Can't return array from php function to ajax request

Question

I have a javascript file in which I am doing an ajax request to foo.php in order to get an array of mysql results. Here is my javascript file.

//foo.js
$(document).ready(function(){
 $.ajax({
 url:"foo.php",
 type:"POST",
 data:{
 action:"test",
 remail:"foo"
 },
 success:function(data){
 var messages = data;
 console.log("Ext req");
 try{
 console.log(JSON.parse(data));
 }catch(e){
 console.log(data);
 }
 },
 failure:function(){
 }
 })
});

In order to receive my array of results from php I do the following:

//foo.php
<?php 
 if(isset($_POST)){
 $action = $_POST['action'];
 if(empty($action)){
 return;
 }
 switch($action){
 case "test":
 $query = "select * from message where seen";
 $ar = [];
 $res = mysqli_query($con,$query);
 while($row = mysqli_fetch_array($res)){
 $ar[] = $row;
 }
 echo json_encode($ar);
 break;
 }
 } 
?>

This returns an array of objects to my ajax request which then I can handle according to my needs. However if I try to move the php code inside the switch statement into a function and return the encoded result of the function I only get an empty array as response. Here is how I am trying to do it:

<?php 
 function test(){
 $query = "select * from message where seen";
 $ar = [];
 $res = mysqli_query($con,$query);
 while($row = mysqli_fetch_array($res)){
 $ar[] = $row;
 }
 return $ar;
 }
 if(isset($_POST)){
 $action = $_POST['action'];
 if(empty($action)){
 return;
 }
 switch($action){
 case "test":
 $result = test();
 echo json_encode($result);
 break;
 }
 } 
?>

Any ideas why this happening?

UPDATE

$con is a variable that comes from another file which I include

• Joseph Astrahan: Where do you call the test function?
• Joseph Astrahan: Also the function does not have return statement... at least for the JSON encode
• Manos Kounelakis: The function does have a return statement

Answer 1

When you moved your query logic into a function, $con, MySQL's connection object is not available. Use GLOBAL $con; inside your function.

Read this to understand Variable Scope

Method 1 Using GLOBAL keyword

function test(){
 GLOBAL $con;
 $query = "select * from message where seen";
 $ar = [];
 $res = mysqli_query($con,$query);
 while($row = mysqli_fetch_array($res)){
 $ar[] = $row;
 }
 return $ar;
}

Method 2 Pass an argument to a function

function test($con){
 $query = "select * from message where seen";
 $ar = [];
 $res = mysqli_query($con,$query);
 while($row = mysqli_fetch_array($res)){
 $ar[] = $row;
 }
 return $ar;
}
Call it like this:
test($con);

• Manos Kounelakis: $con comes from another php file I will update my post
• Samir Selia: That doesn't matter. You can fix it with solution i provided. Your query is failing due to no connection found.
• Manos Kounelakis: I can accept this answer in 5 minutes.However I cannot understand why isn't $con available inside my function
• Samir Selia: Function has its own scope for variables php.net/manual/en/language.variables.scope.php

Answer 2

Global variables if not used carefully can make problems harder to find as other solution suggested. Pass $con as argument to your function:

function test($con){
 $query = "select * from message where seen";
 $ar = [];
 $res = mysqli_query($con,$query);
 while($row = mysqli_fetch_array($res)){
 $ar[] = $row;
 }
 return $ar;
 }

Answer 3

Let us talk about the problems you have:

1. You pass a failure function to $.ajax. You surely wanted to use error instead of failure. See here.

2. You have a messages variable initialized, but unused inside success. Get rid of it.

3. You check for isset($_POST), but that will always be true. You wanted to check isset($_POST["action"]) instead, or you wanted to check whether it is a POST request.

4. $con is not available inside the function. You will need to either initialize it inside the function or pass it to it and use it as a parameter.