输出

Question

我在mysql中执行下面的查询，但它不起作用。给出以下错误，

警告：mysql_num_rows（）期望参数1为资源，第228行/usr/share/info/.info/php/userutil.php中给出布尔值

这是我的代码：

  echo $con = mysql_connect('localhost', 'root', 'root123','qhutm');
  if(!$con){
    echo 'connection failed';
   }else{

    echo 'conn successful';
  }
    $query ="SELECT no_of_license 
             FROM Service_Master AS s,Service_Details AS d
             WHERE s.service_id=d.service_id 
             AND s.description='Internet Access' ";
    $result = mysql_query($query,$con)
    print_r($result);
    echo $result.'--pra';
    echo mysql_num_rows($result);

输出

conn successful

如果我在mysql上回显并执行相同的查询，它可以正常工作如果我回复$con，则会给出Resource id #126

Answer 1

也许你使用普通加入？

select no_of_license 
from   Service_Master as s
join   Service_Details as d on s.service_id = d.service_id 
where  s.description = 'Internet Access'

Answer 2

尝试JOIN喜欢

"SELECT no_of_license 
         FROM Service_Master AS s,
         LEFT JOIN Service_Details AS d
         ON s.service_id=d.service_id 
         WHERE s.description='Internet Access'

Answer 3

$query ="SELECT XXXXX FROM `Service_Master` s, `Service_Details` d WHERE s.service_id=d.service_id  AND s.description='Internet Access' ";   


replace it 
and also XXXXX  if no_of_license field in Service_Master this table write  s.`no_of_license`
else if this field  is in Service_Details than write d.`no_of_license`
i think its work fine

Answer 4

您尚未选择数据库，请使用mysql_select_db（）

您的连接应该是这样的

<?php

$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
    die('Not connected : ' . mysql_error());
}

$db_selected = mysql_select_db('foo', $link);
if (!$db_selected) {
    die ('Can\'t use foo : ' . mysql_error());
}
?>

Answer 5

您在创建数据库连接后忘记选择数据库请在创建与数据库的连接后添加以下行。

mysql_select_db（'qhutm'，$ con）;

Mysql查询失败，一切看起来都很好

输出

5 个答案: