php & mysql 数据库拉取错误

Question

您好，我有一个网站，我正在使用 php 进行编码。我有一个附加的数据库。我试图让它从数据库中将数据读入页面。我将数据库分成几个表。我以为我已经正确完成了代码，但我不断收到以下错误代码：

警告：mysqli_query() 期望参数 1 为 mysqli，整数在第 13 行的 C:\xampp\htdocs\stadium\alpha\a.php 中给出

警告：mysqli_fetch_all() 期望参数 1 为 mysqli_result，在第 14 行的 C:\xampp\htdocs\stadium\alpha\a.php 中给出 null

注意：尝试在第 15 行获取 C:\xampp\htdocs\stadium\alpha\a.php 中非对象的属性

警告：mysqli_fetch_all() 期望参数 1 为 mysqli_result，在第 16 行的 C:\xampp\htdocs\stadium\alpha\a.php 中给出 null

我已经尝试对此进行研究一段时间了，但我无法对此做出正面或反面。我的数据拉取和显示代码如下：

$sql = "SELECT colConferenceName.tblConference, colSchoolName.tblSchool, colClass.tblSchool, colSurfaceName.tblSurface, colSurfaceCompany.tblSurface, colStadiumName.tblStadium, colAddress.tblStadium, colCity.tblStadium, colRegion.tblStadium, colCounty.tblStadium, colCapacity.tblStadium, colSurfaceYear.tblSurface FROM tblStadium, tblConference, tblSurface, tblSchool WHERE tblConference.colConferenceID = tblSchool.colConferenceID AND tblSurface.colSurfaceID = tblStadium.colSurfaceID AND tblStadium.colStadiumID = tblStadiumSchool.colStadiumID AND tblSchool.colSchoolID = tblStadiumSchool.colSchoolID AND colStadiumName LIKE 'A%' ORDER BY colSchoolName";
$schoolinfo = mysqli_query($conn,$sql);
mysqli_fetch_all($schoolinfo,MYSQLI_ASSOC);
while (!$schoolinfo->EOF){ //looping through the recordset (until End Of File) 
while ($row = mysqli_fetch_all($schoolinfo,MYSQLI_ASSOC)){
    echo '<p>School Name: ' . $row['colSchoolName'] . '</br>Conference: ' . $row['colConferenceName'] . '</br>Class: ' . $row['colClass'] . '</br>Stadium Name: ' . $row['colStadiumName'] . '</br>Address: ' . $row['colAddress'] . '</br>City: ' . $row['colCity'] . '</br>County: ' . $row['colCounty'] . '</br>Region: ' . $row['colRegion'] . '</br>Capacity: ' . $row['colCapacity'] . '</br>Surface Type: ' . $row['colSurfaceName'] . '</br>Surface Company: ' . $row['colSurfaceCompany'] . '</br>Year Installed: ' . $row['colSurfaceYear'] . '</p>';
        }
    }

你能为我提供的任何帮助都会很棒。提前致谢。

至于连接，它是在一个单独的连接文件中完成的。它拉取的方式是 $conn = require ('xxx_xxx.xxx');我原则上询问输出以及如何修复编码错误。我 xxx 断开连接，因为我不希望人们访问文件名。我还为此进行了大约 80 到 100 个连接，因此使用一个文件进行连接比必须编辑每个 php 文件的连接部分更有意义。另外，我发现这样可以提高安全性。

Answer 1

我没有看到您连接到 mysql 的时间点。试试mysqli_connect

$conn = mysqli_connect('<your-host>','<your-username>','<your-password>','<your-databasename>').

Answer 2

需要先连接mysql服务器：

$con = mysqli_connect($host,$username,$password);

那么您需要选择您将使用的数据库： $db = mysqli_db_select($database);

您还必须设置正确的参数：$host、$username、$password 和 $database，并且必须首先配置您的 mysql 服务器以接受连接

好吧，你为什么不尝试一个 php 框架？

Answer 3

你可以试试这个：

记得在脚本的开头设置你的数据库登录

<?php

/* connect*/
$mysqli = new mysqli("localhost", "my_user", "my_password", "db_name");

/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

/* your sql query */
$query = "SELECT colConferenceName.tblConference, colSchoolName.tblSchool, ".
         "colClass.tblSchool, colSurfaceName.tblSurface, colSurfaceCompany.tblSurface, ".
         "colStadiumName.tblStadium, colAddress.tblStadium, colCity.tblStadium, ".
         "colRegion.tblStadium, colCounty.tblStadium, colCapacity.tblStadium, ".
         "colSurfaceYear.tblSurface FROM tblStadium, tblConference, tblSurface, tblSchool ".
         "WHERE tblConference.colConferenceID = tblSchool.colConferenceID AND ".
         "tblSurface.colSurfaceID = tblStadium.colSurfaceID AND ".
         "tblStadium.colStadiumID = tblStadiumSchool.colStadiumID AND ".
         "tblSchool.colSchoolID = tblStadiumSchool.colSchoolID AND ".
         "colStadiumName LIKE 'A%' ORDER BY colSchoolName";

/* query your db */
if ($result = $mysqli->query($query)) {

    /* fetch associative array */
    while ($row = $result->fetch_assoc())
    {
        /* display row */
        echo '<p>School Name: ' . 
        $row['colSchoolName'] . '</br>Conference: ' . 
        $row['colConferenceName'] . '</br>Class: ' . 
        $row['colClass'] . '</br>Stadium Name: ' . 
        $row['colStadiumName'] . '</br>Address: ' . 
        $row['colAddress'] . '</br>City: ' . 
        $row['colCity'] . '</br>County: ' . 
        $row['colCounty'] . '</br>Region: ' . 
        $row['colRegion'] . '</br>Capacity: ' . 
        $row['colCapacity'] . '</br>Surface Type: ' . 
        $row['colSurfaceName'] . '</br>Surface Company: ' . 
        $row['colSurfaceCompany'] . '</br>Year Installed: ' . 
        $row['colSurfaceYear'] . '</p>';
    }

    /* free result set */
    $result->free();
}

/* close connection */
$mysqli->close();

?>