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為什么我的數據庫中沒有顯示我的結果?

[英]Why are my results not showing up from my database?

我有一個程序可以使用php和AJAX技術顯示作者的書代碼和書名,但是由於某種原因,數據沒有顯示在表格中。 我知道我的SQL代碼是正確的,因為我們的講師為我們提供了該代碼,但是某些原因阻止了數據出現在表中。 任何提示或建議,將不勝感激!

<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>

可疑的行是:AND writer.author_number =''為什么它為空?

在第二個查詢之后進行檢查:

$r2 = mysqli_query($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);

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