[Resolved] function menu name not displaying correctly.

[lmayer4]

Afternoon,

I am sure there is a simple explanation but it escapes me. I have a funtion that displays a select box. I need the function so I can save the array in a session. Now I need it display like this:

<select name="rowdata[0][Facility]">

what its doing is this:

<select name="rowdata[$cnt][Facility]">

<?
for ($cnt=0;$cnt<$lngNumRows;$cnt++) {
// this will be the name="" value in the html select list
$menuName = 'rowdata[$cnt][Facility]';
if (isSet($POST[$menuName])) {
$SESSION[$menuName] = $_POST[$menuName];
}
?>

for my other input boxes I have it like this:

<select name="rowdata[<?PHP echo $cnt; ?>][TrailerType]">

This one works fine. How can I get the other little bugger to do what I want?

Thanks

Laura

[kingbear]

You just need double quotes.

Replace this:

$menuName = 'rowdata[$cnt][Facility]';

with this:

$menuName = "rowdata[$cnt][Facility]";

[lmayer4]

Ah thanks for that. Now I have a new problem. It won't hold the session value.....Oh well back to the drawing board.

Thanks again

Laura

[rehfeld]

i have a feeling your using my code 🙂

http://codewalkers.com/seecode/507.html

are you trying to use it to do multiple select lists on the same page?

post more of your code and ill help

[lmayer4]

Yours and Thomas' 🙂 He suggested I take the select menu and put it in an array so the session would work. I've been coming back to it every once in a while.

I've attached it below

Thanks for any ideas you may have.

Laura

[tsinka]

Hi,

the problem is this:

The correct syntax to get the post value is:

$_POST['rowdata'][$cnt]['Source']

So you need to change the code so that it uses that syntax.

One question (it's been a while so I don't remember completely):

Why are you storing the values in the session in this case (better: just the Source values) ? Do you need them on another page ?

Thomas

[lmayer4]

Thanks to your brilliant coding 🙂 it saves them on the page then collects them all and submits them together in an insert statement. I'm just trying to get fancy nad have 2 fields remember the values. Source and location.

Laura

[tsinka]

Ok,

replace the complete block with

<?
  $selectMenu = makeSelectMenu("rowdata[".$cnt."][Source]", $arrFacilities, $arrRowData[$cnt]['Source']); 
  echo $selectMenu; 
?> 

If someone submits the form then $arrRowData holds all the form data. Just use $arrRowData[$cnt]['Source'] as function argument.

Thomas

[tsinka]

Note:

Add some form validation code to make sure that only valid values are submitted (and to prevent things like SQL injection).

Thomas

[lmayer4]

As always thank you Thomas.

One more question. Is is possible, and maybe it's not, but when the user clicks on add a row is there a way for the form to remember what the user picked last time? That's what I've been trying to do when I am mangling what you gave me.

Laura

[tsinka]

Hmmm ... did you already try it with the new code ? The code to add a row already is there and should work.

Thomas

[lmayer4]

Yes, it remembers the last one perfectly but and heres my issue, when the user adds a row it goes back to the default row not the one they've previously selected.

so if the last time the selected "Chester"

"Chester" is remembered but when they click on add row "Aderdeen" is now presented.

[tsinka]

I hope I understood completely:

Let's say you have n rows with several different Sources selected and the user clicks on "Add row" then all rows show Aberdeen or do all rows show the previously selected Sources and a new row is added with a default selection of Aberdeen ?

The present code should remember all selections of the already existing rows and then add a row at the end with the default value "Aberdeen".

Thomas

[lmayer4]

To answer:

You asked----
"Let's say you have n rows with several different Sources selected and the user clicks on "Add row" then all rows show Aberdeen or do all rows show the previously selected Sources and a new row is added with a default selection of Aberdeen ?

All rows show the previously selected sources and the new added row shows aberdeen. I want it to be 'smart' and remember the last row added and make that the next row added. That's what I was wondering if that was possible.

"The present code should remember all selections of the already existing rows and then add a row at the end with the default value "Aberdeen"."

Which is exactly what it does.

[tsinka]

Ok,

then try something like this:

......
  if (isset($_POST['addrow'])) {
    // addrow clicked, increase counter by one
    // and add default row to row array to avoid
    // PHP warnings
    $arrNewRow = $arrDefaultRow;

if ($lngNumRows > 0) {
  $val = $arrRowData[$lngNumRows-1]['Source'];
  $arrNewRow['Source'] = $val;
}

$arrRowData[] = $arrNewRow;
$lngNumRows += 1;    
  } else if (isset($_POST['submit'])) {
..........

Thomas

[lmayer4]

That's it!!! 🙂

Thankyou so much.

Laura