php – 根据标题将csv数据导入数据库
发布时间:2020-12-13 16:22:24 所属栏目:PHP教程 来源:网络整理
导读:这是我的脚本. if (isset($_POST['submit'])) { if (is_uploaded_file($_FILES['filename']['tmp_name'])) { echo "h1" . "File ". $_FILES['filename']['name'] ." uploaded successfully." . "br/h1"; echo "h2Displaying contents:/h2"; readfile($_FILES
这是我的脚本.
if (isset($_POST['submit'])) { if (is_uploaded_file($_FILES['filename']['tmp_name'])) { echo "<h1>" . "File ". $_FILES['filename']['name'] ." uploaded successfully." . "<br></h1>"; echo "<h2>Displaying contents:</h2>"; readfile($_FILES['filename']['tmp_name']); echo "<br>"; echo $headers; } $handle = fopen($_FILES['filename']['tmp_name'],"r"); $header = fgetcsv($handle); while(! feof($handle)){ while (($data = fgetcsv($handle,1000,",")) !== FALSE) { $import="INSERT into CSVtest($header[0],$header[1],$header[2],$header[3],$header[4]) values ('$data[0]','$data[1]','$data[2]','$data[3]','$data[4]')"; mysql_query($import) or die(mysql_error()); } } fclose($handle); echo "Done"; } 这是我根据标题将数据插入数据库的方式,即使它与表列的排列方式不同. 实际上我希望有一个函数,即使csv中有很多列,表中不存在,它仍然可以选择正确的列并插入表中. 提前致谢. 解决方法
您可能希望使用mysql_fetch_field来获取当前列,然后遍历每个标头以确保在构建查询时它存在于表中.
http://php.net/manual/en/function.mysql-fetch-field.php (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |