getting error " mysqli_query() expects parameter 1 to be mysqli, resource given in "

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Hi friends

I getting error in my php code

mysqli_query() expects parameter 1 to be mysqli, resource given in my php code

Quote:
Warning: mysqli_query() expects parameter 1 to be mysqli, resource given in C:\xampp\htdocs\Sites\freak\SignUp.php on line 16

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in C:\xampp\htdocs\Sites\freak\SignUp.php on line 17

my code is here

Quote:
if(isset($_POST["usernamecheck"])){
$db_conx = mysqli_connect('localhost','root');
$username = preg_replace('#[^a-z0-9]#i','',$_POST['usernamecheck']);
$sql = "SELECT id FROM users WHERE username = '$username'LIMIT 1";
$query = mysqli_query($db_conx,$sql);
$unamecheck = mysqli_num_rows($query);
if(strlen($username)< 3 || strlen($username) > 16)
{
echo'3-16 characters please';
exit();
}

if(is_numeric($username[0]))
{
echo' Usernames must begin with a letter';
exit();
}

if($unamecheck < 1) {

echo ''.$username. 'is OK';
exit();

}
else

{
echo''.$username.'is taken ';
exit();
}

}

?>

Posted 8-Sep-13 23:07pm

Md Jamaluddin Saiyed

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CPallini · Accepted Answer · 2013-09-08T23:27:00

Solution 1

Your connection attempt failed but you didn't check the mysqli_connect call result, as you should have done (see the documentation: "mysqli_query"[^] (see the code sample).

Posted 8-Sep-13 23:27pm

CPallini

Comments

Md Jamaluddin Saiyed 9-Sep-13 5:36am

Ok thanks for solution first error solved but second still comes

CPallini 9-Sep-13 5:38am

What is the 'second error'?

Member 12663316 20-Mar-18 6:04am

<?php
include 'connection.php';
$conn->set_charset('utf8');

?>
<?php

// php code to search data in mysql database and set it in input text

if(isset($_POST['search']))
{
// id to search
$id = $_POST['sino'];

// connect to mysql

// mysql search query
$query = "SELECT `office`, `promoteddate`, `fixeddarbandi` FROM `empjobdetails` WHERE `sino` = $sino LIMIT 1";

$result = mysqli_query($conn, $query);

// if id exist
// show data in inputs
if(mysqli_num_rows($result) > 0)
{
while ($row = mysqli_fetch_array($result))
{
$office = $row['office'];

$promoteddate = $row['promoteddate'];
} $fixeddarbandi = $row['fixeddarbandi'];
}

// if the id not exist
// show a message and clear inputs
else {
echo "Undifined ID";
$office = "";
$promoteddate = "";
$fixeddarbandi = "";
}

mysqli_free_result($result);
mysqli_close($connect);

}

// in the first time inputs are empty
else{
$office = "";
$promoteddate = "";
$fixeddarbandi = "";
}

?>

<!DOCTYPE html>

<title> PHP FIND DATA

<meta charset="UTF-8">

<meta name="viewport" content="width=device-width, initial-scale=1.0">

सि न:

कार्यलय:

बढुवा मिति :

कायम भएको दरबन्दी :

it shows the error like this

Warning: mysqli_query() expects parameter 1 to be mysqli, resource given in C:\xampp\htdocs\pmis\test2.php on line 20

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in C:\xampp\htdocs\pmis\test2.php on line 22
undefined id
Warning: mysqli_free_result() expects parameter 1 to be mysqli_result, null given in C:\xampp\htdocs\pmis\test2.php on line 37

Warning: mysqli_close() expects parameter 1 to be mysqli, resource given in C:\xampp\htdocs\pmis\test2.php on line 38