Empty array when dd($request->all()) upload file via ajax laravel - javascript

first of all i try to do the file uploading via ajax , when i try dd($request->all())in my controller , it give result empty array
public function uploadFile(Request $request){
dd($request->all());
}
My blade view with ajax
<label for="inputfile">
<a title="Click here to upload record "><i class="fa fa-upload"></i></a>
</label>
<input id="inputfile" name="inputfile" type="file" />
<script>
$('#inputfile').on('change',function(ev){
ev.preventDefault();
var postData=new FormData();
postData.append('file',this.files[0]);
$.ajax({
url:'{{url('reporting/uploadFile')}}',
headers:{'X-CSRF-Token':$('meta[name=csrf_token]').attr('content')},
type:"get",
contentType:false,
data:postData,
processData:false,
dataType:'json',
success: function( data ) {
console.log(data)
},
error: function() {
alert('error');
} }); });
</script>
My laravel version is 5.8 . The flow is when the user upload attachment, it will directly store to file storage without clicking button submit . But when i try to retrieve $request->all() its return empty array which is i can't continue further step. Sorry if my explaination not clear .

Yes ok laravel can be a real pain sometimes especially when it comes to file uploads.
You can try this article for laravel 5.8 give it a try and let me know if it works.
https://www.w3adda.com/blog/laravel-5-8-jquery-ajax-form-submit
I think the main difference with this article is the way it sets the data in the ajax call. However you might need to check the whole article over and compare it to your code.
$.ajax({
url: "{{ url('jquery-ajax-form-submit')}}",
method: 'post',
data: $('#contact_us').serialize(),

Please ensure you are using the form multipart setting correctly.
This is usually the issue in most cases.
<form action="upload.php" method="post" enctype="multipart/form-data">

let files = $('#inputfile');
let image = files[0].files;
let form_data = new FormData();
if (image.length > 0) {
for (let i = 0; i < image.length; i++)
form_data.append('inputfile[]', image[i]);
}
$.ajax({
url:'{{url('reporting/uploadFile')}}',
headers:{'X-CSRF-Token':$('meta[name=csrf_token]').attr('content')},
type:"get",
contentType:false,
data:form_data,
processData:false,
dataType:'json',
success: function( data ) {
console.log(data)
},
error: function() {
alert('error');
}
});
});
try this.

You just need to set "Content-Type" in header.
You also have pass type get, for file upload must require post.
have you console.log(this.files[0]);
<script>
$('#inputfile').on('change',function(ev){
ev.preventDefault();
console.log(this.files[0]);
var postData=new FormData();
postData.append('file',this.files[0]);
$.ajax({
url:'{{url('reporting/uploadFile')}}',
headers:{
'X-CSRF-Token':$('meta[name=csrf_token]').attr('content'),
'Content-Type': undefined
},
type:"POST",
data:postData,
success: function( data ) {
console.log(data)
},
error: function() {
alert('error');
} }); });
</script>

Related

How can I combine two ajax functions in one function?

I have ajax code which can post name and I have another ajax which can post image I need to combine those two functions in order to have one function which can post both name and image
Below codes used to post image
<script>
$(document).ready(function(){
$("#but_upload").click(function(){
var fd = new FormData();
var files = $('#file')[0].files;
// Check file selected or not
if(files.length > 0 ){
fd.append('file',files[0]);
$.ajax({
url: 'upload.php',
type: 'post',
data: fd,
contentType: false,
processData: false,
success: function(response){
if(response != 0){
$("#img").attr("src",response);
$(".preview img").show(); // Display image element
}else{
alert('file not uploaded');
}
},
});
}else{
alert("Please select a file.");
}
});
});
</script>
And below codes used to post name
<script type="text/javascript">
function clickButton(){
var name=document.getElementById('name').value;
$.ajax({
type:"post",
url:"upload.php",
data:
{
'name' :name
},
cache:false,
success: function (html)
{
alert('Data Send');
$('#msg').html(html);
}
});
return false;
}
</script>
How can I combine above codes in order to use only one url "upload.php", this means upload.php will insert name in database and save image in folder while click save button, that's why I need to combine the codes
Please anyone can help me
You literally combine them.
You can use your first function and do the following:
var fd = new FormData();
var files = $('#file')[0].files;
fd.append('name', $("#name").val();
that is it. And on the other side (backend) you just ask for this name:
$name = $_POST['name'];

