Sending a JavaScript variable to PHP, use said variable in an SQL statement, then display PHP results

Question

I'm trying to send a JavaScript variable to a PHP script, use the variable in my SQL statement, and then send the results back to JavaScript.

JavaScript:

// Sending a variable to PHP script
var variableToSend = '30';
$.post('myPHPscript.php', {variable: variableToSend});

// Using the variable from earlier, run PHP script to get results
var data_from_ajax;
$.get('http://www.mywebsite.com/myPHPscript.php', function(data) {
data_from_ajax = data;
});

// Display results on the document
var displayResult = document.getElementById('displayResult');
displayResult.innerHTML = data_from_ajax;

PHP:

<?php
// Connection stuff
$conn = new mysqli($servername, $username, $password, $dbname);

// Get the sent variable
$variable = $_POST['variable'];

// Run SQL statement
$sql = "
    SELECT column
    FROM myDatabase 
    WHERE id=$variable
";

$result = $conn->query($sql);

// Send results back to JavaScript 
while($row = $result->fetch_array(MYSQLI_ASSOC)) {
    echo $row["column"]; 
}
?>

The reason that JS is calling a PHP script is because I used an event listener in JS to determine the variable that I want to send. The variableToSend varies every time depending on what was clicked, but I have set it to some random number here.

I'm aware that the problem lies in sending variables to PHP scripts in the JavaScript, but I'm unsure how else I can do this, besides using .post and .get.

I also considered using an AJAX call, but I wasn't sure how that would work.

Thank you for any pointers.

you are already using 2 ajax calls. You should only be making one — Steve
– Steve, Commented Mar 18, 2017 at 22:39

Steve · Accepted Answer · 2017-03-18 22:43:12Z

1

You are already using ajax, twice, via $.post and $.get.

You only need to make the post request, and display the response:

// Sending a variable to PHP script via ajax post, and display the returned results
var variableToSend = '30';
$.post('http://www.mywebsite.com/myPHPscript.php', {variable: variableToSend}, function(responseFromPhp){

    $('#displayResult').html(responsefromPhp);
});

answered Mar 18, 2017 at 22:43

Steve

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1 Comment

Cait Over a year ago

I understand what you mean now by using ajax twice. Thank you!

Krzysztof Duszczyk · Accepted Answer · 2017-03-18 22:44:04Z

1

The data you want will be returned on this post ajax call. To retrieve it specify success function.

$.post( "test.php", { name: "John", time: "2pm" })
    .done(function( data ) {
        alert( "Data Loaded: " + data );
    });

and also use some validation and escaping before use post values in mysql: http://php.net/manual/en/mysqli.real-escape-string.php

answered Mar 18, 2017 at 22:44

Krzysztof Duszczyk

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