仅显示数据库表中选定的结果

我想在我的 php 页面上仅显示选定的结果。

这是我的 php 代码：

$bookdetsql = "
SELECT b.bookISBN
     , b.bookTitle
     , b.bookYear
     , b.catID
     , b.pubID
     , p.pubName
     , p.location
     , c.catDesc
     , b.bookPrice 
  FROM nbc_book b
  LEFT 
  JOIN nbc_category c 
    ON b.catID = c.catID 
  LEFT 
  JOIN nbc_publisher p 
    ON b.pubID = p.pubID
";


$bookdetrs = mysqli_query($conn, $bookdetsql) or die(mysqli_error($conn));

$bookdetnum = mysqli_num_rows($bookdetrs);

if($bookdetnum >= 1 ){
echo "<div style='margin: 0 0 10px 0; font-weight: bold;'>$bookdetnum record(s) found!</div>";
while ($row = mysqli_fetch_assoc($bookdetrs)) {
    echo "<tr>";
    echo "<td><center>" . $row['bookTitle']."</a></center></td>";
    echo "<td><center>" . $row['bookYear']."</center></td>";
    echo "<td><center>" . $row['catDesc']."</center></td>";
    echo "<td><center>" . $row['bookPrice']."</center></td>";
    echo "<td><center>" . $row['pubName']."</center></td>";
    echo "<td><center>" . $row['location']."</center></td>";
    echo "</tr>";
}

} else {
   echo "<b>Books not found!</b>";
}

实际上上面的代码实际上显示了整个记录列表。我只希望它显示我在第一个 php 页面上单击的选定记录。

您是如何从第一页获取值的？获取还是发布？

当您拥有值时，您可以在 SELECT 语句中添加一个 where 子句

$bookdetsql = "SELECT bookISBN, bookTitle, bookYear, nbc_book.catID, nbc_book.pubID, pubName,  location, catDesc, bookPrice 
                 FROM nbc_book 
            LEFT JOIN nbc_category ON nbc_book.catID = nbc_category.catID 
            LEFT JOIN nbc_publisher ON nbc_book.pubID = nbc_publisher.pubID
                WHERE bookISBN = '" . $_GET["bookISBN"] . "'";