Submit form array fields using Ajax [duplicate]

I have a form with name orderproductForm and an undefined number of inputs.
I want to do some kind of jQuery.get or ajax or anything like that that would call a page through Ajax, and send along all the inputs of the form orderproductForm.
I suppose one way would be to do something like
jQuery.get("myurl",
{action : document.orderproductForm.action.value,
cartproductid : document.orderproductForm.cartproductid.value,
productid : document.orderproductForm.productid.value,
...
However I do not know exactly all the form inputs. Is there a feature, function or something that would just send ALL the form inputs?
This is a simple reference:
// this is the id of the form
$("#idForm").submit(function(e) {
e.preventDefault(); // avoid to execute the actual submit of the form.
var form = $(this);
var actionUrl = form.attr('action');
$.ajax({
type: "POST",
url: actionUrl,
data: form.serialize(), // serializes the form's elements.
success: function(data)
{
alert(data); // show response from the php script.
}
});
});
You can use the ajaxForm/ajaxSubmit functions from Ajax Form Plugin or the jQuery serialize function.
AjaxForm:
$("#theForm").ajaxForm({url: 'server.php', type: 'post'})
or
$("#theForm").ajaxSubmit({url: 'server.php', type: 'post'})
ajaxForm will send when the submit button is pressed. ajaxSubmit sends immediately.
Serialize:
$.get('server.php?' + $('#theForm').serialize())
$.post('server.php', $('#theForm').serialize())
AJAX serialization documentation is here.
Another similar solution using attributes defined on the form element:
<form id="contactForm1" action="/your_url" method="post">
<!-- Form input fields here (do not forget your name attributes). -->
</form>
<script type="text/javascript">
var frm = $('#contactForm1');
frm.submit(function (e) {
e.preventDefault();
$.ajax({
type: frm.attr('method'),
url: frm.attr('action'),
data: frm.serialize(),
success: function (data) {
console.log('Submission was successful.');
console.log(data);
},
error: function (data) {
console.log('An error occurred.');
console.log(data);
},
});
});
</script>
There are a few things you need to bear in mind.
1. There are several ways to submit a form
using the submit button
by pressing enter
by triggering a submit event in JavaScript
possibly more depending on the device or future device.
We should therefore bind to the form submit event, not the button click event. This will ensure our code works on all devices and assistive technologies now and in the future.
2. Hijax
The user may not have JavaScript enabled. A hijax pattern is good here, where we gently take control of the form using JavaScript, but leave it submittable if JavaScript fails.
We should pull the URL and method from the form, so if the HTML changes, we don't need to update the JavaScript.
3. Unobtrusive JavaScript
Using event.preventDefault() instead of return false is good practice as it allows the event to bubble up. This lets other scripts tie into the event, for example analytics scripts which may be monitoring user interactions.
Speed
We should ideally use an external script, rather than inserting our script inline. We can link to this in the head section of the page using a script tag, or link to it at the bottom of the page for speed. The script should quietly enhance the user experience, not get in the way.
Code
Assuming you agree with all the above, and you want to catch the submit event, and handle it via AJAX (a hijax pattern), you could do something like this:
$(function() {
$('form.my_form').submit(function(event) {
event.preventDefault(); // Prevent the form from submitting via the browser
var form = $(this);
$.ajax({
type: form.attr('method'),
url: form.attr('action'),
data: form.serialize()
}).done(function(data) {
// Optionally alert the user of success here...
}).fail(function(data) {
// Optionally alert the user of an error here...
});
});
});
You can manually trigger a form submission whenever you like via JavaScript using something like:
$(function() {
$('form.my_form').trigger('submit');
});
Edit:
I recently had to do this and ended up writing a plugin.
