所以這是管理員頁面的一部分。 若干修改後,我來到了這一點,這是最好的我能得到不幸的是:循環遍歷選定的用戶並動態填充下拉列表
<?php
$sql = "SELECT name
FROM users";
$result = mysqli_query($connection, $sql);
$users = array();
while ($row = mysqli_fetch_assoc($result))
{
$users[] = $row["name"];
}
echo ('</br><form action="" method="POST" style="display:inline!important"></br>
<select name="deleteUser">
<option value="' . $users[1] . '">' . $users[1] . '</option>' .
'<option value="' . $users[2] . '">' . $users[2] . '</option>' .
'<option value="' . $users[3] . '">' . $users[3] . '</option>' .
'<option value="' . $users[4] . '">' . $users[4] . '</option>' .
'<input class="w3-bar-item w3-button w3-hide-small w3-hover-green" type="submit" value="Delete" name="delete"></select></form>');
if (isset($_POST["delete"]))
{
$sql = "DELETE
FROM users
WHERE name = '" . $_POST["deleteUser"] . "'";
mysqli_query($connection, $sql);
}
?>
我試着寫一個循環來生成選擇選項是這樣的:
<?php
$sql = "SELECT name
FROM users";
$result = mysqli_query($connection, $sql);
$users = array();
while ($row = mysqli_fetch_assoc($result))
{
$users[] = $row["name"];
}
echo ('</br><form action="" method="POST" style="display:inline!important"></br>
<select name="deleteUser">');
for ($i = 0; $i < count($users); $i++)
{
echo('<option value="' . $users[$i] . '">' . $users[$i] . '</option>');
}
echo('<input class="w3-bar-item w3-button w3-hide-small w3-hover-green" type="submit" value="Delete"></select></form>');
if (isset($_POST["delete"]))
{
$sql = "DELETE
FROM users
WHERE name = '" . $_POST["deleteUser"] . "'";
mysqli_query($connection, $sql);
}
?>
但瀏覽器沒有按不會渲染它(例如Chrome),也許是因爲它不能在部分?喜歡等待循環完成,然後以表單的最後一部分結束。無論如何,我怎麼能使它工作?感謝隊友......
你正在得到什麼錯誤? – lalithkumar
也許試用foreach?像這樣'foreach(mysqli_fetch_assoc($ result)as $ val){echo'
@小開發者是的。感謝您的建議,它的工作原理。 – udarH3