PHP操作数据库得到数据进行JSON组装

1.php程序

<?php       
mysql_connect("localhost","root","");     
mysql_select_db("myhospitalv2");       
$q=mysql_query("SELECT * FROM user2 ");      
while($e=mysql_fetch_object($q))           $output[]=$e;           
print(json_encode($output));          
mysql_close();  
?>

2.浏览器返回结果

[{"id":"1","name":"Tony","sex":"1","birthyear":"1999"},{"id":"2","name":"Jim","sex":"1","birthyear":"1987"},{"id":"3","name":"Lucy","sex":"0","birthyear":"1988"},{"id":"4","name":"Tom","sex":"1","birthyear":"1980"}]

3.总结

使用php引擎操作MySQL数据库，若返回结果为记录集，将记录集中记录进行重新存储，如上

$q=mysql_query("SELECT * FROM user2 ");      
while($e=mysql_fetch_object($q))           $output[]=$e;    

然后调用json_encode（）进行JSON数据封装，可得到正确的JSON数据格式，用于不同设备间的数据传输。