Php 从SQL检索数据 我的代码怎么了

以下是我经常遇到的错误：

---

警告：mysql_fetch_array（）希望参数1是资源，布尔值在C:\xampp\htdocs\CRUD\CRUD_Act\includes\dbdisp.php第4行中给出

<?php
$sql = mysql_query("SELECT * FROM users;");

    while($row = mysql_fetch_array($sql))
    {
    $dbCon=mysqli_connect("localhost", "root", "", "dbusers")
       or die(mysqli_error()."Connection disconnected");
    echo "<tr>";
        echo "<td>" . $row['UserID'] . "</td>";
        echo "<td>" . $row['Firstname'] . "</td>";
        echo "<td>" . $row['Lastname'] . "</td>";
        echo "<td>" . $row['Gender'] . "</td>";
        echo "<td>" . $row['Email'] . "</td>";
        echo "<td>" . $row['Status'] . "</td>";
        echo "<td>" . $row['Date_joined'] . "</td>";
    echo "</tr>";
    }

?>

---

在连接mysql数据库之前，您正在查询该数据库。在进行任何数据库工作之前，必须先连接到数据库，然后检查是否有连接。错误返回false，因为没有要查询的数据库，因此无法运行查询

<?php
 $dbCon=mysql_connect("localhost", "root", "", "dbusers")
       or die(mysql_error()."Connection disconnected");
$sql = mysql_query("SELECT * FROM users;");

    while($row = mysql_fetch_array($sql))
    {

    echo "<tr>";
        echo "<td>" . $row['UserID'] . "</td>";
        echo "<td>" . $row['Firstname'] . "</td>";
        echo "<td>" . $row['Lastname'] . "</td>";
        echo "<td>" . $row['Gender'] . "</td>";
        echo "<td>" . $row['Email'] . "</td>";
        echo "<td>" . $row['Status'] . "</td>";
        echo "<td>" . $row['Date_joined'] . "</td>";
    echo "</tr>";
    }

?>

---

正如他们所说：您必须连接到数据库

包括您的文件连接：

比如：

<?php
$dbCon=mysql_connect("localhost", "root", "", "dbusers")
   or die(mysql_error()."Connection disconnected");
$sql = mysql_query("SELECT * FROM users;");

while($row = mysql_fetch_array($sql))
{
    echo "<tr>";
    echo "<td>" . $row['UserID'] . "</td>";
    echo "<td>" . $row['Firstname'] . "</td>";
    echo "<td>" . $row['Lastname'] . "</td>";
    echo "<td>" . $row['Gender'] . "</td>";
    echo "<td>" . $row['Email'] . "</td>";
    echo "<td>" . $row['Status'] . "</td>";
    echo "<td>" . $row['Date_joined'] . "</td>";
    echo "</tr>";
}

或者按照他们的建议去做，因为在连接mysql数据库之前需要先查询它

然后：

include("connexion.php");



PS：查询字符串不应以分号结尾，分号是此处的第4行。你能说清楚吗？你把mysql和mysqli混在一起了。用一个或另一个。更可取地mysqli@MarcelKorpel，代码更新，thanksI已经这样做了，但给出了相同的错误：警告：mysql_fetch_array（）希望参数1是resource，在第54行的C:\xampp\htdocs\CRUD\CRUD_Act\displayRecords.php中给出的布尔值要显示错误，请尝试以下操作：$res=mysql_query（$res）或die（“错误：.mysql_error（））并在此处发布错误我已经这样做了，但给出了相同的错误：警告：mysql\u fetch\u array（）希望参数1是resource，布尔值在C:\xampp\htdocs\CRUD\CRUD\u Act\displayRecords.php中在线给出54@JullienNIKA玉兰油：你确定在<代码>用户< /代码>（不是用户）表中有数据吗？
  <?php
   $sql ="SELECT * FROM users";
   $res=mysql_query($sql) or die("erreur");
    while($row=mysql_fetch_row($res))
    {
    echo "<tr>";
        echo "<td>" . $row['UserID'] . "</td>";
        echo "<td>" . $row['Firstname'] . "</td>";
        echo "<td>" . $row['Lastname'] . "</td>";
        echo "<td>" . $row['Gender'] . "</td>";
        echo "<td>" . $row['Email'] . "</td>";
        echo "<td>" . $row['Status'] . "</td>";
        echo "<td>" . $row['Date_joined'] . "</td>";
    echo "</tr>";
    }

?>