Giving different names to a html object in a loop in PHP

Question

How would I be able to give the different things different names (as I am creating them in a loop) so that I can use the data? Could I modify name=\"type\" to maybe include $x within the name so that it is different each time?

For($x=0; $x<=$noQuestions-1; $x++){

    echo "<table>";
    echo "<tr>"."Question ".($x+1).": ".$question[$x]."</tr>";
    echo "<form method=\"post\">";
    echo "<Select class=\"form-control\" type=\"text\" name=\"type\" required>";
    echo "<option value=\"1\">".$optionData[$x][0]."</option>";
    echo "<option value=\"2\">".$optionData[$x][1]."</option>";
    echo "<option value=\"3\">".$optionData[$x][2]."</option>";
    echo "<option value=\"4\">".$optionData[$x][3]."</option>";
    echo "</select>";
    echo "</form>";
    echo "</table>";
}

Did you try? What exactly is the problem?

Dekel
– Dekel

2016-11-13 13:10:05 +00:00
Commented Nov 13, 2016 at 13:10 — Dekel
– Dekel, Commented Nov 13, 2016 at 13:10

Shubhaw Kumar · Accepted Answer · 2016-11-13 14:32:58Z

1

You can simply concatenate variable $x in name like this:

echo "<Select class=\"form-control\" type=\"text\" name=\"type".$x."\" required>";

EDIT:

$type = "dropdown";
for($x=0; $x<=$noQuestions-1; $x++)
{
       echo "<table>";
       echo "<tr>"."Question ".($x+1).": ".$question[$x]."</tr>";
       echo "<form method=\"post\">";
       echo "<Select class=\"form-control\" type=\"text\" name=\"".$type.$x."\" required>";
       echo "<option value=\"1\">".$optionData[$x][0]."</option>";
       echo "<option value=\"2\">".$optionData[$x][1]."</option>";
       echo "<option value=\"3\">".$optionData[$x][2]."</option>";
       echo "<option value=\"4\">".$optionData[$x][3]."</option>";
       echo "</select>"; echo "</form>";
       echo "</table>";

 }

edited Nov 13, 2016 at 14:32

answered Nov 13, 2016 at 13:21

Shubhaw Kumar

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6 Comments

Kurt Görg Over a year ago

I am only able to get $type0 for some reason when echoing it.

Kurt Görg Over a year ago

if (isset($_REQUEST['submit'])){ extract($_REQUEST); if ($_POST['dropdown0']=="1") { echo "1"; } elseif ($_POST['dropdown0']=="2") { echo "2"; } elseif ($_POST['dropdown0']=="3") { echo "3"; } elseif ($_POST['dropdown0']=="4") { echo "4"; } if ($_POST['dropdown1']=="1") { echo "1"; } elseif ($_POST['dropdown1']=="2") { echo "2"; } elseif ($_POST['dropdown1']=="3") { echo "3"; } elseif ($_POST['dropdown1']=="4") { echo "4"; } }

Shubhaw Kumar Over a year ago

What exactly are you trying to do...can you please elaborate?...I tried the answer which I have given, and it's type0, type1, etc are coming.

Kurt Görg Over a year ago

I am making a multiple choice quiz player. I have already created a test maker so the number of questions in the quiz player depends. So I have a html form within a php loop to create a certain amount of questions and options. Did you echo it in a different way than I did? Could you show me please

Shubhaw Kumar Over a year ago

I guess you are creating type a variable. If this is the case, then do it like this: $type = "dropdown"; & in place of name: name=\"".$type.$x."\"

|

sldeb · Accepted Answer · 2016-11-13 13:12:08Z

0

One simple solution would be to add $x in the name like this for example:

echo "< select class=\"form-control\" type=\"text\" name=\"type-$x\" required>";

answered Nov 13, 2016 at 13:12

sldeb

12 bronze badges

2 Comments

Kurt Görg Over a year ago

how would i echo the variable in regular PHP? Would it be "echo $type-0";

sldeb Over a year ago

It depends on how you submit the fields. If your form method is "post" then you will echo the variable with $_POST ["type-0"], if your form method is "get" you will use $_GET ["type-0"]

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