mysqli的问题。 [英] Problem with mysqli.

问题描述

我想从我的桌子上显示一些记录，但是mysqli部分似乎有问题，这些是不断出现的错误;



警告：mysqli_select_db（）要求参数1为mysqli，在 D：\ xampp \\\\\\\\\\\\\\\\\\\\ > 78



警告：mysqli_query（）期望参数1为mysqli，字符串在 D：\\中给出\\

警告：mysqli_fetch_array（）要求参数1为mysqli_result，在 D> \ xampp \ htdocs \ Project：\\ add_employees.php 90
中给出null />
ID姓名电话号码工作日期失业日期



有人可以解释错误的含义，以及我的意思做错了吗？

I want to display a few records from my table, but there seems to be a problem with the mysqli part, these are the erros that keeps popping up;

Warning: mysqli_select_db() expects parameter 1 to be mysqli, string given in D:\xampp\htdocs\Project\add_employees.php on line 78

Warning: mysqli_query() expects parameter 1 to be mysqli, string given in D:\xampp\htdocs\Project\add_employees.php on line 79

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in D:\xampp\htdocs\Project\add_employees.php on line 90
ID Name Phone Number Date Employed Date Unemployed

Can someone explain what the errors mean, and what I'm doing wrong ?

<?PHP
          $conn = new mysqli("localhost","root","");
          mysqli_select_db("kz_komputer",$conn);
          $data = mysqli_query("SELECT * FROM employees",$conn);
          
          echo "<table>
          <tr>
            <th> ID</th>
            <th> Name</th>
            <th> Phone Number</th>
            <th> Date Employed</th>
            <th> Date Unemployed</th>
          </tr>";
          
          while($record = mysqli_fetch_array($data)){
            echo "<tr>";
            echo "<td>" . $record['id'] . "</td>";
            echo "<td>" . $recod['name'] . "</td>";
            echo "<td>" . $record['phone_number'] . "</td>";
            echo "<td>" . $record['date_employed'] . "</td>";
            echo "<td>" . $record['date_unemployed'] . "</td>";
            echo "</tr>";
          }
          echo "</table>";
          
          mysqli_close($conn);
?>

我尝试过：

What I have tried:

$conn = new mysql("localhost","root","","kz_komputer");
$sql = "SELECT * FROM employees";
$data = mysql_query($sql,$conn);

推荐答案

conn = new mysqli（localhost，根，）;
mysqli_select_db（kz_komputer，

conn = new mysqli("localhost","root",""); mysqli_select_db("kz_komputer",

conn）;

data = mysqli_query（SELECT * FROM employees，

data = mysqli_query("SELECT * FROM employees",

这篇关于mysqli的问题。的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！