在php中顯示來自mysql查詢的結果

Question

我正在嘗試為我的網站制定一個即將開設的課程。 我想在mysql數據庫中維護數據。 我已經在msql中設置了3個表。

1.courses

2.category

3.coursedates

所有這些都由course_id鏈接。

基本上，我想使用查詢在PHP中以以下方式顯示課程數據。

課程名稱天數課程日期

我已經嘗試使用以下編碼

<?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);
?> 

Answer 1

您正在混合mysql_ *和mysqli：

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

但：

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

使用mysqli_query代替：

$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);
?> 

Answer 2

將代碼替換為以下內容：

<?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);
?> 

Answer 3

請注意，您的查詢可能更容易重寫，例如：

"
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;
 "

順便說一句，您可能為一個不存在的課程設置了課程日期似乎很奇怪，因此第一個OUTER JOIN可能會被重寫為INNER JOIN！