1

我正在尝试为我的网站开发即将到来的课程日期。我想在 mysql 数据库中维护数据。我在 msql 中设置了 3 个表。

1.课程

2.类别

3.课程日期

所有都由 course_id 链接。

基本上我想在 PHP 中使用查询以下列方式呈现来自 coursedates 的数据。

课程名称 天数 课程日期

我尝试使用以下编码

<?php
$con=mysqli_connect("localhost","root","","mentertraining");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }


$query = "SELECT `coursedates`.`coursedate_id`,`courses`.`course_title`,`courses`.`no_of_days`,`category`.`category_name`,`coursedates`.`date1` FROM coursedates\n"
    . "LEFT JOIN `mentertraining`.`courses` ON `coursedates`.`course_id` = `courses`.`course_id` \n"
    . "LEFT JOIN `mentertraining`.`category` ON `courses`.`category_id` = `category`.`category_id` LIMIT 0, 30 ";

$result = mysql_query($query);
    echo "<table border='1'>
<tr>
<th>Course Title</th>
<th>Course Date</th>

</tr>";

while($row    = mysql_fetch_assoc($result))
  {
  echo "<tr>";
  echo "<td>" . $row['course_title'] . "</td>";
  echo "<td>" . $row['date1'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

mysqli_close($con);
?>
4

3 回答 3

2

用以下代码替换您的代码：

<?php
$con=mysqli_connect("localhost","root","","mentertraining");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }


$query = "SELECT `coursedates`.`coursedate_id`,`courses`.`course_title`,`courses`.`no_of_days`,`category`.`category_name`,`coursedates`.`date1` FROM coursedates "
    . " LEFT JOIN `mentertraining`.`courses` ON `coursedates`.`course_id` = `courses`.`course_id` "
    . " LEFT JOIN `mentertraining`.`category` ON `courses`.`category_id` = `category`.`category_id` LIMIT 0, 30 ";

$result = mysqli_query($con,$query);
    echo "<table border='1'><tr><th>Course Title</th><th>Course Date</th></tr>";

while($row    = mysqli_fetch_assoc($result))
  {
  echo "<tr>";
  echo "<td>" . $row['course_title'] . "</td>";
  echo "<td>" . $row['date1'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

mysqli_close($con);
?>
于 2013-06-30T09:18:29.720 回答
1

您正在混合 mysql_* 和 mysqli：

$con=mysqli_connect("localhost","root","","mentertraining");

但：

$result = mysql_query($query);//?

改用 mysqli_query：

$query = "SELECT 
       `coursedates`.`coursedate_id`,
       `courses`.`course_title`,
       `courses`.`no_of_days`,
       `category`.`category_name`,
       `coursedates`.`date1` 
FROM
    coursedates
LEFT JOIN
      `mentertraining`.`courses` 
ON 
   `coursedates`.`course_id` = `courses`.`course_id`
LEFT JOIN
     `mentertraining`.`category` 
ON 
   `courses`.`category_id` = `category`.`category_id`
LIMIT 0, 30";

$result = mysqli_query($query);
echo "<table border='1'>
<tr>
<th>Course Title</th>
<th>Course Date</th>

</tr>";

/* fetch associative array */
while ($row = mysqli_fetch_assoc($result)) {
    echo "<tr>";
    echo "<td>" . $row['course_title'] . "</td>";
    echo "<td>" . $row['date1'] . "</td>";
    echo "</tr>";
}

/* free result set */
mysqli_free_result($result);

echo "</table>";

mysqli_close($con);
?>
于 2013-06-30T09:06:24.047 回答
0

请注意，您的查询可能会更清晰地改写成这样......

"
SELECT cd.coursedate_id
     , c.course_title
     , c.no_of_days
     , t.category_name
     , cd.date1 
  FROM coursedates cd
  LEFT 
  JOIN courses c
    ON c.course_id = cd.course_id 
  LEFT 
  JOIN category t
    ON t.category_id = c.category_id 
 LIMIT 0, 30;
 "

顺便说一句，您可以为不存在的课程设置课程日期似乎很奇怪，因此第一个 OUTER JOIN 可能会被重写为 INNER JOIN！

于 2013-06-30T09:30:32.243 回答