0
我有一張表,其中包含一些來自我的數據庫的信息。我在表格中的每一行旁邊都有按鈕。如果按鈕被點擊,它應該發送一封電子郵件給用戶。我可以給每個按鈕一個獨特的值,但不能在if isset中有這個值....我應該如何獲得按鈕名稱的值是可變的,我怎樣才能得到我的網址結束的值。
echo "<table>";
echo "<tr><td>Naam</td><td>EmailAdress</td><td>Datum</td><td>Type</td><td>Producent</td><td>SerieNummer</td><td> imei</td><td>Manager</td><td>bevestigd</td></tr>";
while($row = mysqli_fetch_row($result)){
$sqlid = "SELECT Bevestigd FROM tbl_bevestiging WHERE IDbewijs = " . $row[0];
$resultid = mysqli_query($conn, $sqlid);
$row2 = mysqli_fetch_row($resultid);
echo "<tr><td>" . $row[1] . "</td><td>" . $row[2] . "</td><td>" . $row[3] . "</td><td>" . $row[4] . "</td><td>" . $row[5] . "</td><td>" . $row [6] . "</td><td>" . $row[7] . "</td><td>" . $row[8] . "</td><td>" . $row2 [0] . "</td><td><form method=\"post\"><input name=\"" . $row[0] . "\" type=\"submit\" value=\"" . $row[0] . "\"</form></td></tr>";
}
echo "</table>";
if (isset($_POST[$row[0]])){
$admin_email = "[email protected]," . $row[2];
mail($admin_email, "Ontvangstbewijs geleverde hardware herverstuurd", "url?id=" . $row [0] , "From:" . "[email protected]");
mysqli_close($conn);
}
?>
[表單發佈時發送提交按鈕的值可能重複](http://stackoverflow.com/questions/22579616/send-value-of-submit-button-when-form-gets-posted) – Ionut
* * WHOA !!! ** ** **從不**將數據庫查詢放入*任何*種編程循環中!這是一個非常明顯的標誌,您尚未對SQL有足夠的瞭解。 – John