(function($) {
$.fn.autosubmit = function() {
this.submit(function(event) {
event.preventDefault();
var form = $(this);
$.ajax({
type: form.attr('method'),
url: form.attr('action'),
data: form.serialize()
}).done(function(data) {
// Optionally alert the user of success here...
}).fail(function(data) {
// Optionally alert the user of an error here...
});
});
return this;
}
})(jQuery)
Add a data-autosubmit attribute to your form tag and you can then do this:
HTML
<form action="/blah" method="post" data-autosubmit>
<!-- Form goes here -->
</form>
JS
$(function() {
$('form[data-autosubmit]').autosubmit();
});
You can also use FormData (But not available in IE):
var formData = new FormData(document.getElementsByName('yourForm')[0]);// yourForm: form selector
$.ajax({
type: "POST",
url: "yourURL",// where you wanna post
data: formData,
processData: false,
contentType: false,
error: function(jqXHR, textStatus, errorMessage) {
console.log(errorMessage); // Optional
},
success: function(data) {console.log(data)}
});
This is how you use FormData.
Simple version (does not send images)
<form action="/my/ajax/url" class="my-form">
...
</form>
<script>
(function($){
$("body").on("submit", ".my-form", function(e){
e.preventDefault();
var form = $(e.target);
$.post( form.attr("action"), form.serialize(), function(res){
console.log(res);
});
});
)(jQuery);
</script>
Copy and paste ajaxification of a form or all forms on a page
It is a modified version of Alfrekjv's answer
It will work with jQuery >= 1.3.2
You can run this before the document is ready
You can remove and re-add the form and it will still work
It will post to the same location as the normal form, specified in
the form's "action" attribute
JavaScript
jQuery(document).submit(function(e){
var form = jQuery(e.target);
if(form.is("#form-id")){ // check if this is the form that you want (delete this check to apply this to all forms)
e.preventDefault();
jQuery.ajax({
type: "POST",
url: form.attr("action"),
data: form.serialize(), // serializes the form's elements.
success: function(data) {
console.log(data); // show response from the php script. (use the developer toolbar console, firefox firebug or chrome inspector console)
}
});
}
});
I wanted to edit Alfrekjv's answer but deviated too much from it so decided to post this as a separate answer.
Does not send files, does not support buttons, for example clicking a button (including a submit button) sends its value as form data, but because this is an ajax request the button click will not be sent.
To support buttons you can capture the actual button click instead of the submit.
jQuery(document).click(function(e){
var self = jQuery(e.target);
if(self.is("#form-id input[type=submit], #form-id input[type=button], #form-id button")){
e.preventDefault();
var form = self.closest('form'), formdata = form.serialize();
//add the clicked button to the form data
if(self.attr('name')){
formdata += (formdata!=='')? '&':'';
formdata += self.attr('name') + '=' + ((self.is('button'))? self.html(): self.val());
}
jQuery.ajax({
type: "POST",
url: form.attr("action"),
data: formdata,
success: function(data) {
console.log(data);
}
});
}
});
On the server side you can detect an ajax request with this header that jquery sets HTTP_X_REQUESTED_WITH
for php
PHP
if(!empty($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest') {
//is ajax
}
This code works even with file input
$(document).on("submit", "form", function(event)
{
event.preventDefault();
$.ajax({
url: $(this).attr("action"),
type: $(this).attr("method"),
dataType: "JSON",
data: new FormData(this),
processData: false,
contentType: false,
success: function (data, status)
{
},
error: function (xhr, desc, err)
{
}
});
});
I really liked this answer by superluminary and especially the way he wrapped is solution in a jQuery plugin. So thanks to superluminary for a very useful answer. In my case, though, I wanted a plugin that would allow me to define the success and error event handlers by means of options when the plugin is initialized.
So here is what I came up with:
;(function(defaults, $, undefined) {
