“WARNUNG: Mysqli_fetch_array () erwartet Parameter 1 mySQLi_Result, Zeichenfolge in” Code-Antworten

WARNUNG: MySQLI_Fetch_all () erwartet Parameter 1 mySQLi_Result, bool in C: newxammp htdocs Learn index.php in Zeile 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
Nagraj!

WARNUNG: Mysqli_fetch_array () erwartet Parameter 1 mySQLi_Result, Zeichenfolge in

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

WARNUNG: Mysqli_fetch_array () erwartet Parameter 1 mySQLi_Result, Zeichenfolge in

 mysqli_fetch_array($query)
Homeless Hyena

WARNUNG: Mysqli_fetch_array () erwartet Parameter 1 mySQLi_Result, Zeichenfolge in

mysqli_fetch_array($result) 
Homeless Hyena

WARNUNG: Mysqli_fetch_array () erwartet Parameter 1 mySQLi_Result, Zeichenfolge 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
}
Precious Pigeon

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