問題描述
這就是我必須將發送器信息構建到 php 文件以在登錄頁面時發送到登錄頁面的方式
this is how I have to build the transmitters infomation on to a php file to send to login page when it is
<form action="http://,,,.dk/activate_updater.php" method="post" name="formular" onsubmit="return validerform ()">
<table border="0">
<tr>
<td id="tb-w_a"><p>Kategori</p></td>
<td>:</td>
<td>
<select name="kategori">
<?php
$query = "SELECT id_katogori, navn FROM kategori";
$result = $mysqli->query($query);
while(list($id_katogori, $navn) = $result->fetch_row())
{
echo "<option value="$id_katogori">$navn</option>";
}
?>
</select>
</td>
</tr>
<tr>
<td id="tb-w_a"><p>Djnavn</p></td>
<td>:</td>
<td><input type="text" name="djnavn"></td>
</tr>
<tr>
<td id="tb-w_a"><p>Facebook</p></td>
<td>:</td>
<td><input type="text" name="facebook"></td>
</tr>
<tr>
<td id="tb-w_a"><p>Pris</p></td>
<td>:</td>
<td><input type="text" name="pris"></td>
</tr>
<tr>
<td id="tb-w_a"><p>Booking Email</p></td>
<td>:</td>
<td><input type="email" name="booking"></td>
</tr>
<tr>
<td id="tb-w_a"><p>Mobil</p></td>
<td>:</td>
<td><input type="text" name="mobil"></td>
</tr>
<tr>
<td id="tb-w_a"><p>Upload Profil</p></td>
<td>:</td>
<td><input type="file" name="profilbillede" /></td>
</tr>
</table>
<textarea name="profiltekst" style="width:500px; height:170px;"></textarea><br />
<input type="submit" value="Godkend brugere" name="godkendt-brugere">
</form>
所以新的頁面文件是這樣的，頁面上有錯誤..
so the new page file is like this where there are errors on the page ..
<?php
if($stmt = $mysqli->prepare('UPDATE `brugere` SET `rank`=2, `katogori`=?, `djnavn`=?, `profilbillede`=?, `profiltekst`=?, `facebook`=?, `pris`=?, `booking`=?, `mobil`=? WHERE `code`=?'))
{
$stmt->bind_param('iiiiiiiis', $katogori, $djnavn, $profilbillede, $profiltekst, $facebook, $pris, $booking, $mobil, $g_code);
//fra input ting ting..
$katogori = $_POST["kategori"];
$djnavn = $_POST["djnavn"];
$profilbillede = $_POST["profilbillede"];
$profiltekst = $_POST["profiltekst"];
$facebook = $_POST["facebook"];
$pris = $_POST["pris"];
$booking = $_POST["booking"];
$mobil = $_POST["mobil"];
$g_code = $_GET["code"];
$stmt->execute();
$stmt->close();
header('http://....dk/ (...));
}
else
{
echo 'Der opstod en fejl i erkl?|ringen: ' . $mysqli->error;
}
?>
但是當它出現在文件/頁面上時會這樣說......
but when it comes onto the file / page says it like this ...
致命錯誤:在第 2 行的/home/jesperbo/public_html/....dk/activate_updater.php 中的非對象上調用成員函數 prepare()
就是這樣寫的
if($stmt = $mysqli->prepare('UPDATE `brugere` SET `rank`=2, `katogori`=?, `djnavn`=?, `profilbillede`=?, `profiltekst`=?, `facebook`=?, `pris`=?, `booking`=?, `mobil`=? WHERE `code`=?'))
我不是最能看到錯誤的人，但如果你看到我不必要的錯誤，因為它不會出現在數據庫中，那就太好了
希望你能幫我解決這個問題！
推薦答案
發生這種情況是因為 $mysqli 不是您認為的對象.'沒有正確實例化 MySQLi 類，或者它沒有建立連接.執行 var_dump($mysqli) 以獲取有關問題的更多信息.
This is happening because $mysqli isn't the object you think it is. Either you didn't instantiate the MySQLi class properly, or it didn't make a connection. Do a var_dump($mysqli) to get more info as to what's going wrong.
這篇關于致命錯誤:在非對象上調用成員函數 prepare()的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！