PHP Return JSON object to Javascript [AJAX CALL] not working

Question

I have made Ajax calls before, and let them return JSON objects, which worked, but I dont seem to get it working anymore.

This is my ajax call:

function sendContactForm() {
var nameInput = $('#nameInput').val();
var emailInput = $('#emailInput').val();
var subjectInput = $('#subjectInput').val();
var msgInput = $('#msgInput').val();
$.ajax({
 // Make a POST request to getfile 
 url: "/service/contactmail",
 data: {
 nameInput: nameInput,
 emailInput: emailInput,
 subjectInput: subjectInput,
 msgInput: msgInput
 },
 method: "post",
 // And run this on success 
 success: function (data) {
 if (data.send === 1){
 // VERZONDEN
 }else if(data.send === 2){
 // VARS NIET INGEVULT
 }else{
 // IETS ANDERS FOUT
 }
 console.log(data);
 },
 error: function () {
 alert("fout");
 }
});
}

and this is my php function:

private function sendContactForm() {
 $output = array(
 "test" => null,
 "send" => null
 );
 if ($this->fillVariables()) {
 $this->sendMail();
 $output['send'] = 1;
 return true;
 } else {
 $output['send'] = 2;
 return false;
 }
 header("Content-Type: application/json");
 echo json_encode($output);
}

but the variable "data" has a value of: "" (empty string) There are no other echo's in my php class, so that should not be the problem.

Thanks in advance,
Mats de Waard

• RiggsFolly: Remove the return from inside the if's and then it will get to the header and the echo of your results. Currently it never gets to execute anything after the return
• Xenofexs: Yes RiggsFolly. @Mats de Waard, how you call your private php sendContactForm() function ?
• RiggsFolly: @Xenofexs I have to assume that is being done from another method in that class, as I dont know a way to call a class method directly from an AJAX call, even if it is public
• Mats de Waard: I call the function via another function down the class. (I did not show it, because it is irrelevant)

Answer 1

The return's in your if statements are stopping the execution of this function before you can generate the result back to the page.

private function sendContactForm() {
 $output = array(
 "test" => null,
 "send" => null
 );
 if ($this->fillVariables()) {
 $this->sendMail();
 $output['send'] = 1;
 //return true;
 } else {
 $output['send'] = 2;
 //return false;
 }
 header("Content-Type: application/json");
 echo json_encode($output);
}

• Mats de Waard: Oh my god. I feel stupid. Sorry for the nooby mistake, and I think I need to take a break :P Thanks for the help!
• RiggsFolly: We have all been there pal. Dont let it worry you.