passing javascript variable to php in cakephp 2 view - php

Question

I am trying to pass a javascript variable into a php code in the view file of the cakephp 2.
for (id in response) {
var book = response[id];
if (typeof(book.thumbnail_url) != "undefined") {
var x= book.thumbnail_url;
<?php
$file11 = WWW_ROOT . 'img' . DS . 'book_images';
define('DIRECTORY', $file11);
$content = file_get_contents($abc);
file_put_contents(DIRECTORY . '/'.$isbn.'.jpg', $content);
?>
}
}
i am trying to pass the value of x in the file_get_contents function in place of $abc so that it could save the image coming from the javascript's URL accordingly.
EDIT::
for (id in response) {
var book = response[id];
if (typeof(book.thumbnail_url) != "undefined") {
var x= book.thumbnail_url;
$.ajax({
type: "POST",
url: '/BookSearchs/test',
data: {'yourX':x}
}).done(function(result) {
alert("yes");
}).fail(function() {
alert("no");
});
}
}
This is what i wrote after implementing the answers i got . But Everytime it pops up "no". Here BookSearchs is my controller and test is my function inside it.
EDIT 2:
function handleResponse(response) {
var target = '';
for (id in response) {
var book = response[id];
if (typeof(book.thumbnail_url) != "undefined") {
var x = book.thumbnail_url;
$.ajax({
type: 'POST',
url: "BookSearchs/test",
data: {
myVal: x
},
success: function() {
alert('AjaX Success')
},
error: function() {
alert('AjaX Failed')
}
})
.done(function() {
alert('AjaX Done!');
});
}
}
return true;
}
Currently this is what i've have done so far , the form method did not work out . It was redirecting me to another page . Anyways this is my current code . And 'test' is the my function inside the controller where i want to access the myVal value using POST . Also i have this question do i need to create a physical file for test in order to make the ajax function work, because if i delete the test.ctp file then the ajax starts giving the fail message . So for now i have created a physical test.file in the BookSearchs folder in the view , although it's empty in order to make the ajax function work . I am having a doubt whether my Url in Ajax is wrong or i am not accessing the values properly in the controller.

Answer 1

I don't think that this is a proper way to do that in theory. But, sometime we might need this.
Before we proceed to this way, you might need to think other technologies such as NodeJs (e.g fs.readFileSync)
Basically, you can't directly do that. Because, JavaScript run on client side and PHP is run on sever side.
Anyway, there might be a few tweak to do that. But, this approach might be slow and it depends on how many loop you making.
for (id in response) {
var book = response[id];
if (typeof(book.thumbnail_url) != "undefined") {
var x= book.thumbnail_url;
$.ajax({
type: "POST",
url: '/yourcontroller/route',
data: {'yourX':x}
}).done(function(result) {
//if success, execute other code
}).fail(function() {
//DO other if fail
});
}
}
Then, read this value in your controller
$xValue = $_POST['yourX'];
$file11 = WWW_ROOT . 'img' . DS . 'book_images';
define('DIRECTORY', $file11);
$content = file_get_contents($xValue);
file_put_contents(DIRECTORY . '/'.$isbn.'.jpg', $content);
//do some checking success or fail
//I will assume success
$status = 'success';
echo json_encode(['status'=>$status]);

Answer 2

Usally I use a Trick to pass JS variable to a CakePhp Controller(php variable). Actionly I create a form that contain a hidden input, I'll also put the link of the page that will receive the php variable.
.ctp
<?=
$this->Html->link('<i class="fa fa-file-pdf-o"></i>' .__('Export'),
'javascript:myFunction();',
array('escape' => false, 'class' => 'btn btn-app dispatch', 'id' => 'dispatch_packages',
'style'=>'margin-right:0px;background: #f39c12;color:white;',
'disabled' => 'false'));
?>
<form id="sampleForm" name="sampleForm" style="display: none" method="post" action="<?= $this->Url->build([
'controller' => 'YourController',
'action' => 'youraction']) ?>">
<input type="hidden" name="variable" id="variable" value="">
</form>
JS
var jsVar=0;
function myFunction()
{
document.sampleForm.variable.value = jsVar;
document.forms["sampleForm"].submit();
}
I used it with the CakePhp 3.x framework & it's working very fine. You have just to write the url with the CakePhp 2.x Syntax.
Don't hesitate to comment my answer if you'll have any difficulties to apply it.
Good Luck !