$sql5 = "select id,name,type,size,location,hremail from company"; $result5 = mysql_query($sql5); while($row5 = mysql_fetch_array($result5)) { $companyA[] = $row5; } $jsoncompany = json_encode($companyA); echo $jsoncompany;
21,893
社区成员
140,347
社区内容
加载中
试试用AI创作助手写篇文章吧