/PHP/Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given in

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

Warning: mysqli_fetch_all() expects parameter 1 to be mysqli_result, bool given in C:\newxammp\htdocs\learn\index.php on line 11

#where you are running mysqli_query , add 'or die( mysqli_error($db)'
#e.g
$sql = "SELECT * FROM users";
$result = mysqli_query($db, $sql) or die( mysqli_error($db));
#$db being the variable holding the connection to db

Source: codeproject.com

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

mysqli_fetch_array($result)

Source: stackoverflow.com

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

 mysqli_fetch_array($query)

Source: stackoverflow.com

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

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.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}

Source: hrefcode.com

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

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

-1

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