I have a list of hotels in a select I want update the database base on the hotel selected if it update the field I want to alert it on the home page.
this is the ajax function
function showRoom($hotel_plan_id) {
var id = $hotel_plan_id;
if (id !== "") {
$.ajax({
type: "POST",
dataType: 'JSON',
url:'{{ route('home', '') }}/'+id,
data:{_token:'{{ csrf_token() }}'},
success:function(data){
alert(data);
},
error: function (result) {
alert("Error occur in the update()");
}
});
}
}
my controller
public function update(Request $request, $hotel_plan_id)
{
$hotel=plan_hotel::where('hotel_plan_id', $hotel_plan_id)->first();
$hotel->hotel_name = $request->input('hotel_name');
//$hotel ->room_name = $request->input('room_name');
$hotel->save();
// return redirect('/home')->with('success', 'price updated');
}
my route
Route::post('home/{hotel_plan_id}', 'HomeController#update')->name('home');
my select form
{!! Form::select('hotel_name', array($item[4], $item[10]),'S', array('style'=>' Border:none; ', 'id' => 'hotel_id', 'onclick'=>"showRoom(this.value, $item[8])"));!!}
You have error with ajax url and you dont also pass the value of hotel name.
Check this changes.
function showRoom($hotel_plan_id) {
var id = $hotel_plan_id;
if (id !== "") {
$.ajax({
type: "POST",
dataType: 'JSON',
url:'/home/' + id,
data:{_token:'{{ csrf_token() }}', hotel_name: 'Your value here'},
});
}
Please return json response
return response()->json();
You have to return json object like this
return response->json([
'status' => 200,
'result' => true,
'error' => false,
'data' => $hotel
]);
Related
I'm using laravel 8.4 .
My route :
Route::post('test',[\App\Http\Controllers\DataController::class, 'store'])->name('pos-test');
My Controller :
public function store(Request $request)
{
// DB::table('data')->insert('product_code',$request->id);
$badge = explode(' ', $request);
$employee_id = $badge[0];
\DB::table('data')->insert(['product_code'=> $employee_id]);
return response()->json(['success'=>'Product saved successfully.']);
}
Ajax code :
function handleBarcode(scanned_barcode) {
//handle your code here....
console.log(scanned_barcode);
let _token = $('meta[name="csrf-token"]').attr('content');
event.preventDefault();
$.ajaxSetup({
headers: {
'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content')
}
});
$.ajax({
url: "{{ route('pos-test') }}",
type: "POST", // Can change this to get if required
data: {
code : scanned_barcode,
_token: _token
},
success: function(data) {
$("#status").html(data);
},
error: function(jqXHR, textStatus, errorThrown) {
$("#status").text(textStatus);
console.log(jqXHR);
}
});
};
The request is like this "555444 Razif Raziq" , i would like to explode it so I may insert only "555444" into table but in table column product_code is 'POST' .
The question is how to fix it? thank you
you must explode your correct data in request object not request object itself.
