PHP超级基本内容管理建议请

Question

我使用XAMPP和PHP作为基本网站。

我已经完成了大部分内容管理部分。

我有一个基本的形式，您可以上传图片并添加所有必要的详细信息。我需要帮助保持格式一致，因为目前有三间房屋，价格和细节以HTML编码。

如何以类似的格式回显细节？

安全性不是问题，因为没有使用任何敏感信息，也不会公开使用。我只需要知道如何以类似的方式来设置细节和图片的格式？

我附上了照片，以便您可以更好地了解我需要的帮助。下面是代码：

<?php 
    $servername="localhost"; 
    $username="root"; 
    $password=""; 
    $dbname="content_management"; 
    $tbl_name="houses"; 
    $housepic ="housepic"; 
    $houseprice ="houseprice"; 
    $housetype ="housetype"; 
    $houseloc = "houseloc"; 
    $housedesc = "housedesc"; 


$conn = mysqli_connect($servername, $username, $password, $dbname); 
// Check connection 
if (!$conn) { 
    die("Connection failed: " . mysqli_connect_error()); 
} 

    $sql="INSERT INTO $tbl_name (picture, price, type, location, description) VALUES ('$housepic','$houseprice','$housetype','$houseloc','$housedesc')"; 

    if (mysqli_query($conn, $sql)) { 
    echo ""; 
} else { 
    echo "Error: " . $sql . "<br>" . mysqli_error($conn); 
} 
    mysqli_close($conn); 

?> 
<?php 
    $host="localhost"; 
    $username="root"; 
    $password=""; 
    $db_name="content_management"; 
    $tbl_name="houses"; 
    mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
    mysql_select_db("$db_name")or die("cannot select DB"); 
    $sql = "SELECT * FROM $tbl_name"; 
    $result=mysql_query($sql); 
    while ($row = mysql_fetch_assoc($result)) { 
     echo "<img src='" . $row['picture'] . "'>"; 
      echo $row['price']; 
      echo $row['type']; 
     echo $row['location']; 
     echo $row['description']; 
    } 

?> 

     <div class="housepost"> 
      <img src="img/houses/house_01.jpg"> 
      <h2>£350,000</h2> 
      <p>2 bedroom detached house for sale</p> 
      <p>Deanfield Avenue, Henley-on-Thames</p> 
      <p>Set in the heart of Henley-on-Thames and just a short walk from Henley train station is this rarely available and spacious three bedroom apartment. Offered to the market with no onward chain the property benefits from off road parking.</p> 
     </div> 
     <div class="housepost"> 
      <img src="img/houses/house_02.jpg"> 
      <h2>£475,000</h2> 
      <p>2 bedroom detached bungalow for sale</p> 
      <p>Fair Mile, Henley-on-Thames</p> 
      <p>Set in the heart of the town centre in a quiet backwater this delightful single storey detached home is rarely available and well presented.</p> 
     </div> 
     <div class="housepost"> 
      <img src="img/houses/house_03.jpg"> 
      <h2>£600,000</h2> 
      <p>3 bedroom cottage for sale</p> 
      <p>Remenham Row, Henley-on-Thames</p> 
      <p>The English Courtyard Association and The Beechcroft Trust - synonymous with the very best in retirement housing since the 1980s. An extremely attractive three-bedroom cottage with landscaped riverside gardens in this much sought after location.</p> 
     </div> 
     <?php echo '<div class="housepost"> 
      <img>$row <img> 
      <h2></h2> 
      <p></p> 
      <p></p> 
      <p>$row</p> 
     </div>' ?> 
    </div> 

我不知道我用PHP做什么，我很惊讶，我做了这么远，但我有一个初学者理解。您可以在代码中看到三个硬编码的HTML div，最后一个div是我用PHP尝试失败的实验。如果你看看房子的图片，你可以看到我希望PHP脚本显示信息的格式。 （目前刷新页面自动创建一个记录与表格框标题，我需要修复以及）

