jQuery Ajax - $sql is an object Error

Question

UPDATE at bottom of question

I'm getting the error:

Warning: mysqli_query() expects parameter 2 to be string, object given

Questions about this are incredibly common on Stack Overflow - my apologies in advance. I haven't been able to find a good answer for my specific problem. If there is a thread that addresses this, please let me know.

Here is my Ajax code:

    $.ajax({
        url: "get.php",
        type: "post",
        datatype: "json",
        data:{ ajaxid: altCheck }, //this is an integer passed to MySQL statement
        success: function(response){
            console.log(response);
        },
        error: function(){
            console.log("test");
        }
    });

get.php

<?php

$db = mysqli_connect("...", "...", "...", "...");

$value = filter_var($_REQUEST["ajaxid"], FILTER_SANITIZE_STRING);
$value = mysqli_real_escape_string($db, $value);
var_dump($value); //checking to see what $value is at this point

$sql = $db->prepare("SELECT * FROM table WHERE screeningId = ?");
$sql->bind_param("s",$value);


//THIS LINE THROWS THE ERROR
$result = mysqli_query($db, $sql);
$temp = array();
while ($row = mysqli_fetch_array($result)){
    //output data
    array_push($temp,$row['imageURL']);
    }
echo json_encode($temp);
?>

The fourth line of code var_dump($value); outputs string(0).

UPDATE: MySQLi

<?php

$db = mysqli_connect("...", "...", "...", "...");

$value = filter_var($_REQUEST["ajaxid"], FILTER_SANITIZE_STRING);
$value = mysqli_real_escape_string($db, $value);

$query = $db->prepare('SELECT * FROM table WHERE screeningId = ?');
$query->bind_param('s', $_GET[$value]);
$query->execute();

if ($result = mysqli_query($db, $query)) {
    while ($url = mysqli_fetch_object($result, 'imageURL')) {
        echo $url->info()."\n";
    }
}

?>

Screenshot of MySQL table data columns:

Well again I don't do PHP, but I would expect a prepare() function in any language to return an object. You are using it as an object yourself when you call the $sql->bind_param() method on the next line. — nnnnnn
– nnnnnn, Commented May 30, 2017 at 3:40
So what do you want to use mysqli or PDO ? HERE is a comparison article between the two. — Louys Patrice Bessette
– Louys Patrice Bessette, Commented May 30, 2017 at 3:51
You can use both in one site... But not both in the same request to DB ;) It is easier to choose one and get used to it. — Louys Patrice Bessette
– Louys Patrice Bessette, Commented May 30, 2017 at 4:01

Louys Patrice Bessette · Accepted Answer · 2017-05-30 06:32:54Z

1

EDIT

Okay... 8 edits spent on mysqli... Enought!
Here is how I DO using PDO. And it WILL work first shot.
I have a separate file for the database connection info.

dbconnection.php:
(The advantage of the separate definition file is one place to update the user password when needed.)

<?php
// Database connection infos (PDO).
$dsn = 'mysql:dbname=[DATABASE_NAME];host=127.0.0.1';
$user = '[DATABASE_USER]';
$password = '[USER_PASSWORD]';


try {
  $dbh = new PDO($dsn, $user, $password);
} catch (PDOException $e) {
  echo 'Connexion failed : ' . $e->getMessage();
}
?>

Now in your PHP files where a database request has to be done, include the PDO definition file, the just request what you want:

<?php
include('dbconnection.php');

// JUST TO DEBUG!!!
$_REQUEST['ajaxid'] = "1";

// Database request.
$stmt = $dbh->prepare("SELECT * FROM table WHERE screeningId = ?");
$stmt->bindParam(1, $_REQUEST['ajaxid']);
$stmt->execute();

if (!$stmt) {
   echo "\nPDO::errorInfo():\n";
   print_r($dbh->errorInfo());
   die;
}

// Looping through the results.
$result_array =[];
while($row=$stmt->fetch()){
  array_push($result_array,$row['imageURL']);
}

// The result array json encoded.
echo json_encode($result_array);
?>

edited May 30, 2017 at 6:32

answered May 30, 2017 at 4:43

Louys Patrice Bessette

34k7 gold badges41 silver badges70 bronze badges

Sign up to request clarification or add additional context in comments.

