2016-07-22 103 views
0

我收到此错误:警告:mysql_select_db()预计参数1为字符串,

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"); 

?> 
+1

你混合'mysql_'和'mysqli_'功能。上面改为'mysqli_select_db($ con,'staff');' – Cyclonecode

+0

另外:它应该是'mysql_select_db('staff',$ con)' – Arnauld

+0

@Arnauld - 他不应该**使用'mysql_select_db()'at因为他使用'mysqli',所以你不应该同时使用两个扩展。 – Cyclonecode

回答

1

您与mysqli_功能混合mysql_

变化if (!mysql_select_db($con,'staff'))if (!mysqli_select_db($con,'staff'))

+0

由于某种原因,现在当我插入数据时,错误不再出现,但它显示“未插入”。这里可能是什么问题? @The Codesee –

+0

'echo'未插入';'将'echo(“Error description:”。mysqli_error($ con));'然后让我知道它显示的错误 –

0

编写代码象下面这样:

<?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'; 
} 
0

写你这样的代码

<?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'; 
} 
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