Question

编辑：嘿，我已经在使用php，我不是故意要问的。我问如何在8列html表中显示3-columned mysql结果

--work table--                                               --crew table--

date           |   hours                                     date   |    name   
2013-02-03(Sun)     8                                    2013-02-03     john
2013-02-04(Mon)     7                                    2013-02-03     sam 
                                                         2013-02-03     peter
                                                         2013-02-04     john
                                                         2013-02-04     sam

我有这两张表来记录员工的工作时间我需要将这些数据显示为时间表。

但是如何在下面的html表中显示8列？

我的意思是假设您运行此查询：

SELECT hours, name, DAYNAME(work.date) day
FROM work 
LEFT JOIN crew ON work.date = crew.date
WHERE WEEK(work.date) = 5

它将显示3列。那我该如何构建这个html表呢？

此mysql结果为

hours    name    day
  8      john   Sunday
  8      Sam    Sunday
  8      Peter  Sunday
  7      John   Monday
  7      Sam    Monday

此html表格（如何？）

Name    Sunday   Monday  Tuesday..(day name can be written in html)
John      8        7       ...
Sam       8        7       ...
Peter     8        -       ...

Answer 1

修改此.......如果你使用php

<?php
$con = mysql_connect("localhost","peter","abc123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("my_db", $con);

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

while($row = mysql_fetch_array($result))
  {
  echo $row['FirstName'] . " " . $row['LastName'];
  echo "<br />";
  }

mysql_close($con);
?>

Answer 2

警告：我不是PHP程序员，所以以下是最好的伪代码，并且包含很多语法错误（更不用说任何遗漏的错误处理，甚至是html布局）。我希望无论如何都能得到这个想法。任何人都可以随意编辑

$result = mysql_query("query here");

// a double nested array
// $weekdays['Sunday']['Peter'] will be the hours peter working on sunday 
$weekdays = [];
// a list of all persons in the table
$persons = [];

while($row = mysql_fetch_array($result))
{
  $hours = $row[0];
  $person = $row[1];
  $weekday= $row[2];

  if (!$weekdays[$person]) {
     $weekdays[$person] = [ $person => $hours ];
  } else {
     $weekdays[$person][$weekday] = $hours;
  }
  if (! $person in $persons) {
    $persons[] = $person;
  }
}

// then unroll in two loops
for ($person in $persons) {
  print '<tr><td>'.$person.'</td>'
  for ($weekday in [ "Sunday", "Monday", ...]) {
     print '<td>'.$weekdays[$weekday][$person].'</td>'
  }
  print '</tr>'
}

我如何在html表中显示这个mysql结果？

2 个答案: