如何從博客中選擇三個表，包括作者，帖子和視圖，並在php mysql中顯示？

Question

我要為第三個選擇連接的表是viewcount，它通過prints.print_id = totalview.name連接到“ prints”表

因此，此處的選擇有效，但添加第三個表無效。 我的查詢出了什么問題？ 作品！

$q = "SELECT artists.artist_id, CONCAT_WS(' ', first_name, middle_name, last_name) AS artist, print_name, price, description, print_id, image_name FROM artists, prints WHERE artists.artist_id = prints.artist_id ORDER BY artists.last_name ASC, prints.print_id ASC";

錯誤！

$q = "SELECT artists.artist_id, CONCAT_WS(' ', first_name, middle_name, last_name) AS artist, print_name, price, description, print_id, image_name, totalview.totalvisit AS totalvisit FROM artists, prints WHERE artists.artist_id = prints.artist_id, LEFT JOIN totalview ON totalview.print = prints.print_id ORDER BY artists.last_name ASC, prints.print_id ASC";

顯示在表格中：

$r = mysqli_query ($dbc, $q);
while ($row = mysqli_fetch_array ($r, MYSQLI_ASSOC)) {

    // Display each record:
    echo "\t<tr>
        <td align=\"left\"><a href=\"browse_prints.php?aid={$row['artist_id']}\">{$row['artist']}</a></td>
        <td align=\"left\"><a href=\"view_print.php?pid={$row['print_id']}&name={$row['image_name']}\">{$row['print_name']}</a></td>
        <td align=\"left\">{$row['description']}</td>
        <td align=\"left\">{$row['description']}</td>
        <td align=\"right\">\${$row['price']}</td>
    </tr>\n";

Answer 1

左連接子句位置錯誤..您不能在其中添加左連接子句

並且不要混淆顯式和隱式連接使用顯式和最后（希望），在where條件之后您還會有一個錯誤的逗號

$q = "SELECT 
        artists.artist_id
        , CONCAT_WS(' ', first_name, middle_name, last_name) AS artist
        , print_name
        , price
        , description
        , print_id
        , image_name
        , totalview.totalvisit AS totalvisit 
    FROM artists
    INNER JOIN  prints ON rtists.artist_id = prints.artist_id
    LEFT JOIN totalview ON totalview.print = prints.print_id 
    ORDER BY artists.last_name ASC, prints.print_id ASC";

Answer 2

您的SQL結構不正確：

FROM artists, prints 
WHERE artists.artist_id = prints.artist_id, 
LEFT JOIN totalview ON

應該

FROM artists, prints 
LEFT JOIN totalview