Javascript 未捕获类型错误:无法设置属性';行动';未定义的
我知道动作是未定义的,但我不知道如何修复它 这是我的代码,我正在尝试创建一个表来添加、删除和更新所选行。添加工作,但更新和删除我有行动未定义的问题Javascript 未捕获类型错误:无法设置属性';行动';未定义的,javascript,php,html,mysqli,Javascript,Php,Html,Mysqli,我知道动作是未定义的,但我不知道如何修复它 这是我的代码,我正在尝试创建一个表来添加、删除和更新所选行。添加工作,但更新和删除我有行动未定义的问题 <script> function changeRecord() { document.myForm.action='change.php'; document.myForm.submit(); } function deleteRecord()
<script>
function changeRecord()
{
document.myForm.action='change.php';
document.myForm.submit();
}
function deleteRecord()
{
document.myForm.action='delete.php';
document.myForm.submit();
}
</script>
<?php
$servername = "localhost";
$username = "n00832038";
$password = "Fall2015832038";
$db = "n00832038";
$conn = mysqli_connect($servername, $username, $password, $db);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$query = "SELECT * FROM movies ";
$results = mysqli_query($conn, $query);
?>
</head>
<body>
<form action="add.php" method="get">
<table>
<tr>
<td>Movie Title: </td><td><input type='text' name='title'></td>
</tr>
<tr>
<td>Release Type: </td><td><select name='type'>
<option value="theaters">Theaters</option>
<option value="direct">Direct to Video</option>
<option value="television">T.V.</option>
</select>
</td>
</tr>
<tr>
<td>Release Year: </td><td><input type='text' name='year'></td>
</tr>
<tr>
<td>Remake: </td><td><input type='radio' name='remake' value='yes'> Yes</td>
<td><input type='radio' name='remake' value='no'> No</td>
</tr>
<tr>
<td>Based On A Book: </td><td><input type='radio' name='basedbook' value='yes'> Yes</td>
<td><input type='radio' name='basedbook' value='no'> No</td>
</tr>
<tr> <td colspan='2'> <input type='submit' value='Add Record'></td>
</table>
<form action='' name='myForm' method= 'get'>
<?php
echo "<table border='1'>";
while($row = mysqli_fetch_row($results))
{
$MoviesID = $row[0];
$Title = $row[1];
$ReleaseType = $row[2];
$ReleaseYear = $row[3];
$Remake = $row[4];
$BasedonBook = $row[5];
echo "<tr>";
echo "<td><input type='radio' name='movieID' value='$MoviesID'></td>";
echo "<td>$Title</td>";
echo "<td>$ReleaseType</td>";
echo "<td>$ReleaseYear</td>";
echo "<td>$Remake</td>";
echo "<td>$BasedonBook</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($conn);
?>
<input type='button' value='Delete Record' onClick='deleteRecord()'>
<input type='button' value='Update Record' onClick='changeRecord()'>
</form>
</body>
函数changecord()
{
document.myForm.action='change.php';
document.myForm.submit();
}
函数deleteRecord()
{
document.myForm.action='delete.php';
document.myForm.submit();
}
电影名称:
释放类型:
剧院
直接视频
电视
发布年份:
翻拍:是的
不
根据一本书:是的
不
您应该进行以下更改:
JS:
和HTML:
<form action="add.php" method="get" id="myForm">
您是否使用两个表单?是的,一个用于add.php,另一个用于表和按钮。您的问题可能是由嵌套表单引起的,因为
当前嵌套在
中,因为您从未关闭第一个表单。
<form action="add.php" method="get" id="myForm">