Question

我可以连接到数据库，但是它说我的查询出了点问题，但是我确定没有。这是因为在我的phpmyadmin控制台中进行了estest处理，并且它给了我所需的结果。我不确定我是否以正确的方式制作桌子已经一段时间了。

尝试了几种conn方法并使用diff查询

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <meta http-equiv="X-UA-Compatible" content="ie=edge">
    <title>Display customer Data</title>
</head>
<body>

<?php

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";


// Check connection

$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 



    $sql = " SELECT *FROM  customer ";

if ($conn->query($sql) === TRUE) {


    echo "Your customers are being displayed";

    echo"<table border = '1' ";
    echo "<tr><tr>Customer ID</tr><td>Customer First Name</td><td>Customer Last Name</td><td></td><td>Street Address</td><td>State</td><td>City</td><td>ZIP</td><td>Company Name</td><tr>phone</tr><tr>Phone EXT</tr></tr>";

    while($row = mysqli_fetch_assoc($conn)){

        echo "<tr><td>{$row[`customerID`]}</td><td>{$row[`firstName`]}</td><td>{$row[`lastName`]}</td><td>{$row[`streetAddress`]}</td><td>{$row[`State`]}</td><td>{$row[`City`]}</td><td>{$row[`Zip`]}</td><td>{$row[`companyName`]}</td><td>{$row[`phone`]}</td><td>{$row[`phoneExt`]}</td></tr>";
    }
    echo"</table>";


} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();
?>

</body>
</html>`

也许我的mysqli提取有问题吗？

Answer 1

在$ conn-> query（$ sql）中显示面向对象的样式，而mysqli_fetch_assoc（$ conn）显示过程样式。

If you want to go with object oriented then :
  mysqli();
  $result = $conn->query($sql);
  While($row = $result->fetch_assoc()){//code}

If you want to go with procedural style:
  mysqli_connect();
  $result = mysqli_query($conn, $sql);
  while($row = mysqli_fetch_assoc($result)){//code}

为什么我的查询无法在html表中显示信息

1 个答案: