Warnung: mysql_select_db() erwartet Parameter 1 Zeichenfolge zu sein,

Question

ich diese Störung erhalte:

Warning: mysql_select_db() expects parameter 1 to be string, object given in C:\xampp\htdocs\apply\insert.php on line 11 Database not selectedNot inserted

<?php 

$con = mysqli_connect('localhost','root',''); 

if (!$con) 
{ 
    echo 'Not connected to server'; 
} 

if (!mysql_select_db($con,'staff')) 
{ 
    echo "Database not selected"; 
} 

$Name = $_POST['username']; 
$Email = $_POST['email']; 

$sql = "INSERT INTO person (Name, Email) VALUES ('$Name','$Email')"; 

if(!mysqli_query($con,$sql)) 
{ 
    echo 'Not inserted'; 
} 
else { 
    echo 'Inserted'; 
} 

header("refresh:2; url=index.html"); 

?> 

Answer 1

Sie mischen mysql_ mit mysqli_ Funktionen.

ändern if (!mysql_select_db($con,'staff'))-if (!mysqli_select_db($con,'staff'))

Answer 2

schreiben Sie den Code unten wie:

<?php 


$con = mysqli_connect('localhost','root',''); 

if (!$con) 
{ 
    echo 'Not connected to server'; 
} 

if (!mysqli_select_db($con,'staff')) 
{ 
    echo "Database not selected"; 
} 

$Name = $_POST['username']; 
$Email = $_POST['email']; 

$sql = "INSERT INTO person (Name, Email) VALUES ('$Name','$Email')"; 

if(!mysqli_query($con,$sql)) 
{ 
    echo 'Not inserted'; 
} 
else { 
    echo 'Inserted'; 
} 

Answer 3

Code wie diese Funktionen

<?php 
$con = mysqli_connect('localhost','root','','staff'); 
if (!$con) 
{ 
echo 'connection not established'; 
} 

$Name = $_POST['username']; 
$Email = $_POST['email']; 

$sql = "INSERT INTO person (Name, Email) VALUES ('$Name','$Email')"; 

if(!mysqli_query($con,$sql)) 

{ 
echo 'Not inserted'; 
} 

else { 
echo 'Inserted'; 
}