When a user submits the form on my page I use AJAX to submit the information without refreshing the page. After the user submits the information I want to run a PHP function that I have already written that displays the information. Is this possible or do I need to run another ajax function to update after
$(function () {
$('add').on('submit', function (e) {
$.ajax({
type: 'post',
url: 'submit.php',
data: $('this').serialize(),
success: function () {
alert('form was submitted');
}
});
e.preventDefault();
updateWords(); //PHP FUNCTION
});
});
When a user submits the form on my page I use AJAX to submit the information without refreshing the page. After the user submits the information I want to run a PHP function that I have already written that displays the information. Is this possible or do I need to run another ajax function to update after
$(function () {
$('add').on('submit', function (e) {
$.ajax({
type: 'post',
url: 'submit.php',
data: $('this').serialize(),
success: function () {
alert('form was submitted');
}
});
e.preventDefault();
updateWords(); //PHP FUNCTION
});
});
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asked Dec 9, 2013 at 15:32
user934902user934902
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- Then you should do another ajax call to this php function – krackmoe Commented Dec 9, 2013 at 15:34
- Another AJAX call maybe? Also you can add processing of this function in previous AJAX call. If you are not doing something after first AJAX call response which matters. – sulmanpucit Commented Dec 9, 2013 at 15:34
-
2
Just run the php function from your
submit.php
script and return the data to thesuccess
function. That way you can do everything in one request. Unless you want the two to run simultaneously of course... – jeroen Commented Dec 9, 2013 at 15:42 - my answer is based on the fact that when you update something in the database you want to get a succes response and the new values.my code does both.no need for multiple ajax calls. json_decode is just to get the response in a javascript readable array. – cocco Commented Dec 9, 2013 at 16:22
- @cocco, as I've said several times. Your code doesn't stop multiple ajax calls being needed. – Ryan Commented Dec 9, 2013 at 16:24
5 Answers
Reset to default 3You will need to run another AJAX call on success of the first one.
JavaScript cannot interact with PHP directly and therefore you can't call a PHP function from the success/plete function of the AJAX call.
in response of this.
I have multiple forms on the page and probably 5 different ajax calls in which no more then 2 are called at the same time, if json is better do you have a link to some reading material or additional stack example similar to this so i can teach myself – user934902
first of all
jquery was made for old browsers to support basic functions that ie6 does not support
the use of jquery is good if you want to have full support on almost all browser
but there are also many bad sides:
- it's 81kb code wich is insane (without plugins)
- it's very slow pared to native functions.
- it's used by ppl who don't know how to write simple javascript.
- and much more if we start to talk about the plugins.
now we are in a era where most of the ppl use their mobile devices and modern browsers which support standard javascript 1.7.Android,ios,safari,internet explorer 10,chrome,opera & firefox support javascript 1.7
http://caniuse./
the code below is supported by those browsers.
this is a ajax function written by me it handles post & get
you can read more about that function here https://stackoverflow./a/18309057/2450730
function ajax(a,b,e,d,f,g,c){
c=new XMLHttpRequest;
!f||(c.upload.onprogress=f);
!g||(c.onprogress=g);
c.onload=b;
c.open(e||'get',a);
c.send(d||null)
}
// Params:
// Url,callback,method,formdata or {key:val},uploadFunc,downloadFunc,placeholder
a simple get request would be
ajax('example.php',responseFunction);
and a plex post function would be
ajax('example.php',responseFunction,'post',new FormData(form),uploadFunc,dlFunc);
you need that.
so if you have your form
<form id="myForm">
<input name="name"> Name
<input name="surname"> Surname
<input name="mail"> Email
<input name="file" type="file" multiple> File/Files
</form>
you just have to write a function like that
var form=document.getElementsById('myForm');
form.onsubmit=function(e){
e.preventDefault();
ajax('submit.php',SUCCESS,'post',new FormData(this));
}
and here we e to your question :
create the submit.php file for your needs
<?php
// do whatever you need with the posted info
// copy files to a specific folder
// insert/update/delete the database
// check for errors
// lets say no errors
$err=array();
// load extra info from database to an array called $extrainfo
// load some functions... (you can do what you want here)
// like executing the function you have already written and add that info to
// the $extrainfo.
$extrainfo=array('updated correctly','files copied');
$data=array('post'=>$_POST,'files'=>$_FILES,'info'=>$extrainfo,'errors'=>$err);
echo json_encode($data);
?>
this returns a json encoded array to use later in javascript.
now we need to elaborate this json. in the SUCCESS function
function SUCCESS(){
var data=JSON.parse(this.response);
if(data.errors.length>0){
// you have some errors
}else{
// no errors
// display your response in a proper way.
console.log(data);
}
}
inside this function you just have to display based on the response data.
data
contains everything you need.
here is the whole code.
copy and past into a txt file and save it as submit.php
.
i have tested only in chrome for now.
<?php
if($_POST){
$err=array();
$extrainfo=array('updated correctly','files copied');
$data=array('post'=>$_POST,'files'=>$_FILES,'info'=>$extrainfo,'errors'=>$err);
echo json_encode($data);
}else{
?><!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>upload</title>
<script>
var form,result;
function ajax(a,b,e,d,f,g,c){
c=new XMLHttpRequest;
!f||(c.upload.onprogress=f);
!g||(c.onprogress=g);
c.onload=b;
c.open(e||'get',a);
c.send(d||null)
}
function SUCCESS(){
console.log(JSON.parse(this.response));
var data=JSON.parse(this.response);
if(data.errors.length>0){
result.textContent='you have some errors:'+data.errors[0];
}else{
result.textContent=JSON.stringify(data, null, '\t');
}
}
window.onload=function(){
form=document.getElementById('myForm');
result=document.getElementById('response');
form.onsubmit=function(e){
e.preventDefault();
ajax('submit.php',SUCCESS,'post',new FormData(this));
}
}
</script>
</head>
<body>
<form id="myForm">
<input name="name"> Name
<input name="surname"> Surname
<input name="mail"> Email
<input name="file[]" type="file" multiple> File/Files
<input type="submit" value="send">
</form>
<pre id="response"></pre>
</body>
</html>
<?php
}
?>
If the function is dependant on the AJAX call to submit.php
finishing, then create a new AJAX call. If not, then just append the function to submit.php and call it at the end of the file.
You should use the ajax function that performs the update information to update the page itself at the same time.
Just return usable HTML as the return from the ajax, and use that as the HTML content of the page.
Example: test.php
<script>
function updateName() {
var url = './test.php?inpName=' + $('#inpName').val();
$.get( url, function( data ) {
$( "#frmDivUpdate" ).html( data );
alert( "Call was performed." );
});
}
</script>
<div id="frmDivUpdate">
<form>
Enter your name : <input name="inpName" id="inpName">
<br>
<input type="button" onClick="updateName();" value="Update">
</form>
</div>
<?php
if (isset($_GET))
{
$inpName = $_GET["inpName"];
echo "You updated your val to $inpName";
}
?>
PHP is serverside , javascript is clientside. you can't call php from client if the page is already loaded.
so the easy way is ajax
in the submit.php add:
echo json_encode($_POST);
or the
PHP function that I have already written
... the information what you need.
and in the success function
success: function () {
// i don't use jquery but the response contains the json string
console.log('the words you wanna update:',JSON.parse(response));
// updateWords(JSON.parse(response));
}
EDIT
you also don't need more than one ajax function
you say you already wrote a script to display the next information to the client.
add that script to the submit.php
and the succes function will give you what you need as response.
i added echo json_encode($_POST);
because most of the time as answer/update info you need is that one you just posted.