Question

以下是我的PHP代码：

<?php 
  echo "<h1>test</h1>";
  echo "<table border = '1' width = '50%'>\n";
  echo "<tr>";
  echo "<th>date</th>";
  echo "<th>temperature</th>";


  //get data from file
  $fileName = 'sensor.csv';
  $file = fopen($fileName,"r");


  //while not to end of file
  while (!feof($file) ) 
  {

    while (($csv_line = fgetcsv($file)) !== FALSE)
    {

      $date_format = strtotime($csv_line[0]);
      //print_r($csv_line);
      echo "<tr>";    //beginning new row of record
      echo "<td>" . date("Y-m-d", $date_format)."</td>";  
      echo "<td>" . $csv_line[2]."</td>";  
      echo "</tr>"; //new row

    }
    echo "</table>\n";
  }

?>

以下是当前输出

date        temperature
2015-07-20  22.7
2015-07-20  22.7
2015-07-20  22.8
2015-07-19  32.8
2015-07-19  31.9
2015-07-19  32.8
2015-07-19  32.8
2015-07-18  29.1
2015-07-18  28.8
2015-07-18  29.7
2015-07-18  29.9
2015-07-18  29.4
2015-07-18  29.8
2015-07-16  26.4
2015-07-16  25.9
2015-07-16  24.7
2015-07-16  24.9
2015-07-16  25
2015-07-16  26.4
2015-07-16  27
2015-07-16  26.1
2015-07-16  26

我希望它是这样的：

date        temperature
2015-07-20  average temperature of each day
2015-07-19  average temperature of each day
2015-07-18  average temperature of each day
2015-07-16  average temperature of each day

Answer 1

试

SELECT date, AVG(temperature) as AvgTemp
FROM TableName
GROUP BY date

Answer 2

这是编辑过的代码，因为我无法检查它可能包含错误

<?php

echo "<h1>test</h1>";
echo "<table border = '1' width = '50%'>\n";
echo "<tr>";
echo "<th>date</th>";
echo "<th>temperature</th>";


//get data from file
$fileName = 'sensor.csv';
$file = fopen($fileName,"r");


while (!feof($file)) {

    $cs = fgetcsv($file);
    $last_date = $cs[0][0];
    $sum_temp = 0;
    $count = 0;
    $csv_line = array();
    $firsttime = true;

    do  {
        if (!$firsttime) {
            $date_format = strtotime($csv_line[0]);
            $temp = $csv_line[2];
        } else {
            $date_format = $last_date;
            $temp = $cs[0][2];
        }

        if ($date_format == $last_date) {
            $sum_temp += $temp; // Combine tempratures
            $count++;
        } else {
            $avg_temp = $sum_temp / $count; // Get the average temprature

            echo "<tr>";    //beginning new row of record
            echo "<td>" . date("Y-m-d", $last_date) ."</td>";  
            echo "<td>" . $avg_temp ."</td>";  
            echo "</tr>"; //new row

            $sum_temp = 0;
            $count = 0;
        }
        $firsttime = false;
        $last_date = $date_format;
    } while(($csv_line = fgetcsv($file)) !== FALSE);

    echo "</table>\n";

}

?>

相同的日期是一个日期，并希望在PHP的平均温度@one日

2 个答案: