JSON.parse Error after decoding JSON.stringify-ed Array - php

Question

I want to submit an array of arrays through JSON. So I create an object, apply JSON.stringify() on it and set it to a formData...
var sources = new Array();
$('.source-instance').each(function(){
var auxObject = {};
auxObject["sourceTitle"] = $(this).find(".source-title").val();
sources.push(auxObject);
console.log('sources: ' + JSON.stringify(sources));
});
formData.set('sources', JSON.stringify(sources));
When I check the data
console.log(formData.get('sources'))
I get
[{"sourceTitle":"SomeTitle"},{"sourceTitle":"AnotherTitle"}]
... which I then proceed via AJAX to the server.
On server side in php I loop over the array and let it print:
foreach (json_decode($request['sources'], true) as $sourceInstance) {
print_r($sourceInstance);
}
I get
Array
(
[sourceTitle] => SomeTitle
)
Array
(
[sourceTitle] => AnotherTitle
)
which looks pretty fine. But I also get
SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data
How come that? And how can I repair this? I don't see, where it is coming from.
EDIT
the ajax looks like this:
$.ajax({
url: '/piece/store',
type: 'POST',
processData: false,
contentType: false,
dataType: 'JSON',
data: formData,
success: function(response){
if (response.status == 'error'){
// show the erors
}
if (response.status == 'success'){ // "redirect" to success page
window.location.replace('/succes');
}
},
error: function(response){
$(".optional.errors").show();
$('ul.errors').empty();
$.each(response.errors, function (index, error) {
$('ul.errors').append('<li>' + error + '</li>');
});
}
});
the php responds:
$response = [
'status' => 'success',
];
return response()->json($response);

Answer 1

That error message happens when jQuery is trying to parse the response from the ajax call.
In your ajax request, you have set: dataType: 'json', which tells jQuery to expect the response to be a json-string. jQuery will then try to parse that json-string into a json-object (using JSON.parse()).
However, since you have print_r($sourceInstance) in your PHP-code, the response won't be a valid json-string and will make JSON.parse() fail.
Any output on the server side (echo, var_dump, print_r etc) will be a part of the response.