37 Comments

Louys Patrice Bessette Over a year ago

lolll My bad... json.encode($arrayTojson) sould be json_encode($arrayTojson)... Arrgg... so many similar syntax...

Louys Patrice Bessette Over a year ago

Let's give a try with what I got working for years... ;) PDO !!

Louys Patrice Bessette Over a year ago

Right!... lolll var is some JavaScript deformation in my brain. remove var in front of $result_array. I edited.

Louys Patrice Bessette Over a year ago

I added a "JUST TO DEBUG" line... Now accesse get.php directly... Like http://your-domain/get.php

Louys Patrice Bessette Over a year ago

What you don't know is always harder then what you know. ;) Don't feel cheap, it is the same for everyone. ;)

|

Naga · Accepted Answer · 2017-05-30 04:49:27Z

1

Since you are using mysqli_* all other place in your project, update your get.php as below.

<?php
$db = mysqli_connect("...", "...", "...", "...");

$value = filter_var($_REQUEST["ajaxid"], FILTER_SANITIZE_STRING);
$value = mysqli_real_escape_string($db, $value);
//var_dump($value); //checking to see what $value is at this point

$sql = "SELECT * FROM table WHERE screeningId = '$value'";

$result = mysqli_query($db, $sql);
$temp = array();
while ($row = mysqli_fetch_array($result)){
    //output data
    array_push($temp,$row['imageURL']);
    }
echo json_encode($temp);

EDIT

With respect to bind param with mysqli,

<?php
$conn = new mysqli('db_server', 'db_user', 'db_passwd', 'db_name');


$sql = 'SELECT * FROM table WHERE screeningId = ?';
$stmt = $conn->prepare($sql);
$value = filter_var($_REQUEST["ajaxid"], FILTER_SANITIZE_STRING);
$stmt->bind_param('s', $value);
$stmt->execute();
$res = $stmt->get_result();
$temp = array();
while($row = $res->fetch_array(MYSQLI_ASSOC)) {
  array_push($temp,$row['imageURL']);
}
echo json_encode($temp);

edited May 30, 2017 at 4:49

answered May 30, 2017 at 4:03

Naga

2,1683 gold badges18 silver badges21 bronze badges

21 Comments

chris85 Over a year ago

The OP should keep the parameterized query.

Louys Patrice Bessette Over a year ago

No. It means you should keep this line: $sql->bind_param("s",$value); and keep the ? in the $sql string instead of $value.

Louys Patrice Bessette Over a year ago

Look at the "Security" example in the comparison article I gave you. ==> "mysqli, prepared statements"

chris85 Over a year ago

@coffeebot PDO and mysqli both support parameterized queries, you need to use the manual for whichever driver you are using. php.net/manual/en/mysqli.quickstart.prepared-statements.php

Naga Over a year ago

@coffeebot: use the edited code. It should work. If not please let me know.

|

Ali Hesari · Accepted Answer · 2017-05-30 04:33:18Z

0

Select Data With PDO in get.php:

<?php 

    if( isset($_POST['ajaxid']) ) {
      $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
      $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
      $stmt = $conn->prepare("SELECT * FROM table WHERE screeningId = :screeningId"); 
      $stmt->execute(array(':screeningId' => $_POST['ajaxid']));
      $row = $stmt->fetch();
    }
?>

You configure PDO to throw exceptions upon error. You would then get a PDOException if any of the queries fail - No need to check explicitly. To turn on exceptions, call this just after you've created the $conn object:

$stmt->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

answered May 30, 2017 at 4:33

Ali Hesari

1,9595 gold badges28 silver badges56 bronze badges

Collectives™ on Stack Overflow

jQuery Ajax - $sql is an object Error

3 Answers 3

37 Comments

21 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

37 Comments

21 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related