如何创建从数据库驱动的动态菜单和页面？ [英] How can I create dynamic menus and pages driven from database?

问题描述

我在这里是新手，也是php。我不是新来的stackoverflow虽然:)

我正在创建一个简单的报告系统，我想要从数据库生成menues和页面。

我看到 YouTube上的这部影片< a>并设法创建一个包含以下代码的菜单。

我有一个名为 Reports 的数据库表和名为 rep_id ， rep_date ， rep_ledit_date ， rep_by ，部门，位置，报告和 rep_to 。

所以基于上面的方法，我创建了一个使用这个代码的菜单。

 <？php 
 
 require（includes / db.php）; 
 
 mysqli_select_db（$ con，'$ db_name'）; 
 
 $ sql =SELECT * FROM reports; 
 $ result = mysqli_query（$ con，$ sql）或者死（mysqli_error（$ con））; 
 $ represult = mysqli_fetch_assoc（$ result）; 
 $ rep_by = $ represult ['rep_by']; 
 $ report = $ represult ['report']; 
 
？> 
< li>菜单
< ul> 
<？php do 
 {
？> 
< li>< a href =reports.php？rep_id =<？php echo $ represult ['rep_id'];？>> <？php echo $ represult ['rep_by']。 （。$ represult ['rep_date']。）; ？>< / A>< /锂> 
 
<？php 
} while（$ represult = mysqli_fetch_assoc（$ result））; 
 // $ represult ['rep_by']; 
 // $ represult ['report']; 
 // $ represult ['report']; 
？> 
< / ul> 
< / li> 
  

所以我创建了一个名为 reports.php 查看数据库中的内容的详细信息。我想要的是将以下 rep_by（rep_date）作为标题和内容报告。

我可能想要在内容中使用其他列。那么什么样的代码菜单和 reports.php 应该实现我想要的。我所做的是以下内容，只有当所有的菜单链接被点击时才输出第一行。

  require（includes / db.php）; 
 
 mysqli_select_db（$ con，'$ db_name'）; 
 
 $ sql =SELECT * FROM reports; 
 $ result = mysqli_query（$ con，$ sql）或者死（mysqli_error（$ con））; 
 $ represult = mysqli_fetch_assoc（$ result）; 
 
< h1> <？php echo $ represult ['rep_by']。 （。$ represult ['rep_date']。）; ？>< / A>< / H1> 
<？php echo $ represult ['report']; ？> 
  

解决方案

report.php：

 <？php 
 
 require（'includes / db.php'）; 
 
 mysqli_select_db（$ con，$ db_name）; 
 
 $ sql ='SELECT * FROM Reports WHERE rep_id =？'; 
 $ stmt = $ CON> prepare（$ sql）; 
 $ id = $ _GET ['rep_id']; 
 $ stmt-> bind_param（'i'，$ id）; 
 $ stmt-> execute（）; 
 
 $ stmt-> bind_result（$ rep_id，$ rep_date，$ rep_ledit_date，$ rep_by，$ department，$ position，$ report，$ rep_to）; 
 $ stmt-> fetch（）; 
 
？> 
 
< h1><？php echo$ rep_by（$ rep_date）; ？>< / H1> 
<？php echo $ report; ？> 
  

reports.php（调整一下）

 <？php 
 
 require（'includes / db.php'）; 
 
 mysqli_select_db（$ con，$ db_name）; 
 
 $ sql ='SELECT * FROM Reports'; 
 $ stmt = $ CON> prepare（$ sql）; 
 $ stmt-> execute（）; 
 
 $ stmt-> bind_result（$ rep_id，$ rep_date，$ rep_ledit_date，$ rep_by，$ department，$ position，$ report，$ rep_to）; 
 
？> 
菜单
< ul> 
<？php while（$ stmt-> fetch（））{？> 
 
< li>< a href =report.php？rep_id =<？php echo $ rep_id;？>> <？php echo$ rep_by（$ rep_date）; ？>< / A>< /锂> 
 
<？php}？> 
< / ul> 
  

I am a newbie here and for php too. I am not new to stackoverflow though :)

I am creating a simple reporting system and I want the menues and pages to be generated from the database.

I saw this video on YouTube and managed to create a menu with the following code.

I have a database table called Reports and columns called rep_id, rep_date, rep_ledit_date, rep_by, department, position, report, and rep_to.

So based on the above method, I created a menu using this code.

<?php 

require ("includes/db.php");

mysqli_select_db($con, '$db_name');

$sql = "SELECT * FROM reports";
$result = mysqli_query($con, $sql) or die (mysqli_error($con));
$represult=mysqli_fetch_assoc($result);
$rep_by=$represult['rep_by'];
$report=$represult['report']; 

?>  
<li> Menu
<ul>
<?php do 
       {
?>
<li><a href="reports.php?rep_id=<?php echo $represult['rep_id']; ?>"> <?php echo $represult['rep_by'] . " (" .   $represult['rep_date'] . ")"; ?></a></li>

<?php 
     } while ($represult=mysqli_fetch_assoc($result));
    //$represult['rep_by'];
    //$represult['report'];
    //$represult['report'] ;
?>
</ul>
</li>    

So I created a page called reports.php to see the details of the content in the database. What I wanted was to see the following rep_by (rep_date) as a heading and report as a content.

I might want to use other columns in the content too. So what kind of code the menu and reports.php should have to achieve what I want. What I did was the following and it only outputs the first row when all the menu links are clicked.

 require ("includes/db.php");

 mysqli_select_db($con, '$db_name');

 $sql= "SELECT * FROM reports";
 $result = mysqli_query($con, $sql) or die (mysqli_error($con));
 $represult=mysqli_fetch_assoc($result); 

 <h1> <?php echo $represult['rep_by'] . " (" . $represult['rep_date'] . ")"; ?></a></h1>
 <?php echo $represult['report']; ?>

解决方案

report.php:

<?php

require ('includes/db.php');

mysqli_select_db($con, $db_name);

$sql= 'SELECT * FROM Reports WHERE rep_id = ?';
$stmt = $con->prepare($sql);
$id = $_GET['rep_id'];
$stmt->bind_param('i', $id);
$stmt->execute();

$stmt->bind_result($rep_id, $rep_date, $rep_ledit_date, $rep_by, $department, $position, $report, $rep_to);
$stmt->fetch();

?>

<h1><?php echo "$rep_by ($rep_date)"; ?></h1>
<?php echo $report; ?>

reports.php (tweaked a bit)

<?php 

require ('includes/db.php');

mysqli_select_db($con, $db_name);

$sql = 'SELECT * FROM Reports';
$stmt = $con->prepare($sql);
$stmt->execute();

$stmt->bind_result($rep_id, $rep_date, $rep_ledit_date, $rep_by, $department, $position, $report, $rep_to);

?>  
Menu
<ul>
<?php while ($stmt->fetch()) { ?>

<li><a href="report.php?rep_id=<?php echo $rep_id; ?>"> <?php echo "$rep_by ($rep_date)"; ?></a></li>

<?php } ?>
</ul>

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