Appending a variable name with another variable

[Franky]

Is there anyway i can append a variable name with another variable? For example:

$a = 1;

While ($a <= 5){
does something in here
$result$a = $some result from within this loop
$a++;
}

the idea being that at the end i will have $result1, $result2...$res ult5 with values saved in them.

[r035198x]

Is there anyway i can append a variable name with another variable? For example:

$a = 1;

While ($a <= 5){
does something in here
$result$a = $some result from within this loop
$a++;
}

the idea being that at the end i will have $result1, $result2...$res ult5 with values saved in them.

Are you sure that your design is correct? Wouldn't an array be the appropriate data type to use here?

[code green]

Create a CONSTANT
[PHP]define ('RESULT','$res ult');[/PHP]Concatenate with the '.' operator
[PHP]$a = 1;

While ($a <= 5){
does something in here
RESULT.$a = $some result from within this loop
$a++;
}[/PHP]
I think this works

[devsusen]

Hi,

u can use Variable variables in php, though it leads some ambiguity problem. Other way of doing this thing is the associative array. Where variable name cane be the key and value be the value.

susen

[r035198x]

Hi,

u can use Variable variables in php, though it leads some ambiguity problem. Other way of doing this thing is the associative array. Where variable name cane be the key and value be the value.

susen

I'd prefer the associative array for this.

[Franky]

Thanks for the tips. I tried both, and ended up with the code at the bottom. The problem i have now, is i can't seem to access the data in the array ($QuestionNums) outside the function it is created within. I need it for the process_form function in order to update the db (for now i am just printing them to screen to get it working).
As well as the main code at the bottom, i have tried the following layout:
[PHP]if (!isset($_POST['Submit'])){
//all the stuff from show_form function here
} else {
//all the stuff from process_form here[/PHP]

[PHP]
//Function called from another page
function displayQuestion s($ModNum){ //modnum is used in the db query
global $QuestionNums;

if (!isset($_POST['Submit'])){
show_form($ModN um);
}else{
process_form($Q uestionNums);
}//end if
}//end function "displayQuestio ns"


function show_form($ModN um)
{
global $result, $QuestionNums;
dbConnect();
getQuestion($Mo dNum); //call to query in other file. result is $result?>

<p>Introducti on text here</p>
<TABLE>
<FORM method="post" name="Answers" action="<?php echo $PHP_SELF; ?>">

<?php
$QuestionNums = array();
$a=1;
while ($row = mysql_fetch_ass oc($result))
{ ?>
<TABLE>
<TR><TD><b><? echo "$a. ", $row ['Question']. "<br/>" ?></b></TD></TR>
<TR><TD><inpu t type="radio" name="<?php print 'Question'.$a ?>" value="1"> <? print $row ['Opt_1']. "<br/>" ?></TD></TR>
<TR><TD><inpu t type="radio" name="<?php print 'Question'.$a ?>" value="2"> <? print $row ['Opt_2']. "<br/>" ?></TD></TR>
<TR><TD><inpu t type="radio" name="<?php print 'Question'.$a ?>" value="3"> <? print $row ['Opt_3']. "<br/>" ?></TD></TR>
<TR><TD><inpu t type="radio" name="<?php print 'Question'.$a ?>" value="4"> <? print $row ['Opt_4']. "<br/>" ?></TD></TR>
</TABLE>
<?php

$i=$a;
$num.$i = $row['Q_Num'];
$QuestionNums[]=$num.$i;
print "<br/>";
$a++;
}//end while
//This is just printing them on screen to show they are there - it works
foreach ($QuestionNums as $value) {
echo $value."<br>";
}
?>
<TR><TD><inpu t type="submit" name="Submit" value="Submit"> </TD></TR>
</FORM>
</TABLE>
<?php
}//end function "show_form"


function process_form($Q uestionNums)
{
global $result, $QuestionNums;

//Call to ModDBQueries to connect to the database
dbConnect();

//TEST TO SEE IF IT WORKS
foreach ($QuestionNums as $value) {
echo $value."<br>";
}
}//end function "process_fo rm"
[/PHP]

With the code above, i get the array printing fine just above the submit button, but when submit is pressed i get:
Warning: Invalid argument supplied for foreach() in C:\Program Files\xampp\htd ocs\Lilly\PageF iles\ModDisplay Questions2.php on line 78

Line 78 is "foreach ($QuestionNums as $value) {" in the process_form function.


Any ideas why i can not get to the data and how to rectify the issue?
thanks again

[devsusen]

Hi Franky,

u r populating the $QuestionNums array within the function show_form(). Now when u click the submit button then this function show_form() is not called. So u r not populating the array. As a result u r getting the error of - Invalid argument supplied for foreach()

Populate the array using a function called globally, so while processing the form in the server side (after clicking the submit button) u can populate the array once again.

susen

[psrinivas33]

hi franky At first you have to take a variable like

$result1=$_POST['tot1'];
$result2=$_POST['tot2'];
[B].
.
.[/B]

$result="result";
for($i=1 $i<=5 $i++)
{
$sql=mysql_query("INSERT INTO <TAB-NAME> VALUES ('$$result[B].[/B]$i'[B],[/B].....)");

}

enjoyy..