php – 仅显示数据库表中的选定结果

2019年8月4日 116次阅读

我想在我的php页面上只显示一个选定的结果.

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>";
}

实际上这个代码实际上是显示整个记录列表.我只希望它显示我在第一个php页面上点击的所选记录.

最佳答案 你是如何从第一页获得价值的？ GET还是POST？

当您有值时,可以在SELECT语句中添加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"] . "'";
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