请问如果你需要我澄清，如果你能帮助我，那将是非常感激。我相信答案非常简单，我花了一段时间来解释所有这些。

这是我使用的数据库的事情：

这是房子的一个div我需要PHP每个房子被上传时自动创建格式：

Answer 1

这是一种简单快捷的方式来完成您的要求。

我想你的问题是不理解PHP，DB和HTML基础知识，但是把他们放在一起 - 这样：

阅读代码中的注释仔细了解所有这些三件事情之间的所有步骤和关系。

当你得到它后，你可以按照这个链接：php_file_upload学习如何上传图片（这有点复杂，但是对你的学习来说是一个很好的测试）。

这是代码（全部在同一页！）：

所有的

<!-- HTML form to add a new house --> 
<form method="post"> 
    <!-- this is only giving and existing url, notice you should change this 
     if you want to upload an actual pic --> 
    <input type="text" name="housepic"> 
    <br> 
    <input type="text" name="houseprice"> 
    <br> 
    <input type="text" name="housetype"> 
    <br> 
    <input type="text" name="houseloc"> 
    <br> 
    <input type="text" name="housedesc"> 
    <br> 
    <input type="submit" name="submit" value="Submit"> 
</form> 

<?php 
//init DB vars 
$servername="localhost"; 
$username="root"; 
$password=""; 
$dbname="content_management"; 
$tbl_name="houses"; 

//conect to db 
$conn = mysqli_connect($servername, $username, $password, $dbname); 
// Check connection 
if (!$conn) { 
    die("Connection failed: " . mysqli_connect_error()); 
} 

/* if request method is: "post", a form was submitted. 
    get the submitted form params, and update DB */ 
if ($_SERVER["REQUEST_METHOD"] == "POST") 
{ 
    /* notice you should add here error handling, case user didn't fill 
     all params */ 
    $housepic = (isset($_POST["housepic"])) ? $_POST["housepic"] : ""; 
    $housepic = (isset($_POST["houseprice"])) ? $_POST["houseprice"] : ""; 
    $housepic = (isset($_POST["housetype"])) ? $_POST["housetype"] : ""; 
    $housepic = (isset($_POST["houseloc"])) ? $_POST["houseloc"] : ""; 
    $housepic = (isset($_POST["housedesc"])) ? $_POST["housedesc"] : ""; 

    // update DB 
    $sql="INSERT INTO $tbl_name (picture, price, type, location, description) VALUES ('$housepic','$houseprice','$housetype','$houseloc','$housedesc')"; 

    if (mysqli_query($conn, $sql)) { 
    echo ""; 
    } else { 
     echo "Error: " . $sql . "<br>" . mysqli_error($conn); 
    } 
} 

/* we'll always get to this part. 
    notice, if a form was already submitted, we've just updated our DB 
    accordingly, so the houses list is already updated */ 
while ($row = mysql_fetch_assoc($result)) { 
    /* good practice: grab your params before, 
     for a nicer and reusable code */ 
    $picture = $row['picture']; 
    $price = $row['price']; 
    $type = $row['type']; 
    $location = $row['location']; 
    $description = $row['description']; 

    /******************************************************************** 
     this is the part you are asking about. 
     notice that echoing the HTML tags content is done 
     inside the WHILE loop, 
     which of course will be always in the same format... 
    *********************************************************************/ 
    echo " 
      <div class=\"housepost\"> 
     <img src=\"{$picture}\"> 
     <h2>{$price}</h2> 
     <p>{$type}</p> 
     <p>{$location}</p> 
     <p>{$description}</p> 
    </div>"; 

} 

/* don't forget to close your DB connection */ 
mysqli_close($conn); 

Answer 2

首先，我不明白你通过类似的格式是什么意思。但我猜你需要做这样的事情

<?php 

    //Remove these variables and place them in a file like config.php 

    $servername="localhost"; 
    $username="root"; 
    $password=""; 
    $dbname="content_management"; 
    $tbl_name="houses"; 
    $housepic ="housepic"; 
    $houseprice ="houseprice"; 
    $housetype ="housetype"; 
    $houseloc = "houseloc"; 
    $housedesc = "housedesc"; 

