我有一個idnumber從資料庫中獲取資料并將其放入選擇框中的代碼。
如果我idnumber從選擇框中選擇一個并按下確認按鈕，我該如何撰寫代碼以將資料庫中名為“確認”的列從“否”更新為“是”？
下面的代碼是選擇框的代碼
<select>
<?php
$mysqli = new mysqli("localhost","root","","volunteer");
/* check connection */
if (mysqli_connect_errno()){
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$query = "SELECT * FROM registrations";
$result1 = mysqli_query($mysqli,$query);
while($rows = mysqli_fetch_assoc($result1)){
echo "<option>" . $rows['idnumber'] . "</option>" ;
}
?>
</select>
<form action="confirm.php" method="post">
<button type="submit" name="confirm-btn">Confirm</button>
</form>
uj5u.com熱心網友回復：
- 將您的選擇輸入放在表單標簽下
- 為選擇框命名：
- 點擊確認按鈕。
- 將代碼寫入：confirm.php
if (!empty($_POST['registrations'])) {
if (!empty($_POST['registrations'])) {
$sql = "UPDATE TABLE_NAME SET COLUMN_NAME=".$_POST['registrations']." where YOUR_CONDITION";
if ($conn->query($sql) === TRUE) {
echo "Record updated successfully";
} else {
echo "Error updating record: " . $conn->error;
}
uj5u.com熱心網友回復：
表單輸入：
<form action="confirm.php" method="post">
<select name="idnumber">
<?php
$mysqli = new mysqli("localhost","root","","volunteer");
/* check connection */
if (mysqli_connect_errno()){
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$query = "SELECT * FROM registrations";
$result1 = mysqli_query($mysqli,$query);
while($rows = mysqli_fetch_assoc($result1)){
echo "<option value=". $rows['idnumber'] .">" . $rows['idnumber'] . "</option>" ;
}
?>
</select>
<button type="submit" name="confirm-btn">Confirm</button>
</form>
將代碼寫入：confirm.php
$mysqli = new mysqli("localhost","root","","volunteer");
/* check connection */
if (mysqli_connect_errno()){
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$sql = "UPDATE TABLE_NAME SET COLUMN_NAME=".$_POST['idnumber']." where YOUR_CONDITION";
if ($conn->query($sql) === TRUE) {
echo "Updated successfully";
} else {
echo "Updating record: " . $conn->error;
}
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