php & mysql数据库拉错

Question

你好，我有一个网站，我在php中编码。我有一个数据库附在上面。我试图让它从数据库中读取数据到页面中。我把数据库分成几个表。我认为我已经正确地完成了代码，但是我一直在获取以下错误代码：

警告: mysqli_query()期望参数1是mysqli，在第13行的C:\xampp\htdocs\stadium\alpha\a.php中给出的整数 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中为null 注意:尝试在第15行的C:\xampp\htdocs\stadium\alpha\a.php中获取非对象的属性。 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中为null

我试着研究这个问题已经有一段时间了，我就是不能把这件事搞清楚。我的数据拉出和显示代码如下：

$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>';
        }
    }

如果你能帮我的话就太棒了。提前谢谢。

至于连接，它是在一个独立的连接文件中完成的。它的方式是$conn = require ('xxx_xxx.xxx')；我的原则是询问输出和如何修复编码错误。我退出连接，因为我不希望人们访问文件名。我还使用它进行了大约80到100个连接，因此使用一个文件进行连接比编辑每个php文件的连接部分更有意义。另外，我发现这样它可以有更多的安全性。

Answer 1

你能试试这个吗？

记住在脚本开始时设置db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();

?>

Answer 2

我没有看到你连接到mysql的时间点。试试连接

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

Answer 3

您需要首先连接到mysql服务器：

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

然后您需要选择要使用的数据库：$db = mysqli_db_select($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mysql服务器以接受连接。

那么，为什么不尝试使用php框架呢？