Show repeated value from MySQL DB once PHP

Question

I have a table in MySQL DB which stores Menu served by restaurant on each weekday and in time slots.

Table structure is as below:

  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

I want to display menu as below:

<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/>

Currently i am handling it as below:

$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.

Marc B · Accepted Answer · 2015-02-06 15:23:01Z

1

Just keep track of which day you've output, e.g. use a "state machine":

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

answered Feb 6, 2015 at 15:23

Marc B

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1 Comment

Gags Over a year ago

this is perfect and served the purpose

Calabacin · Accepted Answer · 2015-02-06 15:25:42Z

1

It will of course print three Sundays and three Mondays.

You could for example Select all different days first (SELECT DISTINCT (i_day) FROM menu), then for each day select all menus (SELECT * FROM menu WHERE 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
    }
}

edited Feb 6, 2015 at 15:25

answered Feb 6, 2015 at 15:19

Calabacin

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Comments

Pawan Kartik · Accepted Answer · 2015-02-06 16:41:11Z

-1

Try this:

$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"];

}

Format the above data basing on your need. I've just provided you the basic code.

Alternately,

$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];

}

Repeat it for each day.

edited Feb 6, 2015 at 16:41

answered Feb 6, 2015 at 16:35

Pawan Kartik

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1 Comment

Gags Over a year ago

because same has been described in above answer.

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