我的数据库中有一个项目表,我想要一个删除按钮。
目前它正在运行,但按钮显示了ID号,我希望它只是说"删除"。
我仍然希望它基于joke.ID
删除$result = mysqli_query($con, "SELECT joke.id, joketext, jokedate, name, email FROM joke INNER JOIN author ON authorid = author.id");
echo "<form action='delete1.php' method='post'>
<table border='1'>
<tr>
<th>Joke</th>
<th>Date</th>
<th>Name</th>
<th>Email</th>
<th>Delete</th>
</tr>";
while( $row = mysqli_fetch_array( $result ) ) {
echo "<tr>";
echo "<td>" . $row['joketext'] . "</td>";
echo "<td>" . $row['jokedate'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['email'] . "</td>";
echo "<td><input type='submit' name='deleteItem' value='".$row['id']."' /></td>";
echo "</tr>";
}
echo "</table>";
echo "</form></br>";
mysqli_close($con);
?>
这是delete1.php
if ( isset( $_POST['deleteItem'] ) and is_numeric( $_POST['deleteItem'] ) ) {
// here comes your delete query: use $_POST['deleteItem'] as your id
mysqli_query($con,"DELETE FROM joke WHERE id='$_POST[deleteItem]'");
}
mysqli_close( $con );
header('Location: joke1.php');
exit();
?>
我想按一下按钮说&#34;删除&#34;
答案 0 :(得分:7)
走另一条路。创建一个隐藏的输入元素如下:
echo "<input type='hidden' name='deleteItem' value='".$row['id']."'>";
将现有的更改为
echo "<td><input type='submit' value='Delete' /></td>";
就是这样。
EDIT1:你必须为每个笑话保留一个不同的表格。
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['joketext'] . "</td>";
echo "<td>" . $row['jokedate'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['email'] . "</td>";
echo "<td>";
echo "<form action='delete1.php' method='post'>
echo "<input type='hidden' name='deleteItem' value='".$row['id']."'>";
echo "<input type='submit' value='Delete' />";
echo "</form>";
echo "</td>";
echo "</tr>";
}
答案 1 :(得分:0)
使用简单的方法
echo "<td><input type='button' value='Delete' onclick='javascript:location.href=\'delete1.php?id=" .$row['id']. "\'' /></td>";
<强> delete1.php 强>
if(isset($_GET['id']) and is_numeric($_GET['id']))