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所以我做了一个查询，得到以下结果

 +---------------+---------------+---------+---------
 | payID         | payRate       | hours   | Earnings
 +---------------+---------------+---------+--------
 |  entertainment| 12            |      18 | 216
 |        retail | 10            |      28 | 280
 +---------------+---------------+---------+----------

查询和其他一些代码部分如下：

 $query = "SELECT jobId, payRate, SUM(hours) AS 'All_Hours' ,payRate * SUM(hours) AS 'total'
                       FROM users INNER JOIN deposit ON userId = empId
                      WHERE users.email = '" . $_SESSION['email'] ."' 
                      GROUP BY jobId,payRate";



                      $result = mysqli_query($db, $query); //we make the query

                  if (!$result) { //if the query failed
                   echo("<p id = 'greatideadescription'>
                              Error, the query could not be executed: " .
                   mysqli_error($db) . "</p>");
                   mysqli_close($db);}

                if (mysqli_num_rows($result) == 0) { //if no rows returned
                   echo("<tr><td />
                               <td>No Results</td>
                               <td /></tr>");
                   mysqli_close($db); //close the database
                   exit("</table></div></form></div></div>
                               <script>DisplayFooter();</script></body></html>");
                        } 
                       $numRows = mysqli_num_rows($result); //gets number of rows
                       $numFields = mysqli_num_fields($result); //gets number of fields
                       //prints the data in the table
                       PrintTable($result, $numRows, $numFields);

函数PrintTable如下

 function PrintTable($result, $numRows, $numFields) {
  $row = mysqli_fetch_row($result); //fetches the first row
  for ($i = 0; $i < $numRows; $i++) {
  echo("<tr>"); //opens a new row

  for ($j = 0; $j < $numFields; $j++) { //second loop goes through columns
  echo("<td>" . $row[$j] . "</td>"); 
   } //end inner for loop

   $row = mysqli_fetch_row($result); //fetches subsequent rows
   echo("</tr>"); //closes the row
   } //end outer for loop
  }

我想要的是呼应收入的总和，在这种情况下它将是 496 我该怎么做？

4

1 回答 1

0

如果您知道总和列的列索引，则可以尝试以下操作：

function PrintTable($result, $numRows, $numFields) {
    $total = 0; // Initialize variable before going through each row
    $row = mysqli_fetch_row($result); //fetches the first row
    for ($i = 0; $i < $numRows; $i++) {
        echo("<tr>"); //opens a new row

        for ($j = 0; $j < $numFields; $j++) { //second loop goes through columns
            echo("<td>" . $row[$j] . "</td>");
            if($j == 3)
              $total += $row[$j];
        } //end inner for loop

        $row = mysqli_fetch_row($result); //fetches subsequent rows
        echo("</tr>"); //closes the row
    } //end outer for loop
}

之后，您可以使用您的总数或将其返回到您的主代码中以在关闭表格或或或...后回显它...

更好的方法是进行 3 项更改：

  1. payRate * SUM(hours) AS earnings在您的 sql 语句中，您稍后将使用该名称
  2. 用于mysqli_fetch_assoc获取关联数组，您可以在其中按名称访问列 ( $row['earnings'])
  3. 对列使用 foreach 循环，因为您没有可以迭代的索引

这取决于你使用哪种方式...

于 2013-03-20T07:39:40.290 回答