$badge = explode(' ', $request->code);
If the 'code' value is sent correctly, just use this
$request->code
public function store(Request $request)
{
\DB::table('data')->insert(['product_code'=> $request->code]);
return response()->json(['success'=>'Product saved successfully.']);
}
I try to create a relational select option in laravel 8 using ajax but didn't work I get this error in the console: 405 Method Not Allowed
POST http://127.0.0.1:8000/decomptes/%7B%7B%20route(%20'getOperationsChantier'%20)%20%7D%7D 405 (Method Not Allowed)
the route
Route::post('/getOperationsChantier', [DecompteController::class, 'getOperationsChantier'])->name('getOperationsChantier');
the jquery code
$(document).ready(function(){
// When an option is changed, search the above for matching choices
$('#chantiers').change(function(){
$("#mySelect option").remove();
// $("#city option").remove();
var id = $(this).val();
$.ajax({
url : "{{ route( 'getOperationsChantier' ) }}",
data: {
"_token": "{{ csrf_token() }}",
"id": id
},
type: 'post',
dataType: 'json',
success: function( result )
{
$.each( result, function(k, v) {
$('#mySelect').append($('<option>', {value:k, text:v}));
});
},
error: function()
{
//handle errors
alert('error...');
}
});
}).change();
});
the function in the controller :
function getOperationsChantier( Request $request )
{
$this->validate( $request, [ 'id' => 'required|exists:chantiers,id' ] );
$chantierOperations = ChantierOperation::where('chantier_id', $request->get('id') )->get();
$output = [];
foreach( $chantierOperations as $chantierOperation )
{
$output[$chantierOperation->id] = $chantierOperation->operation_id;
}
return $output;
}
can someone help me this is the first time when I use ajax
I am using laravel 6.0 and i am building crud application. I have following jquery code in view file
function updaterecord(id) {
$('#modalupdate').on('submit', function (e) {
e.preventDefault();
$.ajax({
url: 'update/'+id,
method: 'post',
success: function (res) {
console.log(res);
}
})
});
}
And this is the code in controller
public function update(Request $request, $id='') {
$country = $request->input('countryname');
$sortname = $request->input('sortname');
$phonecode = $request->input('phonecode');
//return $country.$sortname.$phonecode;
return $request;
// DB::table('countries')->where('id',$id)->update(
// [
// 'name' => $country,
// 'sortname' => $sortname,
// 'phonecode' => $phonecode,
// ]);
}
The problem is $request returns empty.
If I don't use ajax then I am getting all input values. But I dont know why its not working for ajax request. Also I have added this line in view file
headers: {
'X-CSRF-TOKEN': '{!! csrf_token() !!}'
}
});
Please help me to solve this problem
You are not passing your form data. Try this:
function updaterecord(id) {
$('#modalupdate').on('submit', function (e) {
e.preventDefault();
$.ajax({
url: 'update/' + id,
method: 'post',
data: $(this).serialize();
success: function (res) {
console.log(res);
}
})
});
}
laravel by default does not send raw data , you have to convert your data to json, the best practice is :
return response()->json([
'data' => $request
]);
Just try this code for example and see if you get any hint.
function updaterecord(id) {
$('#modalupdate').on('submit', function (e) {
e.preventDefault();
$.ajax({
url: 'update/' + id,
method: 'post',
data: {'countryname' : 'India','sortname' : 'Sort Name', 'phonecode' : '022'};
success: function (res) {
console.log(res);
}
})
});
}
See if you are getting any response.
I made an Ajax POST request into my Laravel function however i am facing this result:
<script> Sfdump = window.Sfdump || (function (doc) { var refStyle = doc.createElement('style')
This happens when i die and dump my data so as to see what i get from ajax request. I have this jquery method:
$.ajaxSetup({
headers: {
'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content')
}
});
$('#save-person').on('click', function() {
let first_name = $('#first_name').val();
let middle_name = $('#middle_name').val();
let third_name = $('#third_name').val();
let family_name = $('#family_name').val();
$.ajax({
url: urlReq+"/api/employee/customize",
type: "POST",
data: {
first_name: first_name,
middle_name: middle_name,
third_name: third_name,
family_name: family_name,
},
cache: false,
success: function(dataResult){
console.log(dataResult);
let data = dataResult;
if(data.statusCode==200){
$("#success").show();
$('#success').html('Data added successfully !');
}
else if(dataResult.statusCode==201){
alert("Error occured !");
}
}
});
});
and on my php method i have this:
public function customize_store(Request $request){
//dd($request->first_name);
$input = $request->all();
dd($input);
return response()->json(['Person'=>$input]);
}
which result to <script> Sfdump = window.Sfdump || (function (doc) { var refStyle = doc.createElement('style'), rxEsc = /([.*+?^${}()|\[\]\/\\])/g, idRx =... but my input are also present there that looks like this:
#<span class=sf-dump-protected title="Protected property">parameters</span>: <span class=sf-dump-note>array:15</span> [<samp>
"<span class=sf-dump-key>first_name</span>" => "<span class=sf-dump-str title="7 characters">Michael</span>"
"<span class=sf-dump-key>middle_name</span>" => "<span class=sf-dump-str title="6 characters">Sangga</span>"
"<span class=sf-dump-key>third_name</span>" => <span class=sf-dump-const>null</span>
"<span class=sf-dump-key>family_name</span>" => "<span class=sf-dump-str title="7 characters">Smith</span>"
How would i extract those data so that i can persist it on my database?
