Question

因为它没有计算SUM first_payment的整个列。 I tried "SELECT SUM(first_payment) FROM \客户信息``，它作为未识别的索引first_payment返回。有什么理由吗？

    <?php
    error_reporting(E_ALL);
    ini_set("display_errors", 1);


    $pay=$_REQUEST['payment']; 

    // MySQL database connection, username, password, database name
    $con=mysqli_connect("localhost","user","pass","db_name");

    // Check connection
    if (mysqli_connect_errno()) {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    if ($pay=="first") 
      $query="SELECT *, SUM(first_payment) FROM `Customer Information`";
else if ($pay=="second") 
      $query="SELECT *, SUM(second_payment) FROM `Customer Information`";
else if ($pay=="third") 
      $query="SELECT *, SUM(third_payment) FROM `Customer Information`";
else if ($pay=="fourth") 
      $query="SELECT *, SUM(fourth_payment) FROM `Customer Information`";


    $result = mysqli_query($con,$query);

    // Executing and error checking of query
    if (!mysqli_query($con,$query)) {
      die('Error: ' . mysqli_error($con));
    }


    while ($row = mysqli_fetch_array($result)) {
    switch($pay)
    {
       case "first":
          echo $row['first_payment'];
          break;
       case "second":
          echo $row['second_payment'];
          break;
       case "third":
          echo $row['third_payment'];
          break;
       case "fourth":
          echo $row['fourth_payment'];
          break;
       default:
          echo "nothing selected";
    }
    }
    mysqli_close($con);
    ?>

Answer 1

更改您的查询并为您在代码中使用的总金额提供正确的别名。试试这个

if ($pay=="first") 
   $query="SELECT *, SUM(first_payment) AS first_payment FROM `Customer Information`";
else if ($pay=="second") 
   $query="SELECT *, SUM(second_payment) AS second_payment FROM `Customer Information`";
else if ($pay=="third") 
   $query="SELECT *, SUM(third_payment) AS third_payment FROM `Customer Information`";
else if ($pay=="fourth") 
   $query="SELECT *, SUM(fourth_payment) AS fourth_payment FROM `Customer Information`";

Answer 2

从查询中提取数组时使用MYSQLI_ASSOC：

mysqli_fetch_array($result,MYSQLI_ASSOC);

或使用mysqli_fetch_assoc()函数：

mysqli_fetch_assoc($result)
在查询中为摘要列使用别名：

例如：

"SELECT *, SUM(first_payment) AS SUMMARY  FROM `Customer Information`";

此外，最好通过在任何操作之前检查$pay变量来避免与数据库的不必要连接，如下所示：

<?php
error_reporting(E_ALL);
ini_set("display_errors", 1);

$pay=$_REQUEST['payment']; 
// first of all: let's check $pay variable:
switch($pay)
{
   case "first":
      $row_index = 'first_payment';
      break;
   case "second":
      $row_index = 'second_payment';
      break;
   case "third":
      $row_index = 'third_payment';
      break;
   case "fourth":
      $row_index = 'fourth_payment';
      break;
   default:
      echo "nothing selected";
      return;  // return if nothing is selected, no need to run any database manipulations.
}

// MySQL database connection, username, password, database name
$con=mysqli_connect("localhost","user","pass","db_name");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// prepare query:
$query="SELECT *, SUM(".$row_index.") AS SUMMARY  FROM `Customer Information`";

// Executing and error checking of query
if (!$result = mysqli_query($con,$query)) {
  die('Error: ' . mysqli_error($con));
}
while ($row = mysqli_fetch_array($result,MYSQLI_ASSOC)) {
  echo $row['SUMMARY'];
}
mysqli_free_result($result); // don't forget to free query results
mysqli_close($con);
?>

Answer 3

除了上面提供的答案，你必须知道出了什么问题：

在您的select语句中，您选择了字段的总和，而不是您尝试检索其结果的字段本身，因此错误未定义索引...

如上面的答案，给总和一个别名，并在while循环中使用它来检索它。

没有总结整列

3 个答案: