我尝试了什么
$sql="select * from project_status where node_id=".$vid." and ". strtolower($field_progress_flags)."!='' and ". strtolower($field_progress_flags)." IS NOT NULL";
$query=$sql;
$array = array();
$result = mysql_query($query);
while ($object = mysql_fetch_array($result)) {
$array[] = $object;
}
mysql_freeresult($result);
echo "<pre>";
print_r($array);
实际的sql查询是select * from project_status where node_id=385 and draft!='' and draft IS NOT NULL
这里没有得到任何数据 这里有什么问题?
答案 0 :(得分:0)
一些R&amp; D得到解决方案如下:
$sql="select * from project_status where node_id=".$vid." and ". strtolower($field_progress_flags)."!='' and ". strtolower($field_progress_flags)." IS NOT NULL";
$result = db_query($sql);
while($record = $result->fetchAssoc()){
$data[]=$record;
}
它的作品对我来说