Declaring PHP variables with numbers (based on loop and based on num of records returned by a query)

Question

I have this on the first page: returned results

it shows 2 groups with different ID and with different number of members. IT09 has 3 and IT10 has 4. when the "Evaluate" label is clicked, the page then proceeds here.

Those textboxes are declared in such a way like this:

<table>
$i=1;
while ($data = mysql_fetch_array($querys))
{
echo '<tr>
<td width="120">'.$data['thenames'].'</td>
<td width="50">
<input type="text" name="newSG'.$i.'" id="newSG'.$i.'" class="evaluation" value="'.$data[$toshow].'" onkeypress="return isNumber(event)" />
</td>
<td>
<input type="hidden" name="stud'.$i.'" id="stud'.$i.'" value="'.$data['memberID'].'" />
</td>
</tr>';
$i++;
}
</table>

As you can see, it creates a textbox and a hidden input with name and id with a number at the end. That's what i also want to happen on my php part.

I have this code

$ii = 1;
while ($data = mysql_fetch_array($qry))
{
$stud(here) = $_POST['stud'(and here)]; <<<<

$ii++
}

I want to declare variables using this loop . as you can see (here , andhere) I want that to have numbers 1,2,3 or depending on the number of records returned by my query. the problem here is that I want to declare something like $stud1, $stud2 and so on. I dont know how to add the value of the loop after the variable $stud. can someone help?

yah yah yah that's my problem. can you give me an example of a syntax where the variable is concatenated with the number ? — Patrick John Calumno
– Patrick John Calumno, Commented Dec 30, 2015 at 10:40
@Barmar You're right. If this would be the only wrong thing(mysql_*). — Rizier123
– Rizier123, Commented Dec 30, 2015 at 10:44

John Pangilinan · Accepted Answer · 2015-12-30 11:25:51Z

1

make an associative array instead.

$studs = array(); while(...){ $studs['stud'.index] = $row; }

access it :

echo $studs['stud1'][field];

if you really want it as a variable then:

eval('return $stud'.$ii.'=json_decode(\''.json_encode($row).'\');');

then you can:

echo $stud1->field;

Edit:

it seems your assigning your stud variable with the post. then you can:

eval('return $stud'.$ii.'=$_POST[\''.$ii.'\']');

access it:

echo $stud1;

edited Dec 30, 2015 at 11:25

answered Dec 30, 2015 at 10:50

John Pangilinan

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2 Comments

Patrick John Calumno Over a year ago

im finding it a bit hard to understand, shoud I write anything inside array(here)? and the word index, should it be just that word or any index number i would need? and lastly $row, (the value) what if it's from a post method. $_POST['stud'] how should I include the number on that?

John Pangilinan Over a year ago

dont put anything in array() it's just declaring an array. index should be in your case the $ii. $row should be the $data in your mysql_fetch_array

Barmar · Accepted Answer · 2015-12-30 11:04:51Z

0

Don't use separate variables, use an array.

$studs = array();
$i = 1;
while ($data = mysql_fetch_array($querys)) {
    $studs[$i] = $_POST['studs' . $i];
}

I'm not sure why you're filling in a variable from POST data in a loop over database results, but that's not really relevant.

I also suggest that you take $i out of the form inputs. Give them names like studs[] and newSG[], and PHP will automatically put them into an array in $_POST. You can then do:

$studs = $_POST['studs'];
$newSG = $_POST['newSG'];

answered Dec 30, 2015 at 11:04

Barmar

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1 Comment

Patrick John Calumno Over a year ago

There. it's done! :) thanks @barmar. but the part that you suggest, sorry but i have my reasons why I include $i on their name values. hehe.. Thank you so much anyway :) Happy new year

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