//Make a function with all this block like function inserhome 
$conn = mysqli_connect($servername, $username, $password, $dbname); 
// Check connection 
if (!$conn) { 
    die("Connection failed: " . mysqli_connect_error()); 
} 

//Send a variable named $_REQUEST["doinsert"] from the referer page to check if to insert or not 

if($_REQUEST["doinsert"] == true){ 
    $conn = mysqli_connect($servername, $username, $password, $dbname); 
    // Check connection 
    if (!$conn) { 
     die("Connection failed: " . mysqli_connect_error()); 
    } 

     $sql="INSERT INTO $tbl_name (picture, price, type, location, description) VALUES ('$housepic','$houseprice','$housetype','$houseloc','$housedesc')"; 

     if (mysqli_query($conn, $sql)) { 
     echo ""; 
    } else { 
     echo "Error: " . $sql . "<br>" . mysqli_error($conn); 
    } 
     mysqli_close($conn);  
} 


//Make a function like function selecthomes 
    mysql_connect($servername, $username, $password)or die("cannot connect"); 
    mysql_select_db($db_name)or die("cannot select DB"); 

    $sql = "SELECT * FROM $tbl_name"; 

    $result=mysql_query($sql); 


/*Change made to print all tables using mysql_fetch_assoc. I would 
recommend using pdo instedof mysql_fetch_assoc functions since they are 
deprecated. check http://php.net/manual/en/book.pdo.php 
check the deprecation http://php.net/manual/en/function.mysql-fetch-assoc.php*/ 

    while ($row = mysql_fetch_assoc($result)) { ?> 
     <div class="housepost"> 
      <img src="<?php echo $row['picture'] ?>"> 
      <h2><?php echo $row['price'] ?></h2> 
      <p><?php echo $row['type'] ?></p> 
      <p><?php echo $row['location'] ?></p> 
      <p><?php echo $row['description'] ?></p> 
     </div> 
<?php  } ?> 

其次，你也说：

目前刷新页面会自动创建与 形式框标题，我需要修复以及

记录

很明显，这个脚本总是会插入一条新记录，因为每次你在php文件的开始处插入，所以不可避免地会插入一条新记录。

你需要检查的一件事是如果记录已经插入。

我添加了一些功能来演示如何检查插入记录与否。最好你应该传递一个像http://mypage/inserthouse.php?doinsert=true这样的url变量，或者你可以从引用页面创建一个表单并从隐藏字段传递该变量。

您可以制作两个按钮。一个将仅显示房屋记录，另一个将插入并随后显示新插入的记录。 所以第一个按钮将链接到http://mypage/inserthouse.php?doinsert=true 和秒http://mypage/inserthouse.php?doinsert=false或http://mypage/inserthouse.php

我建议你应该为每个动作的功能，并检查枯萎你需要执行插入或没有。提出问题，我应该何时插入新记录？

此外，您不需要变量名称$ tbl_name。在你的代码中并不重要，它会延长脚本。

你还需要使用一个包含文件的变量，

$servername="localhost"; 
    $username="root"; 
    $password=""; 
    $dbname="content_management"; 
    $tbl_name="houses"; 
    $housepic ="housepic"; 
    $houseprice ="houseprice"; 
    $housetype ="housetype"; 
    $houseloc = "houseloc"; 
    $housedesc = "housedesc"; 

你的代码看起来像你需要阅读一些用PHP的面向对象编程。给这个镜头，http://www.codecademy.com/courses/web-beginner-en-bH5s3/0/1

我也建议使用框架，而不是使它成为你自己。 你的技能还没有制作你自己的cms。使用MVC和面向对象的模式。

检查zend框架，它是非常好的，易于使用。

我希望我帮助