PHP一直说无效的列名
PHP一直说无效的列名
提问于 2015-09-17 18:27:06
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，我得到以下结果：
INSERT INTO delivery_comments (commentor, comment, jobno)
VALUES (frontdesk,bvhjfhj, 85450) 然后我得到了错误：
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'. ) )注意，回显的insert语句是第一个。这是正确的。然后，错误的其余部分一直告诉我要插入的数据是无效的列名。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。
回答 2
Stack Overflow用户
回答已采纳
发布于 2015-09-18 14:50:34
这就是我最后使用的：
$query = "INSERT INTO delivery_comments (commentor, comment, jobno) VALUES ('$txtCommentor','$txtComment', '$jobno') ";Stack Overflow用户
发布于 2015-09-17 20:36:17
<?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);?>以上是您查询的净化版本。
你应该在这方面取得更大的成功
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/32637505
复制相关文章
相似问题
腾讯云开发者