var getSubmitHandler = function(onsubmit, success, error) {
return function(event) {
if (typeof onsubmit === 'function') {
onsubmit.call(this, event);
}
var form = $(this);
$.ajax({
type: form.attr('method'),
url: form.attr('action'),
data: form.serialize()
}).done(function() {
if (typeof success === 'function') {
success.apply(this, arguments);
}
}).fail(function() {
if (typeof error === 'function') {
error.apply(this, arguments);
}
});
event.preventDefault();
};
};
$.fn.extend({
// Usage:
// jQuery(selector).ajaxForm({
// onsubmit:function() {},
// success:function() {},
// error: function() {}
// });
ajaxForm : function(options) {
options = $.extend({}, defaults, options);
return $(this).each(function() {
$(this).submit(getSubmitHandler(options['onsubmit'], options['success'], options['error']));
});
}
});
})({}, jQuery);
This plugin allows me to very easily "ajaxify" html forms on the page and provide onsubmit, success and error event handlers for implementing feedback to the user of the status of the form submit. This allowed the plugin to be used as follows:
$('form').ajaxForm({
onsubmit: function(event) {
// User submitted the form
},
success: function(data, textStatus, jqXHR) {
// The form was successfully submitted
},
error: function(jqXHR, textStatus, errorThrown) {
// The submit action failed
}
});
Note that the success and error event handlers receive the same arguments that you would receive from the corresponding events of the jQuery ajax method.
I got the following for me:
formSubmit('#login-form', '/api/user/login', '/members/');
where
function formSubmit(form, url, target) {
$(form).submit(function(event) {
$.post(url, $(form).serialize())
.done(function(res) {
if (res.success) {
window.location = target;
}
else {
alert(res.error);
}
})
.fail(function(res) {
alert("Server Error: " + res.status + " " + res.statusText);
})
event.preventDefault();
});
}
This assumes the post to 'url' returns an ajax in the form of {success: false, error:'my Error to display'}
You can vary this as you like. Feel free to use that snippet.
jQuery AJAX submit form, is nothing but submit a form using form ID when you click on a button
Please follow steps
Step 1 - Form tag must have an ID field
<form method="post" class="form-horizontal" action="test/user/add" id="submitForm">
.....
</form>
Button which you are going to click
<button>Save</button>
Step 2 - submit event is in jQuery which helps to submit a form. in below code we are preparing JSON request from HTML element name.
$("#submitForm").submit(function(e) {
e.preventDefault();
var frm = $("#submitForm");
var data = {};
$.each(this, function(i, v){
var input = $(v);
data[input.attr("name")] = input.val();
delete data["undefined"];
});
$.ajax({
contentType:"application/json; charset=utf-8",
type:frm.attr("method"),
url:frm.attr("action"),
dataType:'json',
data:JSON.stringify(data),
success:function(data) {
alert(data.message);
}
});
});
for live demo click on below link
How to submit a Form using jQuery AJAX?
I know this is a jQuery related question, but now days with JS ES6 things are much easier. Since there is no pure javascript answer, I thought I could add a simple pure javascript solution to this, which in my opinion is much cleaner, by using the fetch() API. This a modern way to implements network requests. In your case, since you already have a form element we can simply use it to build our request.
const form = document.forms["orderproductForm"];
const formInputs = form.getElementsByTagName("input");
let formData = new FormData();
for (let input of formInputs) {
formData.append(input.name, input.value);
}
fetch(form.action,
{
method: form.method,
body: formData
})
.then(response => response.json())
.then(data => console.log(data))
.catch(error => console.log(error.message))
.finally(() => console.log("Done"));
Try
fetch(form.action,{method:'post', body: new FormData(form)});
function send(e,form) {
fetch(form.action,{method:'post', body: new FormData(form)});
console.log('We submit form asynchronously (AJAX)');
e.preventDefault();
}
<form method="POST" action="myapi/send" onsubmit="send(event,this)" name="orderproductForm">
