Question

我想将第一个脚本的总和回显到第二个脚本

两者都是同一页我尝试但不工作示例

echo '<td>' . $row['status'] . '</td>';
echo '<td>' .$final_result4. '</td>';

我怎么能这样做，请帮我解决这个问题，谢谢......

这是我的第一个脚本

     $res1 = (($basicsalary+$allowsalary)*$days*$yeardays1)/$yeardays/$monthdays;
     $res2 = (($basicsalary1+$allowsalary1)*$days*$yeardays2)/$yeardays/$monthdays;
     $res3 = (($basicsalary2+$allowsalary2)*$days*$yeardays3)/$yeardays/$monthdays;
     $res4 = (($basicsalary+$allowsalary)*$days*$yeardays4)/$yeardays/$monthdays;
     $res5 = (($basicsalary1+$allowsalary1)*$days*$yeardays5)/$yeardays/$monthdays;
     $res6 = (($basicsalary2+$allowsalary2)*$days*$yeardays6)/$yeardays/$monthdays;
     $res8 = 221/730*$miscamount +$otheramount;
     $res7 = $startdays -$enddays;

 $result1  = number_format((round($res1, 1)),3);
 $result2  = number_format((round($res2, 1)),3);
 $result3  = number_format((round($res3, 1)),3);
 $result4  = number_format((round($res4, 1)),3);
 $result5  = number_format((round($res5, 1)),3);
 $result6  = number_format((round($res6, 1)),3);
 $result7 = number_format((round($res7, 1)),3);
 $result8 = number_format((round($res8, 1)),3);




 $final_result   = number_format(($result1 +$result2 +$result3 +$result4 +$result5 +$result6 ),3);
     $final_result2   = number_format((round($final_result/730*$result7 ,1 )),3);
      $final_result3   = number_format((round(($final_result2 +$result8) , 1)),3);
       echo $final_result4   = number_format((round(($final_result -$final_result3) , 1)),3);


    //Ending of php
    ?>

这是我的第二个剧本

 echo "<span  align='center' class='style2'>Over All Report For The Period Of $a to $b</span>"; 
           echo "<div id='testTable' id='non-printable'><table class='hovertable' border='1' cellpadding='10'>";
            echo "<tr> <th>Date</th><th>Id</th><th>Name</th> <th>Division</th><th>Payment</th><th>Status</th><th>Total</th></tr>";


     // get results1 from database
          $result1 = mysql_query("SELECT * FROM fullfsaccounts where date BETWEEN '$a' AND '$b' order by date ASC");
                    while($row = mysql_fetch_array($result1))
                    {   



                    // echo out the contents of each row into a table
                    echo "<tr>";
                    echo '<td>' . $row['date'] . '</td>';
                    echo '<td>' . $row['cardno'] . '</td>';
                    echo '<td>' . $row['name'] . '</td>';
                    echo '<td>' . $row['division'] . '</td>';
                    echo '<td>' . $row['typecash'] . '</td>';
                    echo '<td>' . $row['status'] . '</td>';
                    echo '<td>'  '</td>';         
                    echo "</tr>"; 

                //Increment the value of the Total_total variable
                //by the salary value of one row till the while loop finishes
                $Total_rows=$num=mysql_num_rows($result1);
                $Total_total=$Total_total+$row['total'];

        }
        echo "</table>";

        echo "<table class='hovertable' border='1' >";
        echo "<tr width='100%'>";
       echo '<td>Total</td>';
       echo '<td>' . $Total_rows .'</td>';
       echo '<td>' . $Total_total .'</td>';
       echo "</tr>";  



            // close table>
            echo "</table>";

Answer 1

在您的代码中，您在行回显''''中有一个基本错误;你的需要把回声放''。 '';

我认为你也不能很好地解释，因为如果它们和你说的页面相同，你就不应该有任何问题来调用页面局部变量的值。

我建议你加上不同的 - ＆gt; echo'pass';或者echo'pass：'。resultxxx：在控制页面的流程中，我明白如果它们在同一页面上并且没有走出去并且进入它，你应该能够获取局部变量的值而不用任何问题。

如果您需要帮助，请详细说明您的问题。

如何在行中回显sum值

1 个答案: