Code:
$(document).ready(function () {
$(".NotesAccessor").click(function () {
notes_name = $(this).parent().parent().find(".user_table");
run();
});
});
function run(){
var url = '/pcg/popups/grabnotes.php';
showUrlInDialog(url);
sendUserfNotes();
}
function showUrlInDialog(url)
{
var tag = $("#dialog-container");
$.ajax({
url: url,
success: function(data) {
tag.html(data).dialog
({
width: '100%',
modal: true
}).dialog('open');
}
});
}
function sendUserfNotes()
{
$.ajax({
type: "POST",
dataType: "json",
url: '/pcg/popups/getNotes.php',
data:
{
'nameNotes': notes_name.text()
},
success: function(response) {
$('#notes_msg').text(response.the_notes)
}
});
}
function getNewnotes(){
new_notes = $('#notes_msg').val();
update(new_notes);
}
// if user updates notes
function update(new_notes)
{
$.ajax({
type: "POST",
//dataType: "json",
url: '/pcg/popups/updateNotes.php',
data:
{
'nameNotes': notes_name.text(),
'newNotes': new_notes
},
success: function(response) {
alert("Notes Updated.");
var i;
$("#dialog-container").effect( 'fade', 500 );
i = setInterval(function(){
$("#dialog-container").dialog( 'close' );
clearInterval(i);
}, 500);
}
});
}
/******is user closes notes ******/
function closeNotes()
{
var i;
$("#dialog-container").effect( 'fade', 500 );
i = setInterval(function(){
$("#dialog-container").dialog( 'close' );
clearInterval(i);
}, 500);
}
Let me know if you need anything else!
UPDATE:
The basic layout is
<div>
<div>
other stuff...
the table
</div>
</div>
Assuming that #notes_msg is located in #dialog-container, you would have to make sure that the actions happen in the correct order.
The best way to do that, is to wait for both ajax calls to finish and continue then. You can do that using the promises / jqXHR objects that the ajax calls return, see this section of the manual.
You code would look something like (you'd have to test it...):
function run(){
var url = '/pcg/popups/grabnotes.php';
var tag = $("#dialog-container");
var promise1 = showUrlInDialog(url);
var promise2 = sendUserfNotes();
$.when(promise1, promise2).done(function(data1, data2) {
// do something with the data returned from both functions:
// check to see what data1 and data2 contain, possibly the content is found
// in data1[2].responseText and data2[2].responseText
// stuff from first ajax call
tag.html(data1).dialog({
width: '100%',
modal: true
}).dialog('open');
// stuff from second ajax call, will not fail because we just added the correct html
$('#notes_msg').text(data2.the_notes)
});
}
The functions you are calling, should just return the result of the ajax call and do not do anything else:
function showUrlInDialog(url)
{
return $.ajax({
url: url
});
}
function sendUserfNotes()
{
return $.ajax({
type: "POST",
dataType: "json",
url: '/pcg/popups/getNotes.php',
data: {
'nameNotes': notes_name.text()
}
});
}
It's hard to tell from this, especially without the mark up, but both showUrlInDialog and sendUserfNotes are asynchronous actions. If showUrlInDialog finished after sendUserfNotes, then showUrlInDialog overwrites the contents of the dialog container with the data returned. This may or may not overwrite what sendUserfNotes put inside #notes_msg - depending on how the markup is laid out. If that is the case, then it would explains why the notes sometimes do not appear, seemingly randomly. It's a race condition.
There are several ways you can chain your ajax calls to keep sendUserOfNotes() from completing before ShowUrlInDialog(). Try using .ajaxComplete()
jQuery.ajaxComplete
Another ajax chaining technique you can use is to put the next call in the return of the first. The following snippet should get you on track:
function ShowUrlInDialog(url){
$.get(url,function(data){
tag.html(data).dialog({width: '100%',modal: true}).dialog('open');
sendUserOfNotes();
});
}
function sendUserOfNotes(){
$.post('/pcg/popups/getNotes.php',{'nameNotes': notes_name.text()},function(response){
$('#notes_msg').text(response.the_notes)
},"json");
}
James has it right. ShowUrlInDialog() sets the dialog's html and sendUserOfNotes() changes an element's content within the dialog. Everytime sendUserOfNotes() comes back first ShowUrlInDialog() wipes out the notes. The promise example by jeroen should work too.
I have an ajax function in jquery calling a php file to perform some operation on my database, but the result may vary. I want to output a different message whether it succeeded or not
i have this :
echo '<button id="remove_dir" onclick="removed('.$dir_id.')">remove directory</button>';
<script type="text/javascript">
function removed(did){
$.ajax({
type: "POST",
url: "rmdir.php",
data: {dir_id: did},
success: function(rmd){
if(rmd==0)
alert("deleted");
else
alert("not empty");
window.location.reload(true);
}
});
}
</script>
and this
<?php
require('bdd_connect.php');
require('functions/file_operation.php');
if(isset($_POST['dir_id'])){
$rmd=remove_dir($_POST['dir_id'],$bdd);
}
?>
my question is, how to return $rmd so in the $.ajax, i can alert the correct message ?
thank you for your answers
PHP
<?php
require('bdd_connect.php');
require('functions/file_operation.php');
if (isset($_POST['dir_id'])){
$rmd=remove_dir($dir_id,$bdd);
echo $rmd;
}
?>
JS
function removed(did){
$.ajax({
type: "POST",
url: "rmdir.php",
data: {dir_id: did}
}).done(function(rmd) {
if (rmd===0) {
alert("deleted");
}else{
alert("not empty");
window.location.reload(true);
}
});
}
i advice to use json or :
if(isset($_POST['dir_id'])){
$rmd=remove_dir($dir_id,$bdd);
echo $rmd;
}
You need your php file to send something back, then you need the ajax call on the original page to behave based on the response.
php:
if(isset($_POST['dir_id'])){
$rmd=remove_dir($dir_id,$bdd);
echo "{'rmd':$rmd}";
}
which will output one of two things: {"rmd": 0} or {"rmd": 1}
We can simulate this return on jsBin
Then use jquery to get the value and do something based on the response in our callback:
$.ajax({
type: "POST",
dataType: 'json',
url: "http://jsbin.com/iwokag/3",
success: function(data){
alert('rmd = ' + data.rmd)
}
});
View the code, then watch it run.
Only I didn't send any data here, my example page always returns the same response.
Just try echoing $rmd in your ajax file, and then watching the console (try console.log(rmd) in your ajax response block)
$.ajax({
type: "POST",
url: "rmdir.php",
data: {dir_id: did},
success: function(rmd){
console.log(rmd);
}
});
You can then act accordingly based on the response
Try echo the $rmd out in the php code, as an return to the ajax.
if(isset($_POST['dir_id'])){
$rmd=remove_dir($dir_id,$bdd);
//if $rmd = 1 alert('directory not empty');
//if $rmd = 0 alert('directory deleted');
echo $rmd;
}
Your "rmd" in success: function(rmd) should receive the callabck.