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错误：警告：mysqli_fetch_array（）要求参数1为mysqli_result，第15行/var/www/html/data.php中给出布尔值[复制]

分类: 技术问答 • 2022-03-04 19:04:21

错误：警告：mysqli_fetch_array（）要求参数1为mysqli_result，第15行/var/www/html/data.php中给出布尔值[复制]

问题描述：

I am trying to select data from a MySQL table to show data's in my c# datagridview, but I get one of the following error messages:

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /var/www/html/data.php on line 15

This is my PHP code:

<?php 

$servername = "********";
$username = "root";
$password = "pws";
$dbname = "dbs";

$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
$sql = "SELECT userid,password,first_name,last_name,role,active FROM aster_users";
$result = mysql_query($sql);
while($row = mysqli_fetch_array($result)) 
    {
        $userid=$row['userid'];
        $password=$row['password'];
        $first_name=$row['first_name'];
        $last_name=$row['last_name'];
        $role=$row['role'];
        $active=$row['active'];
    }
$conn->close();
?>

Help me,

</div>

此问题已经存在 这里有一个答案： p>

  • mysql_fetch_array（）/ mysql_fetch_assoc（）/ mysql_fetch_row（）/ mysql_num_rows etc ...期望参数1为资源或结果 32 answers span> li> ul> div>

    我试图从MySQL表中选择数据来显示我的c＃datagridview中的数据，但是我收到以下错误消息之一： p>

    警告：mysqli_fetch_array（）要求参数1为mysqli_result，在第15行的/var/www/html/data.php中给出 布尔值 p> blockquote>

    这是我的PHP代码： p> 

     ＆lt;？php 
 
 $ servername =“**  ******“; 
 $ username =”root“; 
 $ password =”pws“; 
 $ dbname =”dbs“; 
 
 $ conn = new mysqli（$ servername，$ username  ，$ password，$ dbname）; 
 
if（$ conn-＆gt; connect_error）{
 die（“连接失败：”。  $ conn-＆gt; connect_error）; 
} 
 $ sql =“SELECT userid，password，first_name，last_name，role，active FROM aster_users”; 
 $ result = mysql_query（$ sql）; 
while（$ row =  mysqli_fetch_array（$ result））
 {
 $ userid = $ row ['userid']; 
 $ password = $ row ['password']; 
 $ first_name = $ row ['first_name']; 
  $ last_name = $ row ['last_name']; 
 $ role = $ row ['role']; 
 $ active = $ row ['active']; 
} 
 $ conn-＆gt; close（）  ; 
？＆gt; 
  code>  pre> 
 
 
    帮帮我， p> 
  div>

答

You cannot mix the mysql and mysqli. For database queries please use mysqli_query($conn,$query). http://php.net/manual/en/mysqli.query.php

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
$sql = "SELECT userid,password,first_name,last_name,role,active FROM aster_users";
$result = mysqli_query($conn,$sql); // NOTICE HOW THIS IS MYSQLI QUERY. NOT MYSQL
while($row = mysqli_fetch_array($result)) 
    {
        $userid=$row['userid'];
        $password=$row['password'];
        $first_name=$row['first_name'];
        $last_name=$row['last_name'];
        $role=$row['role'];
        $active=$row['active'];
    }
mysqli_close($conn); //Use Procedural Code Here. not $conn->close();

答

Mixing mysql and mysqli in your code !! and Object oriented style with Procedural style

Use

  $conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT userid,password,first_name,last_name,role,active FROM aster_users";
if (!$result = $conn->query($sql)) {
    printf("Error: %s
", $conn->error);
}
$rows = $result->num_rows;
if ($rows > 0) {

    while ($row = $result->fetch_array(MYSQLI_ASSOC)) {
        // your code
    }
} else {
    echo "No result found";
}

Read http://php.net/manual/en/mysqli.query.php

答 

$sql = "SELECT userid,password,first_name,last_name,role,active FROM aster_users";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
  while($row = $result->fetch_assoc()) {
    $userid=$row['userid'];
    $password=$row['password'];
    $first_name=$row['first_name'];
    $last_name=$row['last_name'];
    $role=$row['role'];
    $active=$row['active'];
  }
} else {
echo "0 results";
}
$conn->close();

Please don't mix mysql_* & mysqli_*object oried style in php.

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