MYSQL PHP Select Order By

Question

I have example table date_table, as a basis of data filter 2 columns - day and day_of_week I do query in SQL

select distinct day_of_week,day  from date_table
order by day_of_week

No problem

day_of_week        day
1                  Monday
2                  Tuesday
3                  Wednesday
4                  Thursday
5                  Friday
6                  Saturday
7                  Sunday

Then I implement this into select box in PHP , so I can choose value

$query = "
select distinct day_of_week,day  from date_table
order by day_of_week
";
$res = mysql_query($query) or die ("Error in query: $query. ".mysql_error());
$option = '';
while($row = mysql_fetch_array($res))
    {
if(!empty($_REQUEST["sday"]))
   {
  if($row['day_of_week'] == $_REQUEST["sday"])
    $option .= '<option  value = "'.$row['day_of_week'].'"

    selected="selected">'.$row['day'].'</option>';
   else
   $option .= '<option  value =
    "'.$row['day_of_week'].'">'.$row['day'].'</option>';        
}
  else
 $option .= '<option  value = "'.$row['day_of_week'].'">'.$row['day'].'</option>';  
}

And Select Box

<select name="sday" class="sday">
<option value="">Select Day</option>
<?php echo $option; ?>
</select>

Problem 1

the end Result is something Like

Monday
Friday
Saturday
Sunday
Thursday
Tuesday
Wednesday

Problem 2

When I select Value "Monday" , all fine . It passes value "1"

When I select "Tuesday", It selects value 2,3 and 7. Same with wednesday and sunday.

There are few other select boxes, that are made with same idea, works fine. There are no data duplicates.

What could be the problem?

$query = "select day_of_week, day from date_table order by day_of_week"; and also re-write $option instead of $option5. — Rohit Suthar
– Rohit Suthar, Commented Feb 19, 2014 at 12:48
@Strawberry some Calendar systems store days in the database. — bear
– bear, Commented Feb 19, 2014 at 12:50

bear · Accepted Answer · 2014-02-19 12:53:25Z

1

Have you tried to

 ORDER BY day_of_week ASC

You may want to restructure your code a but such that

$selectedString = ($row['day_of_week'] == $_REQUEST["sday"]) ? ' selected ' : '';

<option value = "<?php echo $row['day_of_week'];?>" <?php echo $selectedStr;?> ><?php echo $row['day'];?></option>'

answered Feb 19, 2014 at 12:53

bear

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4 Comments

Mohammad Faisal Over a year ago

+1 for ($row['day_of_week'] == $_REQUEST["sday"]) ? ' selected ' : '';

kaulainais Over a year ago

Problem is in "sday". If I choose Sunday, It gives value 2 or 3 or 7

bear Over a year ago

@kaulainais can you GROUP BY day_of_week? There's no reason why all three of them should be selected, unless your input is incorrect.

kaulainais Over a year ago

I think I know whee the problem is. Thanks Shamil

bear · Accepted Answer · 2014-02-19 12:49:11Z

0

you use variable $option5, instead of $option

edited Feb 19, 2014 at 12:49

bear

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answered Feb 19, 2014 at 12:47

ziollek

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1 Comment

kaulainais Over a year ago

I changed it. that is not an issue

iubema · Accepted Answer · 2014-02-19 12:52:42Z

0

Try this code:

    if(!empty($_REQUEST["sday"]))
{
    if($row['day_of_week'] == $_REQUEST["sday"])
    {
        $option .= '<option  value = "'.$row['day_of_week'].'"
        selected="selected">'.$row['day'].'</option>';
    }
    else
    {
        $option .= '<option  value ="'.$row['day_of_week'].'">'.$row['day'].'</option>';
    }
}
else
{
$option .= '<option  value = "'.$row['day_of_week'].'">'.$row['day'].'</option>';

}

Its cleaner and you will dont have problems with bracktes

answered Feb 19, 2014 at 12:52

iubema

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Comments

Strawberry · Accepted Answer · 2014-02-19 12:58:14Z

0

I don't get it. Why not do this instead (the array isn't even necessary, but hey)...

<select>
<?php
$days = array('monday','tuesday','wednesday','thursday','friday','saturday','sunday');

for($i=0;$i<count($days);$i++){

echo "<option>$days[$i]</option>\n";

}

?>

</select>

answered Feb 19, 2014 at 12:58

Strawberry

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