所以我目前遇到了我的代码问题,我试图改变的csv文件没有填充我的SQL数据库中的数据。 csv文件正在被改变,因为每次运行我的提交按钮时,csv文件会与最新提交同时更新。因此,我怀疑我的问题与我与sql数据库通信的方式或数据被解析为csv文件的方式有关。我的代码链接如下。谢谢你的时间和期待。注意我没有收到任何错误。
<?php
if(isset($_POST['submit']))
{
$db = get_db_connection('swcrc');
$db->connect();
$filename = 'uploads/testfile.csv';
$fp = fopen($filename,"w");
$db -> query("SELECT * FROM dbo.Response");
$q = $db;
$result = mysql_query($q) or die ('SQL Error :: '.mysql_error());
$num_rows = $result->num_rows;
$row = mysql_fetch_assoc($q);
$seperator = "";
$comma = "";
foreach($row as $name => $value)
{
$seperator .= $comma . '' .str_replace('','""',$name);
$comma = ",";
}
$seperator = "\n";
echo $seperator;
fputs($fp,$seperator);
mysql_data_seek($q,0);
while($row = mysql_fetch_assoc($q))
{
$seperator = "";
$comma = "";
foreach($row as $name => $value)
{
$seperator .= $comma . '' .str_replace('','""',$value);
$comma = ",";
}
$seperator = "\n";
fputs($fp,$seperator);
}
fclose($fp);
}
?>
新代码:
$conn = get_db_connection('swcrc');
$conn->connect();
$sql = query("SELECT [request_ID], [user_ID], [complete], [comment], [response_date], [animalClass],[animalCommon], [bioSurveyStatus] FROM dbo.Response");
$results = $conn->query($sql);
$filename = 'uploads/testfile.csv';
$handle = fopen($filename, 'w+');
fputcsv($handle, array('request_ID','user_ID','complete','comment','response_date','animalClass',
'animalCommon','bioSurveyStatus'));
foreach($result as $row)
{
fputcsv($handle, array($row['request_ID'], $row['user_ID'],$row['complete'],$row['comment'],
$row['response_date'],$row['animalClass'],$row['animalCommon'],
$row['bioSurveyStatus']));
}
fclose($handle);