<input hidden name="csrfToken" value="$0meh#$h">
<input name="email" value="aa#bb.com">
<input name="phone" value="123-456-666">
<input type="submit">
</form>
Look on Chrome Console > Network after/before 'submit'
consider using closest
$('table+table form').closest('tr').filter(':not(:last-child)').submit(function (ev, frm) {
frm = $(ev.target).closest('form');
$.ajax({
type: frm.attr('method'),
url: frm.attr('action'),
data: frm.serialize(),
success: function (data) {
alert(data);
}
})
ev.preventDefault();
});
You may use this on submit function like below.
HTML Form
<form class="form" action="" method="post">
<input type="text" name="name" id="name" >
<textarea name="text" id="message" placeholder="Write something to us"> </textarea>
<input type="button" onclick="return formSubmit();" value="Send">
</form>
jQuery function:
<script>
function formSubmit(){
var name = document.getElementById("name").value;
var message = document.getElementById("message").value;
var dataString = 'name='+ name + '&message=' + message;
jQuery.ajax({
url: "submit.php",
data: dataString,
type: "POST",
success: function(data){
$("#myForm").html(data);
},
error: function (){}
});
return true;
}
</script>
For more details and sample Visit:
http://www.spiderscode.com/simple-ajax-contact-form/
To avoid multiple formdata sends:
Don't forget to unbind submit event, before the form submited again,
User can call sumbit function more than one time, maybe he forgot something, or was a validation error.
$("#idForm").unbind().submit( function(e) {
....
If you're using form.serialize() - you need to give each form element a name like this:
<input id="firstName" name="firstName" ...
And the form gets serialized like this:
firstName=Chris&lastName=Halcrow ...
I find it surprising that no one mentions data as an object. For me it's the cleanest and easiest way to pass data:
$('form#foo').submit(function () {
$.ajax({
url: 'http://foo.bar/some-ajax-script',
type: 'POST',
dataType: 'json',
data: {
'foo': 'some-foo-value',
'bar': $('#bar').val()
}
}).always(function (response) {
console.log(response);
});
return false;
});
Then, in the backend:
// Example in PHP
$_POST['foo'] // some-foo-value
$_POST['bar'] // value in #bar
This is not the answer to OP's question,
but in case if you can't use static form DOM, you can also try like this.
var $form = $('<form/>').append(
$('<input/>', {name: 'username'}).val('John Doe'),
$('<input/>', {name: 'user_id'}).val('john.1234')
);
$.ajax({
url: 'api/user/search',
type: 'POST',
contentType: 'application/x-www-form-urlencoded',
data: $form.serialize(),
success: function(data, textStatus, jqXHR) {
console.info(data);
},
error: function(jqXHR, textStatus, errorThrown) {
var errorMessage = jqXHR.responseText;
if (errorMessage.length > 0) {
alert(errorMessage);
}
}
});
JavaScript
(function ($) {
var form= $('#add-form'),
input = $('#exampleFormControlTextarea1');
form.submit(function(event) {
event.preventDefault();
var req = $.ajax({
url: form.attr('action'),
type: 'POST',
data: form.serialize()
});
req.done(function(data) {
if (data === 'success') {
var li = $('<li class="list-group-item">'+ input.val() +'</li>');
li.hide()
.appendTo('.list-group')
.fadeIn();
$('input[type="text"],textarea').val('');
}
});
});
}(jQuery));
HTML
<ul class="list-group col-sm-6 float-left">
<?php
foreach ($data as $item) {
echo '<li class="list-group-item">'.$item.'</li>';
}
?>
</ul>
<form id="add-form" class="col-sm-6 float-right" action="_inc/add-new.php" method="post">
<p class="form-group">
<textarea class="form-control" name="message" id="exampleFormControlTextarea1" rows="3" placeholder="Is there something new?"></textarea>
</p>
<button type="submit" class="btn btn-danger">Add new item</button>
</form>
There's also the submit event, which can be triggered like this $("#form_id").submit(). You'd use this method if the form is well represented in HTML already. You'd just read in the page, populate the form inputs with stuff, then call .submit(). It'll use the method and action defined in the form's declaration, so you don't need to copy it into your javascript.
examples

