Question

我有一个存储活动类型(swimming, running, soccer, tennis and basketball)under task_cstm的数据库。

另一个数据库存储任务中存储的salesman name，date_start和date_due。

当我在phpMyAdmin上运行时，

SELECT COUNT( tasks_cstm.activity_type_c ) FROM tasks_cstm
LEFT JOIN tasks ON tasks.id = tasks_cstm.id_c
where tasks_cstm.activity_type_c ="swimming" and assigned_user_id="abcdefg"

我得到了我想要的结果：因为它会为我计算游泳总数。

但是我将其放入 .PHP 文件中，该文件允许用户选择SALESMAN和DATE以查看销售人员在此期间未展示的活动数量：

    <?php
$result0 = $GLOBALS['db']->query("SELECT tasks_cstm.activity_type_c,
    COUNT( tasks_cstm.activity_type_c ) FROM tasks_cstm
    LEFT JOIN tasks ON tasks.id = tasks_cstm.id_c
    WHERE activity_type_c='swimming' AND date_start>='$st'
    AND date_due<='$dt' AND assigned_user_id='$salesman' AND t.id=tc.id_c")

$result1 = $GLOBALS['db']->query("SELECT tasks_cstm.activity_type_c,
    COUNT( tasks_cstm.activity_type_c ) FROM tasks_cstm
    LEFT JOIN tasks ON tasks.id = tasks_cstm.id_c
    WHERE activity_type_c='running' AND date_start>='$st'
    AND date_due<='$dt' AND assigned_user_id='$salesman' AND t.id=tc.id_c")

    $result2 = $GLOBALS['db']->query("SELECT tasks_cstm.activity_type_c,
    COUNT( tasks_cstm.activity_type_c ) FROM tasks_cstm
    LEFT JOIN tasks ON tasks.id = tasks_cstm.id_c
    WHERE activity_type_c='tennis' AND date_start>='$st'
    AND date_due<='$dt' AND assigned_user_id='$salesman' AND t.id=tc.id_c")

    $result3 = $GLOBALS['db']->query("SELECT tasks_cstm.activity_type_c,
    COUNT( tasks_cstm.activity_type_c ) FROM tasks_cstm
    LEFT JOIN tasks ON tasks.id = tasks_cstm.id_c
    WHERE activity_type_c='soccer' AND date_start>='$st'
    AND date_due<='$dt' AND assigned_user_id='$salesman' AND t.id=tc.id_c")

    $result4 = $GLOBALS['db']->query("SELECT tasks_cstm.activity_type_c,
    COUNT( tasks_cstm.activity_type_c ) FROM tasks_cstm
    LEFT JOIN tasks ON tasks.id = tasks_cstm.id_c
    WHERE activity_type_c='basketball' AND date_start>='$st'
    AND date_due<='$dt' AND assigned_user_id='$salesman' AND t.id=tc.id_c")


    echo "<tr><td>" . $result0 . "</td><td>" . $result1 . "</td><td>" . $result2 . "</td><td>" . $result3 . "</td><td>" . $result4 . "</td></tr>";

有人可以帮助我并给我一些指导吗？

PS：由于我无法编辑我以前的帖子，我在这里发布。

谢谢。

Answer 1

每次查询后都缺少半冒号。

$result2 = $GLOBALS['db']->query("SELECT tasks_cstm.activity_type_c,
  COUNT( tasks_cstm.activity_type_c ) FROM tasks_cstm
  LEFT JOIN tasks ON tasks.id = tasks_cstm.id_c
  WHERE activity_type_c='tennis' AND date_start>='$st'
  AND date_due<='$dt' AND assigned_user_id='$salesman' AND t.id=tc.id_c");

Answer 2

我不知道你在那里有$ GLOBALS [＆＃39; db＆＃39;]对象。但是通常来说，方法query（）仍然不会产生结果。你必须执行（）它然后它会产生一个数组。然后迭代数组并回显值。

$query  = $db->query('SQL select query goes here');
$result = $query->execute(); 
foreach($result as $s) {
   echo $r[0];
   echo $r[1];
   ...
}

Answer 3

谢谢大家的帮助。我得到了我想要的结果。

我做的是添加 ... SELECT tasks_cstm.activity_type_c，COUNT（tasks_cstm.activity_type_c）每个结果的AS TOTALCOUNT（0-4）。

并且对于每个结果（0-4）我需要另外一个fetchbyassoc，所以我输入

$r0= $GLOBALS['db']->fetchByAssoc($result0);
$r1= $GLOBALS['db']->fetchByAssoc($result1);
$r2= $GLOBALS['db']->fetchByAssoc($result2);
$r3= $GLOBALS['db']->fetchByAssoc($result3);
$r4= $GLOBALS['db']->fetchByAssoc($result4);

我也回应了这个

echo "<tr>
<td>".$r0['totalCount0']."</td>
<td>".$r1['totalCount1']."</td>
<td>".$r2['totalCount2']."</td>
<td>".$r3['totalCount3']."</td>
<td>".$r4['totalCount4']."</td>
</tr>";

是的，我得到了我需要的答案。非常感谢所有UR帮助（^ 3 ^）

如何回显/打印声明

3 个答案: