Question

当我使用此代码时。它说：

数据库访问失败：未选择数据库＆＃34;

它还给出了：

不推荐使用：mysql_query（）：不推荐使用mysql扩展名将来删除：使用mysqli或PDO代替

错误。以下是我使用的代码。

$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);

Answer 1

您使用 mysql 代替 mysqli ，而mysqli需要与查询建立连接。

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>';
}

Answer 2

你混合了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>";

Answer 3

您正在使用 mysqli

构建数据库连接

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

在您的查询中，您使用 mysql ，您需要在查询中使用 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>';
}

将$ result更改为MySQLi或PDO PHP

3 个答案: