选择随机行，更新列如图所示

Question

我希望有一行名为“'see'的列，然后此列将具有默认值no值或”yes“值。

所以..

1. 抓住一行之前没有“看到”的地方。
2. 更新上面的“已见列”以表示“是”。
3. 如果所有行的值都为“yes”，则会显示通知/错误： 您已成功完成所有号码。

我已经尽了最大努力实现它，但它无法正常工作。我认为我解决这个问题的逻辑可能不正确？

    include 'DB.php';
    $con = mysqli_connect($host,$user,$pass);
    $dbs = mysqli_select_db($databaseName, $con);

    // Grabs one row where it hasn't been seen before
    $query = mysqli_query("SELECT number, association, image_file, skeleton, sound, colour, comments FROM num_image WHERE seen='' ORDER by rand() LIMIT 1");

    // Updates the above row with the 'seen' column saying 'yes''
    $query = mysqli_query("UPDATE num_image SET seen = yes");

    // Fetches Result
    $thestuff = mysqli_fetch_row($query);


$seenme=$_POST['seen']; // get value of 'seen' column

$result = mysqli_query("SELECT * FROM num_image where seen=$seenme");

// Trying to delivery a message if the enitre 'seen' column is ALL yes.
while($row = mysqli_fetch_row($result))
{
    if($row['seen'] == 'yes')
    { // All numbers seen
      echo 'You have successfully completed all numbers.';
      echo json_encode($thestuff); 
    }
    else
    { // Show numbers
      echo json_encode($thestuff); 
    }
}

SELECT和UPDATE行是否也必须是if语句？ 干杯

Answer 1

您必须转义更新语句中的值：

$query = mysqli_query("UPDATE num_image SET seen = 'yes' ");