Unable to view JSON data from PHP file

Question

I am using MySQL and created 2 tables in it; Users and Activity. User table has following data:

For now I am using the above data only. I want this data to show in JSON format and then I will use it in my Android app. I have tried to convert it but I am getting the below result:

( ! ) Warning: mysqli_query() expects at least 2 parameters, 1 given in D:\xampp\htdocs\MobileApp\index.php on line 18 Call Stack #TimeMemoryFunctionLocation 10.0015134480{main}( )...\index.php:0 20.0034141920getUserName( )...\index.php:38 30.0034142272http://www.php.net/function.mysqli-query' target='_new'>mysqli_query ( )...\index.php:18

( ! ) Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in D:\xampp\htdocs\MobileApp\index.php on line 20 Call Stack #TimeMemoryFunctionLocation 10.0015134480{main}( )...\index.php:0 20.0034141920getUserName( )...\index.php:38 30.0050142240http://www.php.net/function.mysqli-fetch-array' target='_new'>mysqli_fetch_array ( )...\index.php:20

{"users":[]}

I have posted the above result in 3 steps so it can be clearly identified what mistake I am making.

Below is my PHP file:

require_once ('config.php');

function getUserName()
{
// array for json response
$response = array();
$response["users"] = array();

// Mysql select query
$result = mysqli_query("SELECT * FROM Users");

while ($row = mysqli_fetch_array($result))
{
    // temporary array to create single category
    $tmp = array();
    $tmp["Id"] = $row["Id"];
    $tmp["Name"] = $row["Name"];

    // push category to final json array
    array_push($response["Users"], $tmp);
}
// keeping response header to json
header('Content-Type: application/json');

// echoing json result
echo json_encode($response);
}

getUserName();

And my config file:

<?php 
 define("HOST","localhost");
 define("DATABASE","app");
 define("USERNAME","root");
 define("PASSWORD","");

 $con=mysqli_connect(HOST,USERNAME,PASSWORD,DATABASE);
 if(!$con){
    die("Database Connection Error: " . mysqli_connect_error());
 }
 else{
  echo "Connection successful";
 }

I am getting ‘Connection successful’ from config.

I don't know what wrong I am doing.

JazZ · Accepted Answer · 2017-02-14 06:54:06Z

1

The problem comes from the way you use mysqli_query.

As said in error message mysqli_query expects two parameters. The first one will be the connection object and the second one, the query.

i.e. :

$result = mysqli_query($con, "SELECT * FROM Users");

And then, build you json data like this with json_encode() php function.

require_once ('config.php');

function getUserName()
{
    // defines global since $con not in the scope of the function variables
    global $con;

    $response = array();
    while ($row = mysqli_fetch_array($result))
    {
        // temporary array to create single category
        $tmp = array();
        $tmp["Id"] = $row["Id"];
        $tmp["Name"] = $row["Name"];

        // build response array
        $response['users'][] = $tmp;
    }

    // convert $response array to json and return it
    return json_encode($response);
}

// get function return 
$users_name = getUserName();

Hope it helps.

edited Feb 14, 2017 at 6:54

answered Feb 14, 2017 at 5:59

JazZ

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9 Comments

Moeez Over a year ago

While adding $con in the $result = mysqli_query($con, "SELECT * FROM Users"); it's giving me error unidentified variable con

JazZ Over a year ago

Are you sure you included the right connection file ?

Moeez Over a year ago

I am using a function so is there something to do with it?

JazZ Over a year ago

What if you try to access config.php directly ? Do you see the "Connection successful" string or a mysqli_connection error ?

Moeez Over a year ago

Working like a champ :)

|

Dhruv Tyagi · Accepted Answer · 2017-02-14 06:41:00Z

0

Add this line After sql Query

$sql="insert sql query";

$res=mysqli_query($con,$sql);

Add in while loop use this one

$row=mysqli_fetch_array($res)

updated

    <?php
require_once ('config.php');

$result= "SELECT * FROM Users";
$res=mysqli_query($con,$result);

$response = array();

while ($row=mysqli_fetch_array($res)){
        array_push($response,array('Id' =>$row['Id'] ,
    'Name' =>$row['Name']

        //      add element on your array

    ));
    }
    echo json_encode(($response));

    mysqli_close($con);

?>

Put config and php file in the same folder

edited Feb 14, 2017 at 6:41

answered Feb 14, 2017 at 5:55

Dhruv Tyagi

8121 gold badge9 silver badges28 bronze badges

6 Comments

Moeez Over a year ago

While adding $con in the $result = mysqli_query($con, "SELECT * FROM Users"); it's giving me error unidentified variable con

Dhruv Tyagi Over a year ago

add this in your config file: $con = mysqli_connect(HOST,USERNAME,PASSWORD,DATABASE)or die('Unable to Connect'); and remove if else loop

Moeez Over a year ago

Do i have to exclude my function ?

Dhruv Tyagi Over a year ago

there is no need to add your function.. but if u want u can

Moeez Over a year ago

The problem is still there as i can find $con in my intellesence

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