Question

我有一个程序，使用php和显示作者的书籍代码和书名 AJAX技术，但由于某种原因，数据没有出现在表中。我知道我的SQL代码是正确的，因为我们的教师为我们提供了代码，但有些东西阻止了数据出现在表中。任何提示或建议将不胜感激！

<body>
        <?php
            $authorid = 0;
            $authorid = (int) $_GET['authorid'];
            if ($authorid > 0) {
                require_once('dbtest.php');
                $query = "SELECT * FROM author";
                $r = mysqli_query($dbc, $query);
                if (mysqli_num_rows($r) > 0) {
                    $row = mysqli_fetch_array($r);
                } else {
                echo "Title Not Returned<br>";
                }
                echo "<table border='1'><caption>Titles for </caption>";
                echo "<tr>";
                echo "<th>Book Code</th>";
                echo "<th>Book Title</th>";
                echo "</tr>";
                $q2 ="SELECT wrote.author_number As ANo, wrote.book_code As BookCd, book.book_title As Title ";
                $q2 .= " FROM wrote, book ";
                $q2 .= " WHERE wrote.book_code=book.book_code ";
                $q2 .= " AND wrote.author_number = '  ' ";
                $q2 .= " ORDER BY book.book_title";
                $r2 = mysqli_query($dbc, $q2);
               $row = mysqli_fetch_array($r2);
                while ($row) {
                    echo "<tr>";
                    echo "<td>" .$row['BookCd']. "</td>";
                    echo "<td>" .$row['Title']. "</td>";
                    echo "</tr>";
                   $row = mysqli_fetch_array($r2);
                }
                echo "</table>";

            } else {
               echo "<p>No Author ID from prior page</p>"; 
            }
        ?>
        </form>
    </body>

Answer 1

可疑线是：AND writ.author_number =＆＃39; ＆＃39; 为什么它是空的？

在第二个查询后进行检查：

$r2 = mysqli_query($dbc, $q2);

if (mysqli_num_rows($r2) > 0) {
    echo "rows are Returned<br>";
} else {
    echo "rows are Not Returned<br>";
}

$row = mysqli_fetch_array($r2);

为什么我的结果没有显示在我的数据库中？

1 个答案: