我只是编程方面的初学者,所以我只是通过教程引用我的所有代码。幸运的是,我在youtube上发现了这个在线教程,允许用户使用php在mysql中添加,更新和删除数据。我遵循他的所有指示,我得到了它的工作,但当我添加css时停止了。MySQL - 更新和删除查询不起作用
这不是一个普遍问题,我只是需要一些帮助。如果有人能帮助我,非常感谢。非常感谢。
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "rssfeed";
$connect = mysql_connect($servername, $username, $password, $dbname);
if (!$connect) {
die("Connection failed. Error" . mysql_error());
}
$database = mysql_select_db($dbname, $connect);
if (!$database) {
die("Can't select database");
}
$sql = "SELECT * FROM record";
$data = mysql_query($sql, $connect);
if (isset($_POST['update'])){
$updateQuery = "UPDATE record SET name = '$_POST[name]', url = '$_POST[url]', description = '$_POST[desc]' WHERE name = '$_POST[hidden]'";
mysql_query($updateQuery, $connect);
header("Location: maintenance.php");
};
if (isset($_POST['delete'])){
$deleteQuery = "DELETE FROM record WHERE name = '$_POST[hidden]'";
mysql_query($deleteQuery, $connect);
header("Location: maintenance.php");
};
if (isset($_POST['add'])){
$addQuery = "INSERT INTO record (name, url, description) VALUES ('$_POST[iName]', '$_POST[iUrl]', '$_POST[iDesc]')";
mysql_query($addQuery, $connect);
header("Location: maintenance.php");
};
echo "<div class=center>
<table id=myTable border=1>
<tr>
<th> Name </th>
<th> URL </th>
<th> Description </th>
</tr>";
while($record = mysql_fetch_array($data)) {
echo "<form method=post action=maintenance.php>";
echo "<tr>";
echo "<td>" . "<input type=text name=name id=name value=" . $record['name'] . " </td>";
echo "<td>" . "<input type=text name=url id=url value=" . $record['url'] . " </td>";
echo "<td>" . "<textarea rows=1 cols=50 wrap=physical name=desc id=desc>" . strip_tags($record['description']) . "</textarea></td>";
echo "<input type=hidden name=hidden value=" . $record['name'] . ">";
echo "<td>" . "<input type=submit name=update id=update value=update" . " </td>";
echo "<td>" . "<input type=submit name=delete id=delete value=delete" . " </td>";
echo "</tr>";
echo "</form>";
}
echo "</table>";
echo "<table border=1>";
echo "<form method=post action=maintenance.php>";
echo "<tr>";
echo "<td><input type=text name=iName></td>";
echo "<td><input type=text id=url name=iUrl></input></td>";
echo "<td><textarea rows=1 cols=50 name=iDesc></textarea></td>";
echo "<td>" . "<input type=submit name=add id=add value=add" . " </td>";
echo "</tr>";
echo "</form>";
echo "</table>";
echo "</div>";
mysql_close($connect);
?>
你有2个提交的1表格,PHP不喜欢那样。我建议你看看这里:http://stackoverflow.com/questions/9054363/form-with-two-button-submit-with-different-value – Matheno 2014-12-04 16:59:58
@ User8889仍然无法工作:( – Kamilah 2014-12-04 17:09:36
然后尝试例如使用url而不是像这样的按钮,然后获取像这样的动作:$ _GET ['action'],然后执行您的查询 – Matheno 2014-12-05 08:37:20