将$result更改为MySQLi或PDO PHP


Change $result to MySQLi or PDO PHP

当我使用此代码时。上面写着:

数据库访问失败:未选择数据库"

它还提供了:

已弃用:mysql_query():mysql扩展已弃用,将将来删除:使用mysqli或PDO代替

错误。以下是我使用过的代码。

$con = mysqli_connect("localhost", "root", "","radian")
    or die ("Couldn't connect to  mySQL");  

$query = "SELECT Staff_id,Fname,Sname,Gender,username FROM staff  ";
$result = mysql_query($query);
if (!$result) die ("Database access failed: " . mysql_error());
$rows = mysql_num_rows($result);
for ($j = 0 ; $j < $rows ; ++$j)

{
    $row = mysql_fetch_array($result);
    $staff_id = $row['Staff_id'];
    $Fname = $row['Fname'];
    $Sname = $row['Sname'];
    $Gender = $row['Gender'];
    $username = $row['username'];
    echo '<tr> <td>'.$staff_id.'</td> <td>'.$Fname.'</td> <td>'.$Sname.'</td> <td>'.$Gender.'</td> <td>'.$username.'</td> </tr>';
}

它说错误来自这一行:

$result = mysql_query($query);

您使用mysql来代替mysql,并且mysqli需要与查询连接。

只需用以下代码替换您的代码:

$con = mysqli_connect("localhost", "root", "","radian")
    or die ("Couldn't connect to  mySQL");  

$query = "SELECT Staff_id,Fname,Sname,Gender,username FROM staff  ";
$result = mysqli_query($con, $query);
if (!$result) die ("Database access failed: " . mysqli_error($con));
$rows = mysqli_num_rows($result);
for ($j = 0 ; $j < $rows ; ++$j)

{
    $row = mysqli_fetch_array($result);
    $staff_id = $row['Staff_id'];
    $Fname = $row['Fname'];
    $Sname = $row['Sname'];
    $Gender = $row['Gender'];
    $username = $row['username'];
    echo '<tr> <td>'.$staff_id.'</td> <td>'.$Fname.'</td> <td>'.$Sname.'</td> <td>'.$Gender.'</td> <td>'.$username.'</td> </tr>';
}

您混合了mysql/mysqli。

$con = mysqli_connect("localhost", "root", "","radian");
if(!$con)
{
  exit("Couldn't connect: ".mysqli_connect_error());
}
mysqli_set_charset($con, "utf8");
echo "<table>";
echo "<tbody>";
$query = "SELECT Staff_id,Fname,Sname,Gender,username FROM staff";
$result = mysqli_query($con, $query);
while ($row = mysqli_fetch_assoc($result))
{
    $staff_id = $row['Staff_id'];
    $Fname = $row['Fname'];
    $Sname = $row['Sname'];
    $Gender = $row['Gender'];
    $username = $row['username'];
    echo '<tr><td>'.$staff_id.'</td><td>'.$Fname.'</td><td>'.$Sname.'</td><td>'.$Gender.'</td><td>'.$username.'</td></tr>';
}
echo "</tbody>";
echo "</table>";

您正在使用mysqli建立数据库连接

$con = mysqli_connect("localhost", "root", "","radian")
or die ("Couldn't connect to  mySQL");

当您在查询中使用mysql时,您需要在您的查询中使用mysqli,而不是使用mysql

$query = "SELECT Staff_id,Fname,Sname,Gender,username FROM staff  ";
 $result = mysqli_query($query);
 if (!$result) die ("Database access failed: " . mysql_error());
  $rows = mysqli_num_rows($result);
 for ($j = 0 ; $j < $rows ; ++$j)
 {
$row = mysqli_fetch_array($result);
$staff_id = $row['Staff_id'];
$Fname = $row['Fname'];
$Sname = $row['Sname'];
$Gender = $row['Gender'];
$username = $row['username'];
echo '<tr> <td>'.$staff_id.'</td> <td>'.$Fname.'</td> <td>'.$Sname.'</td>    <td>'.$Gender.'</td> <td>'.$username.'</td> </tr>';
}