jQuery $.ajax() with PHP - php

Question

I'm trying to use jQuery $.ajax() but I'm facing some difficulties.
Here's the textbox field that I want to use for POST:
<input name="url" class="url" type="text" >
Here's the code:
$.ajax({
type: "post",
url: "file.php",
data: $(this).serialize(),
success: function(data) { ...............
Now this is the file.php:
<?php
if( $_REQUEST['url'] )
{
$url = $_REQUEST['url'];
$url = file_get_contents($url);
// I would need now to return something here but not sure how??!!
}
?>
Now, my question is, how to return variables in this PHP code and use them in my code above, I mean in the success part of $.ajax(). Also if I want to perform some additional stuff on the $url variable, how to do it? How to return them? :/

Answer 1

If you want to return some variables/fields, the best way would be to echo a JSON string. Here is a small example:
PHP Code:
<?php
if( $_REQUEST['url'] )
{
$url = $_REQUEST['url'];
$url = file_get_contents($url);
$result['var1'] = 'something';
$result['var2'] = 500;
echo json_encode($result);
}
?>
JS Code:
$.ajax({
type: "post",
url: "file.php",
data: $(this).serialize(),
dataType: 'json', // maybe not needed? I do not know if jQuery autodetects it
success: function(data) {
// here you can use data.var1 and data.var2 to read the fields
}
});

Answer 2

You just print/echo your 'return' value.
file.php
<?php
if( $_REQUEST['url'] )
{
$url = $_REQUEST['url'];
$url = file_get_contents($url);
// I would need now to return something here but not sure how??!!
echo "something";
}
?>
Then in your JS:
$.ajax({
type: "post",
url: "file.php",
data: $(this).serialize(),
success: function(data) {
console.log(data); // "something"
}
});
As a side note. Your script looks like it accepts any URL and fetches it. It is possible to abuse scripts like that. Make sure you are aware that.