MySQL使用Ajax返回表名而不是值

Question

I'm trying to make an AJAX request to fetch some data from the database but the only thing returned is the name of the column. Can somebody explain why? Here is my code:

=> Table Name :- tblstudents

id = 0
fname = john
lname = doe
tel = 555-564-1585

id = 1 
fname = paul
lname = smith
tel = 555-134-5644

id = 2
fname = laura
lname = mcdo
tel = 555-465-7512

=> AJAX method:

function fetchFromDBPHP(column, fname, id, tel) {
    $.ajax({
        type: "post",
        url: "./php/fetchFromDB.php",
        dataType: 'json',
        data: { column: column, fname: fname, id: id },
        success: function(data) {
            localStorage.setItem(tel, data);
        },
        error:function(request, status, error) {
            console.log("** Error from fetchFromDBPHP **");
            console.log("Error: " + error + "\nMessage: " + request.responseText);
        }
    });
}

=> Javascript :

fetchFromDBPHP(column, fname, id, "one");
var result = localStorage.getItem("one");
console.log("Result: " + result);

=> PHP :

<?php
    $column = $_POST['column'];
    $fname = $_POST['fname'];
    $id = $_POST['id'];

    if (isset($column)) {
        $sql = "SELECT '$column' FROM tblstudents WHERE fname = '$fname' AND id = '" . intval($id) . "'";
        $con = mysqli_connect("localhost", "root", "", "test");
        if (!$con) {
            die("Connection failed: " . mysqli_error($con));
        }
        $result = mysqli_query($con, $sql);
        $to_encode = array();
        while($row = mysqli_fetch_array($result, MYSQLI_NUM)) {
            $to_encode[] = $row;
        }
        echo json_encode($to_encode);
        mysqli_close($con);
    }
?>

As you might know, column , fname and id have values in the Javascript code. As my database is way longer, I tried to be as much close as possible to my real code. The only thing is, the result of the AJAX request gives me a JSON object result containing the name of the column, and not its content. Anybody can help? Thanks in advance :)

Answer 1

You're using the wrong quotes in your query.

"SELECT '$column' FROM ..."

...will simply return the value of the variable $column instead of the value of the actual database column.

Changing it to (back ticks):

"SELECT `$column` FROM ..."

will work.

An important note...

...the posted code is wide open to SQL Injections and should use parameterized Prepared Statements instead of concatenation of the variables in the query. Specially since the user inputs aren't escaped at all.

Rule of thumb, never ever trust user inputs.

Regarding the column name, which you can't parameterize, you should create a white list with allowed column names.