How to pass resource from php to javascript

Question

So I have three different separate files:

functions.php (all functions for the database)
main.html (my main program)
main.js (all javascript functions)

Now, I want to call a function in PHP through AJAX. To do that, I need to pass $conn.

$conn = sqlsrv_connect($serverName, $connectionInfo);

It's a resource, so I can't use json_encode.

The way I set the everything up now is that the php-file is required in the html so I can use the functions and when I change the value of a dropdown, the js is called.

How can I pass the $conn variable to Javascript?

Regards

how can I call $conn if this variable is in a different file than the ajax? — Taccoman
– Taccoman, Commented Aug 30, 2018 at 10:42
or better yet - have a class for db handling and call that class (file include still necessary unless you have namespaces etc.) — treyBake
– treyBake, Commented Aug 30, 2018 at 10:45

Michael Curry · Accepted Answer · 2018-08-30 11:10:34Z

2

It doesn't work like that.

You should never be directly making calls to the database from the front-end.

Think of it as three separate levels. Your HTML/JS is the front-end, your PHP is your server, and your Database is on its own level.

So when the user does something on the front-end, say changes the value of a field and you want to update that in the database the following actions should happen:

Event triggers on JS
AJAX is called as a result of the event being triggered
PHP server receives the AJAX request and executes code to modify database
(optional) PHP server sends something back to the front-end to tell it that the request was successful

Read up on the concept of MVC: https://developer.mozilla.org/en-US/docs/Web/Apps/Fundamentals/Modern_web_app_architecture/MVC_architecture

edited Aug 30, 2018 at 11:10

answered Aug 30, 2018 at 11:04

Michael Curry

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Comments

Zain Farooq · Accepted Answer · 2018-08-30 10:44:01Z

0

Try this in php code as I assume functions.php

$conn = sqlsrv_connect($serverName, $connectionInfo);
echo $conn;//Don't try echo anything other

In Javascript

$.ajax({
           type: "POST",
           url: "functions.php",

           success: function(data)
           {
               var conn = data; // here is your conn which comes from php file
           }
      });

answered Aug 30, 2018 at 10:44

Zain Farooq

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Comments

user8033142 · Accepted Answer · 2018-08-30 11:02:20Z

0

First of all include jquery latest version from cdn

Create an API Url, and use POST method

site.com/api/insert.php // to insert into table

Use $.post() api of jquery to send data

var url = ""; // enter your URL HERE
var postData = {};  // object of post data with table name, cols and values
$.post(url, postData, function(data, status) {
       // do what ever you want with data        
})

ps: you can also create diff insertion / selection / update / delete api for different table. (recommended)

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