CODEWITHSUNDEEP
phpmysqli
age saputra
age saputra

Reputation: 51

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, object given in

I've got a little bit of a problem when I try to run this code. I receive this error:

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, object given in

My problem is in the while statement on line 32:

$connection = mysqli_connect($host, $username, $password, $db_name);
$sql = 'SELECT * FROM provinsi';
while ($r = mysqli_fetch_array($connection, $sql)) { // line 32
    $id = $r['id'];
}

Upvotes: 3

Views: 99228

Answers (3)

Sammy
Sammy

Reputation: 59

$sql = "select * from privinsi";
$result = mysqli_query($connection,$sql);
if (!$result) {
    printf("Error: %s\n", mysqli_error($connection));
    exit();
}else{
   while($row = mysqli_fetch_array($result)){
    // your code here
   }
}

This helps you to display your error so you would be able to find out the exact reason a query is not working. This is helpful when you are using conditions or trying to find specific columns in a table. Tired eyes can make you skip the fact that a particular column is not on your current database and other factors that you can casually overlook when writing a query. If your error is not detailed, you would inevitably run around in circles looking for a solution that is not beyond you.

Upvotes: -1

Davinder Kumar
Davinder Kumar

Reputation: 662

You wrote SQL query but you didn't execute it.

$sql = "select * from privinsi";
$results = mysqli_query($connection, $sql);
while($r = mysqli_fetch_array($results){
   //enter your code here
}

Upvotes: 0

Lorence Hernandez
Lorence Hernandez

Reputation: 1189

mysqli_fetch_array()'s 1st parameter must be a result of a query. What you are doing is you are passing the connection (which doesn't make sense) and the query command itself.

Read the doc here: http://php.net/manual/en/mysqli-result.fetch-array.php

To fix this, execute the query first, then store the result to a variable then later fetch that variable.

$sql = "select * from privinsi";
$result = mysqli_query($connection,$sql);
while($r = mysqli_fetch_array($result))
{
    // your code here
}

Upvotes: 8

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