使用COUNT（*）和SUM（）时如何打印或回显行

Question

我使用PHP-MYADMIN为php生成了以下查询， 我的问题是如何打印它生成的30行？当我使用“看到引用”它只是错误。

我正在尝试用搜索词回显行并计算＆lt; div＆gt;＆lt; / div＆gt;每个标签都有自己的标签。

1. Facebook 38搜索
2. 另一个Feed 25搜索
3. 时间轴18搜索

等等到第30行。

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$结果= mysql_query（$ SQL）

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$sql = 'SELECT COUNT(*) AS `Rows`, `search`, SUM(searched) FROM `af_timeline_search` GROUP BY `search` ORDER BY SUM(searched) DESC LIMIT 0, 30 ';

---

        /* top searches */
        $sqlthis = mysql_query('SELECT COUNT(*) AS `Rows`, `search` FROM `af_timeline_search` GROUP BY `search` ORDER BY `Rows` DESC');
        $num=mysql_num_rows($sqlthis);
        $arrS = mysql_fetch_array($sqlthis); 
        $i=0;
        while ($i < $num){
        echo 'Search '.$arrS[$i].'';
            $i++;
        }

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PHPmyadmin中的查询屏幕截图。

Answer 1

我认为你能做的最好的事情就是看看这个：mysql-fetch-array

你会找到关于如何循环该数组的答案：）

Answer 2

$result = mysql_query("SELECT * FROM YOURDATABASE");

while($row = mysql_fetch_array($result))
  {
  echo $row['search'] . " " . $row['somenumber'];
  echo " searches";
  }

mysql_close($con);

Answer 3

我找到了解决方法。谢谢大家的指导。 :-)“你们都摇滚”

---

$sql = "SELECT COUNT(*) AS Rows, search, SUM(searched) FROM anotherfeed.af_timeline_search GROUP BY search ORDER BY SUM(searched) DESC";
$result=mysql_query($sql) or die ('Error! yo.');
$row = mysql_fetch_array($result);
            echo '<pre>';
            print_r($row);
            echo '</pre>';

---

        /* top searches with loop */
        $sql = "SELECT COUNT(*) AS Rows, search, SUM(searched) FROM anotherfeed.af_timeline_search GROUP BY search ORDER BY SUM(searched) DESC LIMIT 0, 20";
        $result=mysql_query($sql) or die ('Error! yo.');
        while($row = mysql_fetch_array($result))
          {
          echo '  (';
         echo urldecode($row['search']) . " " . $row['SUM(searched)'] . "";
         echo ')  ';
          }

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我能够打印数组，希望我能用while语句得到循环。如果你有或者知道一个更好的话，请随意添加循环...谢谢。