如何從數據庫中獲取10條最新記錄

Question

如何像這樣顯示數據庫中的 10 個最新作業。 請點擊此處查看圖片

到目前為止，我已經編寫了以下代碼，不知道下一步該怎么做。

$query = "SELECT * FROM jobs ORDER BY posting_date DESC LIMIT 10";

$query = mysqli_query($conn, $query);

while ($row = mysqli_fetch_assoc($query) ) {


  $job_title = row["job_title"];
  $company_name = $row["company_name"];
  $department = $row["department"];
  $location = $row["location"];   
  $job_type = $row["job_type"];   
  $job_description = $row["job_description"];
  $posting_date = date('d-m-y');

}

$row 返回數據庫中的所有列。 我只想要四個職位、公司名稱、地點和日期

Answer 1

在選擇查詢中使用列名。

$query = "SELECT job_title,company_name,location,posting_date FROM jobs ORDER BY posting_date DESC LIMIT 10";
$query = mysqli_query($conn, $query);

echo "<table>";
echo "<tr> <th>Job Title</th> <th>Company Name</th> <th>Location</th> <th>Date Posted</th> </tr>";
echo "<tbody>";  
while($row = mysqli_fetch_assoc($query) ) {
  $job_title = $row["job_title"];
  $company_name = $row["company_name"];
  $department = $row["department"];
  $location = $row["location"];   
  $posting_date = date('d-m-y', strtotime($row['posting_date']));

  echo "<tr>";
  echo "<td>".$job_title."</td>";
  echo "<td>".$company_name."</td>";
  echo "<td>".$location."</td>";
  echo "<td>".$posting_date."</td>";
  echo "<tr>";
}
echo "</tbody>";
echo "</table>";

Answer 2

嗯改變查詢

$query = "SELECT job_title, company_name, location,  posting_date FROM jobs ORDER BY posting_date DESC LIMIT 10";

Answer 3

您可以為 rownum 訂購

$query = "SELECT job_title, company_name, location,  posting_date FROM jobs ORDER BY rownum DESC LIMIT 10";

Answer 4

以下是如何構建具有 4 列的 htrml 表的示例：

$query = "SELECT * FROM jobs ORDER BY posting_date DESC LIMIT 10";
$query = mysqli_query($conn, $query);

echo "<table>";
while ($row = mysqli_fetch_assoc($query) ) {
  echo "<tr>";
  echo "<td>$row['job_title']</td>";
  echo "<td>$row['company_name']</td>";
  echo "<td>$row['location']</td>";
  echo "<td>$row['posting_date']</td>";
  echo "<tr>";
}
echo "<table>";