我有一个包含以下内容的SQL服务器中的一个简单的表:SQL服务器 - ID主键不自动递增
- ID(INT) - 主键(标识规范:是的,标识增量:1)
- 名字 - VARCHAR(50)
- 名字 - VARCHAR(50)
- 电子邮件 - VARCHAR(50)
我设置的HTML表格下方插件ERT的数据到这个非常简单的表,其呈现以下错误:
<?php
require ('connect.inc.php');
if(ISSET($_POST['FirstName'])&&ISSET($_POST['LastName'])&&ISSET($_POST['email'])){
if(!empty($_POST['FirstName'])&&!empty($_POST['LastName'])&&!empty($_POST['email'])){
$FirstName = $_POST['FirstName'];
$LastName = $_POST['LastName'];
$email = $_POST['email'];
$sql_insert = "SET IDENTITY_INSERT test.dbo.clients ON
INSERT INTO test.dbo.clients (id, FirstName, LastName, email)
VALUES('', '$FirstName', '$LastName', '$email')";
$sql_select = "SELECT * FROM test.dbo.clients";
$sql_insert_query = odbc_exec($con, $sql_insert);
}
else{
echo "Please input fields";
}
}
else{
echo "Not set";
}
/*$sql = "INSERT INTO clients()
VALUES()"
*/
?>
<html>
<body>
<table border="2">
<form action="inputform.php" method="POST">
<tr>
<td>First name:</td>
<td><input type="text" name="FirstName"></td>
</tr>
<tr>
<td>Last name:</td>
<td><input type="text" name="LastName"></td>
</tr>
<tr>
<td>e-mail:</td>
<td><input type="text" name="email"></td>
</tr>
<tr>
<td><input type="submit" name="submit" value="submit"></td>
</tr>
</form>
</table>
</body>
</html>
下错误指示ID(PK),不automincrementing:
Violation of PRIMARY KEY constraint 'PK_clients'. Cannot insert duplicate key in object 'dbo.clients'.
附件是示出了图像如何id密钥已配置。
在此先感谢。 J
这就是“SET IDENTITY_INSERT test.dbo.clients ON”所做的。 –