php mysql 多表查询 查询不出结果?
我的代码如下:
- PHP code
<!-- Code highlighting produced by Actipro CodeHighlighter (freeware) http://www.CodeHighlighter.com/ --> <?php /*connect database*/ $link=mysql_connect("192.168.XX.XX:3306","root","root") or die("Could not connect"); print "Connected successfully"; mysql_select_db("jiradb") or die ("Could not select database"); mysql_query("set names 'utf8'"); /*execte SQL query*/ $query="SELECT b.pname,a.pkey,a.summary,a.assignee FROM jiraissue a, issuetype b,project c WHERE a.project=c.ID and b.ID=a.issuetype and c.pname='TEST';"; $result=mysql_query($query,$link) or die("Query failed"); /*print result in html*/ MySQL_num_rows($result); while($row=mysql_fetch_array($result)) {echo $row[0]; } /*release resource */ mysql_free_result($result); /*close link*/ mysql_close($link); ?>
结果就只显示
Connected successfully
如果用
- PHP code
<!-- Code highlighting produced by Actipro CodeHighlighter (freeware) http://www.CodeHighlighter.com/ -->$query="select * from jiraissue";
就会显示相应的内容
php mysql多表查询该怎么写?上面哪里出错了?
------解决方案--------------------
没有表结构和真实数据,没有真相。
你装一个phpmyadmin,在里面执行看看。
------解决方案--------------------
echo $row[0];
这里改成 var_dump($row); 看看是否返回数据集正确
------解决方案--------------------
c.pname='TEST';";
不用把这个分号也复制进来
------解决方案--------------------
给点提示:
select co_balance,co_isstaff,staff_group_id,count from
(select co_balance,co_isstaff from dt_co where co_id=156) a,
(select staff_group_id from dt_staff where co_id = 156) b,
(select count(*) as count from dt_staff where staff_shouyetuijian = 1 and co_id = 156) c
这是链接用了
dt_co 和 dt_staff 做的查询
------解决方案--------------------
把你的sql指令在phpmyadmin执行能查出结果吗?
------解决方案--------------------
你把你这个语句拿到数据库里面跑一下不就知道有不有问题了
------解决方案--------------------
先用mysql_error()和mysql_errno()看看错误提示!
------解决方案--------------------
把php文件的编码统一为数据库的编码试试看。 可能是编码不一致导致的误差。
------解决方案--------------------
- PHP code
$result=mysql_query($query,$link) or die("Query failed"); echo "1 ".mysql_error(); /*print result in html*/ MySQL_num_rows($result); echo "2 ".mysql_error(); while($row=mysql_fetch_array($result)) {echo $row[0]; echo "3 ".mysql_error(); } <div class="clear"> </div>

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