有一个提交“更新”的表单。表单已从数据库中设置值。它有效(专辑总是改变)。但是如果我更改表单中的值并不重要,我单击“更新”,它将抛出
致命错误:在布尔值中调用成员函数fetch_assoc() 第35行/www/doc/www.mudreslova.sk/www/admin/edit-galeriu.php
您可以尝试链接“http://www.mudreslova.sk/admin/edit-galeriu.php?url1=fsdfds”,它现在正在运行。 请问代码有什么问题?
<?php
$titulok='Edit ALBUM';
include('header.php');
require_once('connect.php');
if (!empty($_POST["submit"]) && isset($_GET[url1])) { // all entries are required in FORM
$sql="
UPDATE gallery
SET namesk='{$_POST[namesk]}', nameen='{$_POST[nameen]}', descriptionsk='{$_POST[descriptionsk]}',
descriptionen='{$_POST[descriptionen]}', date='{$_POST[date]}', url1='{$_POST[url1]}'
WHERE url1='{$_GET[url1]}';
UPDATE photos
SET url1_gallery='{$_POST[url1]}'
WHERE url1_gallery='{$_GET[url1]}';
";
$success = rename("../photos/$_GET[url1]","../photos/$_POST[url1]");
if ($mysqli->multi_query($sql) && $success===true) {
$messEditAlb="Album was changed :)";
} else {
$messEditAlb="Error, Album wasn't changed.";
}
$_GET["url1"] = $_POST["url1"];
}
?>
<div id="mobile-margin">
<div class="container center">
<h1 class="tshadow">Editovať ALBUM</h1>
<p><?php echo $messEditAlb ?></p>
<?php
if (isset($_GET[url1])) {
$result = $mysqli->query("SELECT * FROM gallery WHERE url1='$_GET[url1]'");
while ($row = $result->fetch_assoc()) { // line 35!
echo
"
<form action='' method='post'>
<div class='form-fotky widexs'>
<label for='namesk' class='left'>Názov albumu - SK:</label>
<input type='text' name='namesk' id='namesk' value='$row[namesk]' required>
</div>
<div class='form-fotky widexs'>
<label for='url1' class='left'>url adresa: napr. michalsky-jarmok2</label>
<input type='text' name='url1' id='url1' value='$row[url1]' required>
</div>
<div class='form-fotky widexs'>
<input type='submit' name='submit' value='Update' class='pure-button pure-button-primary bggreen'>
</div>
</form>";
}};
include('footer.php'); ?>