post binary code not working from Jquery ajax

I made this code to make image upload function but the variable is not getting posted by ajax please help me !!
Jquery()
<script type="text/JavaScript">
$(document).ready(function() {
$("#btn").click(function() {
$.get("i.png", function(response) {
var img = response;
});
$.ajax({
type: "POST",
url: 'lol.php',
data: {r: img},
success: function(data){
$("#ol").html(data)
}
});
return false;
});
});
</script>
PHP Code
<?php
$conn = mysqli_connect("localhost","root","","image");
$r = $_POST['r'];
echo $r;
?>
If you only want to make an image upload (and not exactly match "binary upload")
I suggest you to use the proper functional and native input type file.
In a html form, put an :
<input type="file" id="upload">
<img src="blank.png" title="upload-preview">
and in you javascript:
this will load the preview thumbnail selected
fileInput.addEventListener("change", function(event)
{
var file = $("#upload")[0].files[0];
$(".upload-preview").attr('src', window.URL.createObjectURL(file));
});
and when you click the button, that will send the image
with a "multipart/form-data" Content-Type
$("#btn").on("click", function()
{
var inputs = new FormData();
inputs.append("upload", $("#upload")[0].files[0]);
$.ajax
({
url:"your.php",
type:"POST",
data: inputs,
cache: false, // don't cache the image
processData: false, // Don't process the files
contentType: false,
success: function(response)
{
// next instructions
}
});
});

jQuery file Upload and ajax

So, I have the following form and js/php:
php
<form enctype="multipart/form-data">
<input id="fileupload" type="file" name="files[]" class="files " onChange="UploadImage" accept='image/*'/>
<input type="button" class="submit_form" value="submit">
</form>
<?php
add_action( 'wp_ajax_UploadImage', 'UploadImage' );
function UploadImage()
{
$upload_dir = wp_upload_dir();
$files = $_FILES['files'];
//Some function
}
?>
JS
function UploadImage(e)
{
jQuery('#fileupload').fileupload({
url: upload_image.ajax_url,
});
if(jQuery('#fileupload')) {
var form = document.forms.namedItem("upload_video");
var formdata = new FormData(form);
formdata.append('action', 'UploadImage');
jQuery.ajax({
success : function(data){
alert('sddsf');
}
})
}
};
As you can see from here, when an image is selected using Blueimp jQuery File upload (which the js is not properly written), I want the image file to be handled by the php function.
In other words, the js is not correctly written and I am not sure how to initiate the plugin then when the image is selected, it is processed by the php function via ajax (meaning, how do I parse the file info to php function via ajax?)
Don't use $.ajax directly. The plugin already does that behind the scenes.
Here's a working example, based on your code, but adapted to run on JSFiddle:
$(document).ready(function(){
var url = '/echo/json/';
var formdata = {json: JSON.stringify({field1: 'value1'})};
jQuery('#fileupload').fileupload({
url: url,
formData : formdata,
dataType: 'json',
type: "POST",
contentType:false,
processData:false,
success : function(data){
alert('success...');
console.dir(data);
}
});
});
Demo: http://jsfiddle.net/pottersky/8usb1sn3/3/

pass data($post) to php file using javascript without callback

I need to pass data from HTML page to PHP page But without data callback ....
i'm used two method but One of them did not succeed
1)
$.ajax({
type: "POST",
url: 'phpexample.php',
data: {voteid: x },
success: function(data)
{
alert("success! X:" + data);
}
});
2)
$.post("getClassStudent.php",
{
},
function(data){
$("#div_id.php").html(data);
}
);
as i can understand, you just want to send info to a php script and don't need the response, is that right?
try this
$.post("phpexample.php", {voteid:x});
or simply remove the "succes" function from the equation if you feel more confortable using $.ajax instead of $.post
$.ajax({
type: "POST",
url: 'phpexample.php',
data: {voteid: x }
});
your fisrt example is correct, the second is not well formed.
more info:
http://api.jquery.com/jquery.post/
EDIT: to help you some more :)
<button type="button" id="element-id">click</button>
<button type="button" class="class-name">Click</button>
$(document).ready(function(){
//if you are marking ans element by class use '.class-name'
$(".class-name").click(function(){
$.post("getClassStudent.php");
});
//if marking by id element use '#id-name'
$("#element-id").click(function(){
$.post("getClassStudent.php");
});
});
be carefful with the markings, for debuggin try to use "console.log()" or "alert()" so you can see where is the problem and where the code crushes.
var formData = {
'voteid' : 'x',
};
$.ajax({
type : 'POST',
url : 'phpexample.php',
data : formData, // our data object
dataType : 'json',
encode : true
}).done(function(data) {
console.log(data);
});

Categories

Resources