PHP如何基于列值显示db表

Question

我有一个PHP脚本来回显db中存在的db表的名称，但现在我只想回复那些记录在表中的用户。

这是我目前的剧本：

        if ( !$conn ) {
        die("Connection failed: " . mysqli_connect_error());
    }

    $tablesquerys = "SHOW TABLES";

    $tableresults = mysqli_query($conn, $tablesquerys) or die(mysqli_error($conn));

    if ( mysqli_num_rows($tableresults) > 0 ) {

        echo "<select id='dbtable_selector' class='dbtable-selector'>";
        echo "<option value='Current Tables' selected='selected'>Current Files</option>";

        while( $table = mysqli_fetch_array($tableresults, MYSQLI_NUM) ) {
            echo("<option value='" . $table[0] . "'>" . $table[0] . "</option>");
        }

        echo "</select>";
    }

    mysqli_close($conn);

要比较的用户名存储在$currentuser

中

我的第二次尝试：

    if ( !$conn ) {
        die("Connection failed: " . mysqli_connect_error());
    }

    $tablesquerys = "SHOW TABLES";

    $tableresults = mysqli_query($conn, $tablesquerys) or die(mysqli_error($conn));

    if ( mysqli_num_rows($tableresults) > 0 ) {

        echo "<select id='dbtable_selector' class='dbtable-selector'>";
        echo "<option value='Current Tables' selected='selected'>Current Files</option>";

        while( $table = mysqli_fetch_array($tableresults, MYSQLI_NUM) ) {
            $userquerys = "SELECT username FROM `" . $table[0] . "`";

            if ( mysqli_query($conn, $userquerys) ) {
                if ( $userquerys === $currentuser ) {
                    echo("<option value='" . $table[0] . "'>" . $table[0] . "</option>");
                }
            } else {
                echo "Error updating record: " . mysqli_error($conn);
            }
        }

        echo "</select>";
    }

    mysqli_close($conn);

仍然没有回音或任何错误。

好的，我看到上面尝试的缺陷，所以这是我的第三次尝试：

        if ( !$conn ) {
        die("Connection failed: " . mysqli_connect_error());
    }

    $tablesquerys = "SHOW TABLES";

    $tableresults = mysqli_query($conn, $tablesquerys) or die(mysqli_error($conn));

    if ( mysqli_num_rows($tableresults) > 0 ) {

        echo "<select id='dbtable_selector' class='dbtable-selector'>";
        echo "<option value='Current Tables' selected='selected'>Current Files</option>";

        while( $table = mysqli_fetch_array($tableresults, MYSQLI_NUM) ) {
            $userquerys = "SELECT username FROM `" . $table[0] . "`";

                if ( mysqli_query($conn, $userquerys) === $currentuser ) {
                    echo("<option value='" . $table[0] . "'>" . $table[0] . "</option>");
                }
        }

        echo "</select>";
    }

    mysqli_close($conn);

仍然没有。