php - guest_post() 是如何工作的，输出中的显示问题 - 代码末尾附加并提到的图像

Question

My Issue:

<?php //sqltest.php

//Part 01 - The first part of the code will establish connection with DB using mysqli method

require_once 'login.php';

$conn = new mysqli($hn,$un,$pw,$db);

if ($conn->connect_error) die ($conn->connect_error);

// Part - 02 - Here is the method to delete some data using query by taking input and later checking using isset

if (isset($_POST['delete']) && isset ($_POST['isbn'])){

    $isbn = get_post($conn,'isbn');
    $query ="DELETE FROM classics WHERE isbn = '$isbn'";
    $result = $conn->query($query);

    if (!$result) echo "DELETE failed: $query<br>". $conn->error . "<br><br>";
}

//Part 04 - Here is the method to insert some data using query by taking input by get_post method-(see the last code) and checking using isset

if (isset($_POST['author']) &&
    isset($_POST['title']) &&
    isset($_POST['category']) &&
    isset($_POST['year']) &&
    isset($_POST['isbn'])){

        $author = get_post($conn,'author');
        $title = get_post($conn,'title');
        $category = get_post($conn,'category');
        $year = get_post($conn,'year');
        $isbn = get_post($conn,'isbn');
        $query = "INSERT INTO classics VALUES" . "('$author','$title','$category','$year','$isbn')";

        $result = $conn->query($query);

        if (!$result) echo "INSERT failed:  . $query<br> ". $conn->error. "<br><br>"; 
    }

    //Part - 05 - FORM handler 

    echo <<<_END

    <form action="sqltest.php" 
        method="post">
        <pre> 

        Author <input type = "text" name ="author">
        Title <input type = "text" name = "title">
        Category <input type = "text" name = "category">
        Year <input type = "text" name = "year">
        ISBN <input type = "text" name = "isbn">

        <input type = "submit" value = "ADD RECORD">

        </pre>
        </form>
_END;

// Part - 06 -A new query for showing the whole classics table from DB

$query = "SELECT * FROM classics";

$result = $conn->query($query);

if(!$result) die ("Database access failed: ". $conn->error);

    $rows = $result->num_rows;

    for ($j=0; $j<$rows; ++$j){

        $result->data_seek($j);
        $row = $result->fetch_array(MYSQLI_NUM);

        // Part - 07 The following html code will take the iput for deleting any entry using isbn - refers to 1st part of the code 

        echo <<<_END

        <pre>

        Author $row[0]
        Title $row[1]
        Category $row[2]
        Year $row[3]
        ISBN $row[4]

        </pre>

        <form action = "sqltest.php" method = "post">

        <input type ="hidden" name = "delete" value = "yes">
        <input type = "hiddden" name = "isbn" value = "$row[4]">
        <input type="submit" value = "DELETE RECORD">

        </form>
_END;

    }

    $result->close();
    $conn->close();

    //Part 08 - actually the code begins from here

    function get_post($conn,$var)

    {
        return $conn->real_eascape_string($_POST[$var]); 

        //to avoid special charecter
    }



    ?>

/** 代码运行得很好。除了两件事:1.在代码的第 7 部分中，我提到要隐藏 isbn 编号，只显示删除按钮。但在输出中它同时显示数字和按钮。 2. 带有记录字段的框没有设置，这看起来不像预期的那么好 - 我使用了 pre，但仍然显示损坏的输出。**/

Answer 1

对于#1，您在 hiddden 中存在拼写错误(正确的应该是 hidden)。

对于#2，学习如何使用 css style the form 。还学习如何使用html label tag 。

有些人建议使用表格进行格式化，这不是最佳实践，应该避免。

一般来说，HTML 应该只包含有关内容的信息，CSS 负责内容的呈现。这称为Separation of Concerns .