php mysql select 來自三個沒有關系的表並創建鏈接

Question

我有以下功能。 我正在嘗試從 3 個不同的表中獲取沒有任何關系的數據。 除了a href屬性外，一切正常。 我需要根據表格設置鏈接。 IE: If insert_into_index = 1 in table performances I need link to performance.php, if the insert_into_index = 1 in table education, link should be education.php and if insert_into_index = 1 in artists_works the link should be art.php

    function intro_news($lang) {
    global $connection;
    $lang = $_GET['lang'];  
    
    $sql = "
    SELECT id, title, title_en, insert_into_index, ordering FROM performances
    WHERE insert_into_index=1
    UNION
    SELECT id, title, title_en, insert_into_index, ordering FROM education
    WHERE insert_into_index=1
    UNION
    SELECT id, title, title_en, insert_into_index, ordering FROM artists_works
    WHERE insert_into_index=1
    ORDER BY ordering ASC
    ";  
    
    $query = mysqli_query($connection,$sql) or die(mysqli_error($connection));
    $count = mysqli_num_rows($query);
    if ($count > 0) {
        echo '<div class="grid">';
        echo '<div class="row justify-content-center">';
        while ($row = mysqli_fetch_array($query)) {
            echo '<div class="col-sm-12 col-md-6 col-lg-3 margin-wrapp rotate my-auto">';

            // HERE IS THE ISSUE
            if ($row['insert_into_index'] == '1') {
                echo '<a href="';
                echo 'performance';
                echo '.php?lang='.$lang.'&name='.$row['id'].'">';
            }
            if ($row['insert_into_index'] == '1') {
                echo '<a href="';
                echo 'education';
                echo '.php?lang='.$lang.'&name='.$row['id'].'">';
            }
            if ($row['insert_into_index'] == '1') {
                echo '<a href="';
                echo 'art';
                echo '.php?lang='.$lang.'&name='.$row['id'].'">';
            }
            // ISSUE


            echo '<span class="name">';
            if ($lang === "sk") {
                echo $row['title'];
            }
            if ($lang === "en") {
                echo $row['title_en'];
            }
            echo '</span>';
            echo '</a>';
            echo '</div>';
        } // while loop
        echo '</div>';
        echo '</div>';
    }
}

現在的結果是：

<a href=art.php?lang=sk&name=1>bla bla</a>
<a href=art.php?lang=sk&name=1>bla bla</a>
<a href=art.php?lang=sk&name=4>bla bla</a>

預期的結果應該是：

<a href=performance.php?lang=sk&name=1>bla bla</a>
<a href=education.php?lang=sk&name=1>bla bla</a>
<a href=art.php?lang=sk&name=4>bla bla</a>

我很感激任何幫助。 非常感謝。

Answer 1

在您的 SQL 查詢中添加類型：

SELECT id, title, title_en, insert_into_index, ordering, 'performance' as type FROM performances
    WHERE insert_into_index=1
    UNION
    SELECT id, title, title_en, insert_into_index, ordering, 'education' as type FROM education
    WHERE insert_into_index=1
    UNION
    SELECT id, title, title_en, insert_into_index, ordering, 'art' as type FROM artists_works
    WHERE insert_into_index=1
    ORDER BY ordering ASC

然后您將能夠在 php 代碼中使用類型

echo '<a href="';
echo $row['type'];
echo '.php?lang='.$lang.'&name='.$row['id'].'">';