Question

这是我的代码。我希望通过从一个数据库中获取名称和家庭服务提供者和客户，首先获取名称和家庭service_provider，然后获取姓名和家庭客户。

  <?php
$id=$fgmembersite->UserID(); 

/* echo "$id"; */


$db_host = 'localhost';
$db_name= 'site';
$db_table= 'action';
$db_user = 'root';
$db_pass = '';


$con = mysql_connect($db_host,$db_user,$db_pass) or die("خطا در اتصال به پايگاه داده");
$selected=mysql_select_db($db_name, $con) or die("خطا در انتخاب پايگاه داده");
mysql_query("SET CHARACTER SET  utf8");

$dbresult=mysql_query("SELECT tablesite.name as service_name,
                              tablesite.family as service_family,
                              tablesite.username,
                              tablesite.phone_number,
                              tablesite.email,
                              action.service_provider_comment,
                              action.customer_comment,
                              action.price,
                              action.date,
                              job_list.job_name,
                              action.ind
                       FROM  $db_table
                       INNER JOIN job_list
                       on job_list.job_id=action.job_id 
                       INNER JOIN tablesite
                       on tablesite.id_user=action.service_provider_id
                       WHERE vote!=''

                       UNION


                       SELECT tablesite.name as customer_name,
                              tablesite.family as customer_family
                       FROM  $db_table
                       INNER JOIN job_list
                       on job_list.job_id=action.job_id 
                       INNER JOIN tablesite
                       on tablesite.id_user=action.customer_id
                       WHERE vote!=''",$con);                      


   $i = 1;

                       while($amch=mysql_fetch_assoc($dbresult))

{?>
  <?php

echo "<form id='form_$i' method='post' action='{$_SERVER['PHP_SELF']}' accept-charset='UTF-8'>\r\n";
echo'<div dir="rtl">';
echo "نام خدمت دهنده: "."&nbsp&nbsp&nbsp".$amch["service_name"]." ".$amch["service_family"]."&nbsp&nbsp&nbsp"."شماره تماس: ".$amch["phone_number"]."&nbsp&nbsp&nbsp"."ایمیل: ".$amch["email"].'<br>'.

"شغل انجام شده: ".$amch["job_name"].'<br>'
."تاریخ انجام عملیات: ".$amch["date"].'<br>'
."هزینه ی کار: ".$amch["price"]." تومان".'<br>'
."توضیحات خدمت دهنده".'<br>'."- ".$amch["service_provider_comment"].'<br>';

echo "نام خدمت گیرنده: ".$amch["customer_name"].$amch["customer_family"];
echo "پاسخ خدمت گیرنده".'<br>'."- ".$amch["customer_comment"].'<hr/>';
}
?>

</fieldset>

运行此代码后我遇到了这个问题：

（！）警告：mysql_fetch_assoc（）要求参数1为资源，在第250行的C：\ wamp \ www \ source \ JobList.php中给出布尔值

表格动作：

包含服务提供商信息的表操作的一部分

SELECT tablesite.name as 'service_name',
                              tablesite.family as 'service_family',
                              tablesite.username,
                              tablesite.phone_number,
                              tablesite.email,
                              action.service_provider_comment,
                              action.customer_comment,
                              action.price,
                              action.date,
                              job_list.job_name,
                              action.ind
                       FROM  $db_table
                       INNER JOIN job_list
                       on job_list.job_id=action.job_id 
                       INNER JOIN tablesite
                       on tablesite.id_user=action.service_provider_id
                       WHERE vote!=''

包含客户信息的表操作的一部分：

SELECT tablesite.name as 'customer_name',
                              tablesite.family as 'customer_family'
                       FROM  action
                       INNER JOIN job_list
                       on job_list.job_id=action.job_id 
                       INNER JOIN tablesite
                       on tablesite.id_user=action.customer_id
                       WHERE vote!=''

UNION两个从一个数据库中选择PHP SQL

0 个答案: