多表查询碰到的难题
多表查询遇到的难题
要实现 一个需求:通过硬件设备向服务器发出一个请求,
请求带3个参数:
1.用户编码
2.经度
3.纬度
服务器操作:
1.通过经纬度去百度地图获取具体地理地址
2.通过用户编码进行多表查询。获取用户的手机号码。
3.通过取得的地址使用短信网关发送到查询到的手机中
4.短信网关返回发送成功则保存相关数据到mysql
现在问题出在多表查询没有返回手机号码。但是在操作面板则能查询到。是什么问题
流程图:
php代码:
<?php<br /> include_once "config.php";<br /> header("Content-type: text/html; charset=utf8");<br /> $mysql=mysqli_connect($db_host,$db_user,$db_password,$db_name);<br /> mysql_select_db($db_name,$mysql);<br /> mysql_query("SET NAMES 'utf8'");<br /> mysql_query("set character_set_client='utf8'");<br /> mysql_query("set character_results='utf8'");<br /> mysql_query("set collation_connection='utf8'");<br /> $user=$_GET['user'];<br /> $lat=$_GET['lat'];<br /> $lng=$_GET['lng'];<br /><br /> //获取地理地址<br /> date_default_timezone_set('PRC');<br /> $into_time=Date("Y-m-d H:i:s");<br /> $url="http://api.map.baidu.com/geocoder/v2/?ak=E6f2d2e9e662917d0e6eee74ced3db24&callback=renderReverse&location=".$lat.",".$lng."&output=json&pois=1";<br /><br /><br /><br /> $json=file_get_contents($url);<br /> $xjson = substr($json,29,-1);<br /> $point = $lat.'_'.$lng;<br /> $data=json_decode($xjson,true);<br /> $dress = $data['result']['formatted_address'];<br /> file_put_contents("localjson.txt",$data.'--dress--'.$dress);<br /><br />// $g="select akw_user.mobile from akw_user where akw_user.id in (select user_group.userid from user_group where user_group.sqlid='".$user."')";<br /> //多表查询<br /> $g="select akw_user.mobile from akw_user where akw_user.id in (select user_group.userid from user_group where user_group.sqlid = '".$user."')";<br /> $ids= mysql_query($g);<br /> [email protected]_fetch_assoc($ids);<br /> $f=$n['id'];<br /> echo '--f--'.$f.'--g--'.$g;<br />?>
------解决思路----------------------
晕死,我是让你修改SQL语句,增加对id字段引用:
新增后提示出错:select akw_user.mobile from akw_user where akw_user.id in (select user_group.userid from user_group where user_group.sqlid = '1409203388')<br />Access denied for user 'root'@'localhost' (using password: NO)Copy after login
------解决思路----------------------
<br />$g="select akw_user.mobile,akw_user.id from akw_user where akw_user.id in (select user_group.userid from user_group where user_group.sqlid = '".$user."')";<br /> $ids= mysql_query($g);<br /> [email protected]_fetch_assoc($ids);<br /> $f=$n['id'];<br />
改这样试试。
------解决思路----------------------
我的#15 就被你忽视了
------解决思路----------------------
mysqliiiiiii_connect
mysql_query
说来惭愧一直用框架,加了i的能通用吗
------解决思路----------------------
把你修改后的代码贴出来看看。

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