-3
所以我的代碼工作完美,我能夠刪除行。在我用新數據更新數據庫之後,我想刪除它們以再次測試代碼,並且我已啓用刪除。 我不知道是否在xampp問題,因爲我的代碼工作。不知道該怎麼努力重新工作。我想刪除行表格datebase,我的代碼工作和刪除行然後停止,我不改變任何代碼?
這是我的delete.php文件:
<?php
//connecting to base
// include "spoj.php";
$conn = mysqli_connect('localhost', 'root', '', 'kolegij') or die('Connection error!');
//Defining rows
$id = $_GET['id'];
$query = "DELETE FROM kolegij WHERE id = '$id'";
//delete rows
mysqli_query ($conn, $query);
if (mysqli_affected_rows($conn)==1) {
//if rows delete
?>
<strong>Kolegij je izbrisan</strong><br /><br />
<?php
} else {
//rows not deleted
?>
<strong>Niste uspjeli izbrisati kolegij</strong><br /><br />
<?php
}
?>
這是我的html代碼:
<html>
<head>
<title>Evidencija Kolegija</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
</head>
<body>
<h1>Evidencija Kolegija</h1>
<hr>
<a href="kolegiji.php">Dodaj novi kolegij</a>
<hr>
<?php
//connect to datebase
// include "spoj.php";
$conn = mysqli_connect('localhost', 'root', '', 'kolegij') or die('Connection error!');
//ispis svih studenata u bazi sortiranih po id-u
$sql_upit="SELECT * FROM kolegij ORDER BY id ASC";
if (!$q=mysqli_query($conn, $sql_upit))
{
echo "Nismo uspjeli u čitati kolegije iz baze"."<br>". mysqli_query($conn, $sql_upit);
die();
}
//if rows is === 0 then there is no datea in datebase
if (mysqli_num_rows($q)==0)
{
echo "Nema kolegija u bazi.";
}
else {
echo '<table width="1000" border="1px" cellpadding="2" cellspacing="2">';
echo '<tr><td><b>Id</b></td>';
echo '<td><b>Predmeti</b></td>';
echo '<td><b>Profesor</b></td>';
echo '<td><b>Status</b></td>';
//sve dok ima kolegija u bazi
while ($redak=mysqli_fetch_array($q))
{
echo '<tr><td>'.$redak["id"].'</td>';
echo '<td>'.$redak["Predmeti"].'</td>';
echo '<td>'.$redak["Profesor"].'</td>';
echo '<td>'.$redak["Status"].'</td>';
echo '<td><a href="izmjeni.php?id='.$redak["id"].' ">Uredi kolegij</a></td>';
//刪除按鈕 回聲 'Obriši '; } echo''; } ?>
這將是非常有用的,如果你能翻譯您可以將您的註釋文本翻譯成英文 – Koen