一次显示来自MySQL DB的重复值

Question

我在MySQL DB有一张桌子，它存储餐厅在每个工作日和时段提供的菜单。

表的结构如下：

  i_type      i_name    i_cost   i_day   i_start     i_end
  --------------------------------------------------------
  Breakfast   Prantha   20       Sunday   07:00      11:00
  Lunch       Special   80       Sunday   11:01      15:00
  Dinner      Special   100      Sunday   15:01      21:00
  Breakfast   Prantha1  50       Monday   07:00      11:00
  Lunch       Special1  70       Monday   11:01      15:00
  Dinner      Special1  130      Monday   15:01      21:00

我想按以下方式显示菜单：

<b>Sunday</b><br/>
    Breakfast<hr/>Prantha       20     Add<br/><br/>
    Lunch<br/><hr/>Special      80     Add<br/><br/>
    Dinner<br/><hr/>Special     100    Add<br/><hr/>
<b>Monday</b><br/>
   Breakfast<hr/>Prantha       20     Add<br/><br/>
   Lunch<br/><hr/>Special      80     Add<br/><br/>
   Dinner<br/><hr/>Special     100    Add<br/><hr/>

我现正处理以下事项：

$sql = SELECT * FROM `tabl1` WHERE 1;      
$data = DB::instance()->prepare($sql)->execute()->fetchAll();
foreach($data as $row){
       //print values
 }

but this approach print Sunday and Monday multiple times. 

Answer 1

只需跟踪您输出的日期，例如使用“状态机”：

$previousday = null;
while($row = ... fetch from db... ) {
    if ($row['day'] != $previousday) {
       ... output day header
       $previousday = $row['day'];
    }
    ... output row data ...
}

Answer 2

当然，它会打印三个星期天和三个星期一。

例如，您可以首先选择所有不同的日子(从菜单中选择DISTINCT ( i_day ) )，然后为每一天选择所有的菜单(选择*i_day= ?)

$sql = "SELECT DISTINCT (i_day) FROM menu";      
$data = DB::instance()->prepare($sql)->execute()->fetchAll();
foreach($data as $row){
  // Add all i_day's to an associative array $i_days
}
foreach($i_days as $day) {
  $sql = "SELECT * FROM menu WHERE i_day = $day";      
  $data = DB::instance()->prepare($sql)->execute()->fetchAll();
    foreach($data as $row){
      // Print values
    }
}

Answer 3

试试这个：

$stat = "SELECT * from TABLE_NAME";
$conn -> query ($stat);
foreach($conn -> query($stat) as $menu) {

echo $menu["i_day"];
echo $menu["i_type"];
echo $menu["i_name"];
echo $menu["i_cost"];

}

根据您的需要对上述数据进行格式化。我刚给你提供了基本代码。

交替地，

$stat = "SELECT * from TABLE_NAME where i_day = 'DAY_GOES_HERE'";
$conn -> query ($stat);
foreach($conn -> query($stat) as $menu) {

echo $menu["i_type"];
echo $menu["i_name"];
echo $menu[i_cost];

}

每天重复一遍。