php - 仅当 mysql 中不存在时才显示

Question

我有一张 table ，上面有滑雪者和他们滑雪的滑雪站。

例如:

---------------------   skier     station---------------------|    1     |    2|    1     |    3|    2     |    2 ---------------------

If I have three stations I want to show a text that say that the skier wasn't in this station.

I have:

 $result="SELECT * FROM skiers_and_stations WHERE skier = '$id_skier';";
 $makeResult=mysql_query($result, $conexion);
 while ($row = mysql_fetch_array($makeResult, MYSQL_ASSOC)) {
 if ($row['station'] == '1') {   } else {echo "No here before, station 1"}; 
 if ($row['station'] == '2') {   } else {echo "No here before, station 2"}; 
 if ($row['station'] == '3') {   } else {echo "No here before, station 3"}; 
 }

但是这不起作用。

我想显示:

Skier 1 No here before, station 1

Answer 1

问题是在每次迭代中，有两个 if 语句将为 false，并转到 else 语句。如果您建立一个包含所有可能性的数组，并删除他们去过的车站，最后您将获得他们未去过的车站列表。

此外，您确实不应该使用 mysql_query 函数，因为它们已被弃用，并将很快从 PHP 中删除。查看mysqli 。另外，对您的查询进行 SQL 注入(inject)。为了解决您的问题，我没有进行这些更改。

$result = "SELECT * FROM skiers_and_stations WHERE skier = '$id_skier';";
$makeResult = mysql_query($result, $conexion);
$stations = array(1 => true, 2 => true, 3 => true);
while ($row = mysql_fetch_array($makeResult, MYSQL_ASSOC)) {
    unset($stations[$row['station']]);
}
foreach ($stations as $stationId => $val) {
    echo "Not here before, station $stationId";
}