Question

我想插入数据，其中从php代码返回到现有的html表（name =“userDetails”）。

根据从mysql数据库接收的结果，html表中的行数会有所不同。我怎么能这样做？

代码如下，

<?php
    $connection = mysqli_connect('localhost', 'root', '', 'users');

    $sql = "SELECT id, firstname,email FROM usertable";
    $result = mysqli_query($connection, $sql);

    $userDetailsArray = array();

    if(mysqli_num_rows($result) > 0){

        $index = 0;
        while ($row = mysqli_fetch_assoc($result)) {
            $userDataArray[$index] = $row;
            $row = $userDataArray[$index];

            $userID = $row['id'];
            $firstName = $row['firstname'];
            $email = $row['email'];

            $index++;
        }
    }
?>

<html>
<head></head>

<body>
    <table name = "userDetails">
        <tr>
            <th>User ID</th>
            <th>First Name</th>
            <th>email</th>
        </tr>
    </table>

</body>
<html/>

Answer 1

看起来你正在寻找这样的东西：

<html>
<head></head>

<body>
    <table name = "userDetails">
        <tr>
            <th>User ID</th>
            <th>First Name</th>
            <th>email</th>
        </tr>
        <?php
            $connection = mysqli_connect('localhost', 'root', '', 'users');
            $sql = "SELECT id, firstname,email FROM usertable";
            $result = mysqli_query($connection, $sql);
            if(mysqli_num_rows($result) > 0){

                while ($row = mysqli_fetch_assoc($result)) {
                    echo '<tr>';
                    echo '<td>'. $row['id'] .'</td>';
                    echo '<td>'. $row['firstname'] .'</td>';
                    echo '<td>'. $row['email'] .'</td>';
                    echo '</tr>';
                }
            }
        ?>
    </table>

</body>
<html/>

您只需在while循环中生成HTML即可。不是最好的策略，但肯定会产生结果。

Answer 2

使用相同的页面意味着它将起作用。试试这个

public sealed class CompareCustomAttribute : System.Web.Mvc.CompareAttribute
{
    public CompareCustomAttribute(string otherProperty)
        : base(otherProperty)
    {
    }

    public string ResourceKey { get; set; }

    public string ClassKey { get; set; }

    public override string FormatErrorMessage(string name)
    {
        return Convert.ToString(HttpContext.GetGlobalResourceObject(this.ClassKey, this.ResourceKey));
    }

}

Answer 3

您的代码中这些行的要求是什么？ $ userDetailsArray = array（）; $ userDataArray [$ index] = $ row; $ row = $ userDataArray [$ index]; ??甚至是$ index变量的要求是什么？

我正在为您添加更改程序。

<?php
    $connection = mysqli_connect('localhost', 'root', '', 'users');
?>
<html>
<head></head>
<body>
<table name = "userDetails">
        <tr>
            <th>User ID</th>
            <th>First Name</th>
            <th>email</th>
        </tr>
        <?php
            $sql = "SELECT id, firstname,email FROM usertable";
            $result = mysqli_query($connection, $sql);
            if(mysqli_num_rows($result) > 0){
                while ($row = mysqli_fetch_assoc($result)) {
        ?>
                <tr>
                    <td>$row['id'];</td>
                    <td>$row['firstname'];</td>
                    <td>$row['email'];</td>
                </tr>
        <?php      
               }
            }
            else
            {
        ?>
                <tr>
                    <td colspan=3>Data is not available</td>
                </tr>
        <?php       
            }
        ?>
    </table>
</body>
<html/>

如何将数据插入到PHP中的现有html表中？

3 个答案: