Question

这是我的代码：

<?php $txtCommentor = $_POST['txtCommentor'] //from form?>
<?php $txtComment = $_POST['txtComment'] //from form?>
<?php $jobno = $_GET['jobno'] //from URL?>

<?php
//Query and connect
$query = "INSERT INTO delivery_comments (commentor, comment, jobno) VALUES ($txtCommentor,$txtComment, $jobno) ";?>
<?php $results = sqlsrv_query($conn, $query);?>
<?php echo $query?>
<?php
if( $results === false ) {
     die( print_r( sqlsrv_errors(), true));
}
?>

当我运行页面时，我得到了以下结果：

INSERT INTO delivery_comments (commentor, comment, jobno) 
VALUES (frontdesk,bvhjfhj, 85450)

然后我收到错误：

Array ( [0] => Array ( [0] => 42S22 [SQLSTATE] => 42S22 [1] => 207 [code] => 207 [2] => [Microsoft][ODBC Driver 11 for SQL Server][SQL Server]Invalid column name 'frontdesk'. [message] => [Microsoft][ODBC Driver 11 for SQL Server][SQL Server]Invalid column name 'frontdesk'. ) [1] => Array ( [0] => 42S22 [SQLSTATE] => 42S22 [1] => 207 [code] => 207 [2] => [Microsoft][ODBC Driver 11 for SQL Server][SQL Server]Invalid column name 'bvhjfhj'. [message] => [Microsoft][ODBC Driver 11 for SQL Server][SQL Server]Invalid column name 'bvhjfhj'. ) )

请注意，echoed insert语句是第一个。这是对的。然后剩下的错误告诉我要插入的数据是一个无效的列名。这没有任何意义。该信息不应该是列名，而是插入到列中的数据。

Answer 1

<?php $txtCommentor = $_POST['txtCommentor'] //from form?>
<?php $txtComment = $_POST['txtComment'] //from form?>
<?php $jobno = $_GET['jobno'] //from URL?>

<?php
$params = array( $txtCommentor, $txtComment, $jobno);
//Query and connect
$query = "INSERT INTO delivery_comments (commentor, comment, jobno) VALUES (?,?,?) ";?>
<?php $results = sqlsrv_query($conn, $query,$params);?>

以上是您查询的已清理版本。

你应该取得更大的成功

Answer 2

这是我最终使用的：

$query = "INSERT INTO delivery_comments (commentor, comment, jobno) VALUES ('$txtCommentor','$txtComment', '$jobno') ";

PHP一直说无效的列名

2 个答案: