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Php 将SQL数据转换为网格/表_Php_Sql_Datagrid_Tabular - Fatal编程技术网

Php 将SQL数据转换为网格/表

Php 将SQL数据转换为网格/表,php,sql,datagrid,tabular,Php,Sql,Datagrid,Tabular,我想知道在4x4网格/表格中显示数据行的最简单方法是什么。我环顾四周已经有一段时间了,似乎找不到解决办法/不明白怎么做 下面是我用来从SQL数据库提取数据的代码 <?php require('json.php'); error_reporting(E_ALL); mysql_connect("localhost", "root", "") or die("Could not connect: " . mysql_error()); mysql_select_db("timetable")

我想知道在4x4网格/表格中显示数据行的最简单方法是什么。我环顾四周已经有一段时间了,似乎找不到解决办法/不明白怎么做

下面是我用来从SQL数据库提取数据的代码

<?php
require('json.php');
error_reporting(E_ALL);
mysql_connect("localhost", "root", "") or
die("Could not connect: " . mysql_error());
mysql_select_db("timetable") or die("No such database");
$sql = sprintf("
SELECT *
 FROM `event`
  WHERE `module` = 'SET08101'");

$result = mysql_query($sql)
 or die(mysql_error());
  while ($row = mysql_fetch_array($result)){
   print "<div>$row[id] $row[module] $row[day] $row[start]</div>\n";
}
?>

像这样的东西可以工作:

<?php
require('json.php');
error_reporting(E_ALL);
mysql_connect("localhost", "root", "") or
die("Could not connect: " . mysql_error());
mysql_select_db("timetable") or die("No such database");
$sql = sprintf("
SELECT *
 FROM `event`
  WHERE `module` = 'SET08101'");

$result = mysql_query($sql)
 or die(mysql_error());
 ?>
 <table>
 <?php
  while ($row = mysql_fetch_array($result)){
   ?>
   <tr>
      <td><?php echo $row["id"];?></td>
      <td><?php echo $row["module"];?></td>
      <td><?php echo $row["day"];?></td>
      <td><?php echo $row["start"];?></td>
   </tr>
   <?php
  }
 ?>
 </table>
 <?php
?>

我会这样做:

<?php
error_reporting(E_ALL);

require('json.php');

mysql_connect("localhost", "root", "") or die("Could not connect: " . mysql_error());
mysql_select_db("timetable") or die("No such database");


$sql = sprintf("SELECT *
                FROM `event`
                WHERE `module` = '%s'", 'SET08101');

echo "<table>";

$result = mysql_query($sql) or die(mysql_error());

while ($row = mysql_fetch_array($result)){
   echo "<tr>
             <td>{$row['id']}</td>
             <td>{$row['module']}</td>
             <td>{$row['day']}</td>
             <td>{$row['start']}</td>
         </tr>";
}
echo "</table>";
?>

使用电源,卢克。。。用于grid@Sergey我试图用这些标记实现一个表,但我似乎总是在编码时出错。每个标记都会出现此错误“注意:在C:\wamp\www\ModuleInfo.php的第42行使用未定义的常量id-假定为'id'”,当我用该标记替换代码时。将$row[id]更改为$row['id']等等。。。(这不是错误,而是通知)