php和mysql数据库拉错误

问题描述：

你好，我有一个网站，我在PHP编码。我有一个数据库附加到它。我试图从数据库中读取数据到页面中。我把数据库分成几个表格。我以为我有代码得当，但我一直在得到以下的错误代码：php和mysql数据库拉错误

Warning: mysqli_query() expects parameter 1 to be mysqli, integer given in C:\xampp\htdocs\stadium\alpha\a.php on line 13

Warning: mysqli_fetch_all() expects parameter 1 to be mysqli_result, null given in C:\xampp\htdocs\stadium\alpha\a.php on line 14

Notice: Trying to get property of non-object in C:\xampp\htdocs\stadium\alpha\a.php on line 15

Warning: mysqli_fetch_all() expects parameter 1 to be mysqli_result, null given in C:\xampp\htdocs\stadium\alpha\a.php on line 16

我一直在努力，现在研究这个了一会儿，我不能让这个正面或反面。我的代码为数据拉和显示如下：

$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>'; 
     } 
    }

什么永远不会帮助你可以提供我会很棒。提前致谢。

至于连接它是在一个单独的连接文件中完成。它的方式是$ conn = require（'xxx_xxx.xxx'）;我原则上询问了输出以及如何解决编码错误。由于我不希望人们有权访问文件名，因此我将xxx连接出去。我也做了大约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。另外我发现它可以有更多的安全性。

你在哪里设置你的'$ conn' – CodeBird

$ conn = mysqli_conect（）..... –

**** connect **** mysqli_connect double n – CodeBird

答

你可以试试这个：

记得在脚本

<?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(); 

?>

嗯，我试过了，但它给了我错误，注意：未定义变量：mysqli在C：\ xampp \ htdocs \ stadium \ alpha \ a.php上线29 致命错误：调用成员函数query（）on在第29行的C：\ xampp \ htdocs \ stadium \ alpha \ a.php中有一个非对象。我应该提到，我有一个连接的单独文件，因为我在大约80到100个独立的php中连接到数据库页面。 – mdhof

答

我没有看到你连接到mysql的时间点。尝试mysqli_connect

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

谢谢，但连接不是问题。我不想显示所有背景数据，原因很明显。 – mdhof

答

您需要连接到MySQL服务器第一的开始设置你的数据库登录：

$ con = mysqli_connect（$ host，$ username，$ password）;那么你需要选择你要使用的数据库： $ db = mysqli_db_select（$ database）;

你也必须设置正确paramenters：$主机，$的用户名，$密码和$数据库和MySQL服务器都必须首先配置为接受连接

很好，你为什么不尝试一下PHP框架？

php和mysql数据库拉错误

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