php-获取2个数组之间的差

概述我有2个搜索查询-一个将显示最近7天的内容.另一个将显示2周前的内容.两者都很好.但是我想从第一个查询中获取结果,并从第二个查询中获取差异.然后显示带有差异的第一个查询.$result_account = $db->query(' SELECT nid , COUNT(cat) AS qty , dte , descrip

我有2个搜索查询-一个将显示最近7天的内容.另一个将显示2周前的内容.
两者都很好.但是我想从第一个查询中获取结果,并从第二个查询中获取差异.然后显示带有差异的第一个查询.

$result_account = $db->query("SELECT nID,COUNT(cat) AS qty,dte,descript,cat,name,user   FROM clIEnt_note AS cn   JOIN clIEnt_note_tag_items AS cnti     ON cnti.note_ID = cn.nID   JOIN clIEnt_note_Tags AS cnt     ON cnt.tag_ID = cnti.tag_ID  WHERE dte >= DATE_SUB(CURDATE(),INTERVAL 7 DAY)    AND name NOT liKE 'Resolution%'  GROUP     BY cat  ORDER     BY qty DESC  liMIT 5");       if($count_account = $result_account->num_rows) {               while($row = $result_account->fetch_object()){          echo "<tr>";          echo "<td><h6>".$row->cat."</h6></td><td><h3 class='text-primary'>".$row->qty."</h3></td>";          echo "</tr>";          }       }$result_prevIoUs = $db->query("SELECT nID,user FROM clIEnt_note AS cn JOIN clIEnt_note_tag_items AS cnti ON cnti.note_ID = cn.nID JOIN clIEnt_note_Tags AS cnt ON cnt.tag_ID = cnti.tag_ID WHERE (dte BETWEEN DATE_SUB(CURDATE(),INTERVAL 21 DAY) AND DATE_SUB(CURDATE(),INTERVAL 14 DAY)) AND name NOT liKE 'Resolution%' GROUP BY cat ORDER BY qty DESC liMIT 5");    if($count_prevIoUs = $result_prevIoUs->num_rows) {            while($row_p = $result_prevIoUs->fetch_object()){          echo "<tr>";            echo "<td><h6>".$row_p->cat."</h6></td><td><h3 class='text-primary'>".$row_p->qty."</h3></td>";          echo "</tr>";            }    }

第一个查询将导致：

category   - QtyBaseball   - 45Football   - 33Soccer     - 21Hockey     - 7Basketball - 3

第二个查询将导致：

category   - QtyBasketball - 38Soccer     - 28Hockey     - 16Football   - 12Baseball   - 12

现在我要这样显示

category   - Qty DifferenceBaseball   - 45  +33Football   - 33  +21Soccer     - 21  -7Hockey     - 7   -9Basketball - 3   -35

最佳答案我要做的是创建另一个数组,该数组在键-值的基础上保存来自第一个查询的值,而在下一个查询中,仅从数组中相应的条目中减去值.

我认为这是使用2个查询的最简单,最快的方法.

注意：我没有测试此代码.

注意2：我假设第一类运动缺失将被计为0

注意3：我编写的代码仅用于显示计算差异的方法.让我知道是否应该更新它,以提供您在那里所需的完全相同的输出.

$results = array();$result_account = $db->query("SELECT nID,user FROM clIEnt_note AS cn JOIN clIEnt_note_tag_items AS cnti ON cnti.note_ID = cn.nID JOIN clIEnt_note_Tags AS cnt ON cnt.tag_ID = cnti.tag_ID WHERE `dte` >= DATE_SUB(CURDATE(),INTERVAL 7 DAY) AND name NOT liKE 'Resolution%' GROUP BY cat ORDER BY qty DESC liMIT 5");   if($count_account = $result_account->num_rows) {      while($row = $result_account->fetch_object()){        //output this query results here        $results[$row->cat] = $row->qty;      }   }$result_prevIoUs = $db->query("SELECT nID,INTERVAL 14 DAY)) AND name NOT liKE 'Resolution%' GROUP BY cat ORDER BY qty DESC liMIT 5");if($count_prevIoUs = $result_prevIoUs->num_rows) {    while($row_p = $result_prevIoUs->fetch_object()){      //output this query results here      $results[$row_p->cat] = ((isset($results[$row_p->cat])) ? $results[$row_p->cat] : 0 ) - $row_p->qty;    }}foreach( $results as $key => $result) {  echo "<tr>";    echo "<td><h6>".$key."</h6></td><td><h3 class='text-primary'>".$result."</h3></td>";  echo "</tr>";}

更新-显示第一周及其旁边的差异

你可以试试看我摆脱了foreach,仅使用查询循环来完成所有 *** 作.

注意：同样,这是未经测试的代码

$results = array();$result_account = $db->query("SELECT nID,INTERVAL 7 DAY) AND name NOT liKE 'Resolution%' GROUP BY cat ORDER BY qty DESC liMIT 5");$result_prevIoUs = $db->query("SELECT nID,INTERVAL 14 DAY)) AND name NOT liKE 'Resolution%' GROUP BY cat ORDER BY qty DESC");if($count_prevIoUs = $result_prevIoUs->num_rows) {    while($row_p = $result_prevIoUs->fetch_object()){      //output this query results here      $results[$row_p->cat] = $row_p->qty;    }}if($count_account = $result_account->num_rows) {    while($row = $result_account->fetch_object()){        $difference = $row->qty - ((isset($results[$row->cat])) ? $results[$row->cat] : 0 );        echo "<tr>";          echo "<td><h6>".$row->cat."</h6></td><td><h3 class='text-primary'>".$row->qty."</h3></td><td><h3 class='text-primary'>".$difference."</h3></td>";        echo "</tr>";    }}

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