Try this code.. check the URL, Routes on which you are sending your
data..
public function addPersonData(Request $request){
$save_person = new Person(); // Initialize your model here..
$save_person->first_name = $request->get('first_name');
$save_person->middle_name = $request->get('middle_name');
$save_person->third_name = $request->get('third_name');
$save_person->family_name = $request->get('family_name');
$save_person->save();
return 'ok';
}
I think I got it. First add a name to your route (see here) and the ajax part in your jQuery (assuming you use a form to submit the user data):
in your Route.php add:
Route::post('api/employee/customize', 'PersonController#customize_store')->name('api.employee.customize');
Your jQuery ajax request:
$('#save-person').submit(function(e) {
e.preventDefault();
let first_name = $('#first_name').val();
let middle_name = $('#middle_name').val();
let third_name = $('#third_name').val();
let family_name = $('#family_name').val();
$.ajax({
url: "{{ route('api.employee.customize') }}",
type: "POST",
data: { first_name, middle_name, third_name, family_name },
cache: false,
success: function(data){
console.log(data);
if(data.status === 'success'){
$("#success").show();
$('#success').html('Data added successfully !');
//the person's details are in data.person.first_name etc
//you already knew that, but added is the new data.person.id you may use
}
else {
alert("Error occured !");
}
}
});
});
and your controller, assuming the model linked to this data is Person:
public function customize_store(Request $request){
$person = new Person($request->all());
if ($person->save()) {
return response()->json(['status' => 'success', 'person'=>$person]);
}
return response()->json(['status' => 'fail']);
}
i want to set condition, if teacher already exist in database then update record, if id doesn't exist then add record in database. how can i achieve it using ajax in laravel?
Update.js:
jQuery(document).ready(function($) {
$('#update-data').on('click',function(){
alert("ok");
$.ajax({
type: "POST",
url: "teachers/" + $('#update-data').attr("value"),
dataType: 'json',
headers: { 'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content') },
data : $(this).serialize(),
beforeSend: function() {
},
success: function (data) {
alert("ok");
},
});
});
});
Store.Js:
jQuery(document).ready(function($) {
$("#add-data").submit(function (e) {
$.ajax({
type: "POST",
url: "teachers",
dataType: 'json',
headers: { 'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content') },
data: $(this).serialize(),
success: function (data) {
alert("Added");
data.responseJSON;
refreshTable();
},
});
});
});
Update Controller:
public function update(TeacherRequest $request, $id)
{
$teacher = Teacher::find($id);
if($teacher->save()){
return response()->json([
'status' => 'success',
'msg' => 'esecond has been updated'
]);
}
}
Store Controller:
public function store(Request $request)
{
$teacher = new Teacher;
$teacher=teacher::create($request);
}
There's a custom method for this
$teacher = Teacher::firstOrCreate($id, [
// Pass data here
]);
And if you want to check manually and traverse the request to another method
public function update(TeacherRequest $request, $id)
{
$teacher = Teacher::find($id);
if (is_null($teacher)) { // If model not found, pass request to store method
$this->store($request);
}
if($teacher->save()){
return response()->json([
'status' => 'success',
'msg' => 'esecond has been updated'
]);
}
}
From the docs
Hope this helps
Eloquent provides a updateOrCreate method so you can update or create a record.
$teacher = Teacher::updateOrCreate([
// attributes to search for
'name' => 'Test Teacher',
], [
// values
'grade' => 6,
]);
This would find a teacher with name 'Test Teacher' and update grade to 6 or create a new teacher with name 'Test Teacher' and grade set to 6.
"You may also come across situations where you want to update an existing model or create a new model if none exists. Laravel provides an updateOrCreate method to do this in one step. Like the firstOrCreate method, updateOrCreate persists the model ..." - Laravel 6.0 Docs - Eloquent - Other Creation Methods - updateOrCreate