PHP: $_POST of input that has a variable name

Question

So, I'm having trouble figuring this one out. I populate a select box with values from Excel, and give the select box the name of the worksheet.

function populateDropDown($excelSheet) {
$objPHPExcel = PHPExcel_IOFactory::load($excelSheet, $num);
foreach ($objPHPExcel->getWorksheetIterator() as $worksheet) {
    $worksheetTitle     = $worksheet->getTitle();
    $highestRow         = $worksheet->getHighestRow(); // e.g. 10
    $highestColumn      = $worksheet->getHighestColumn(); // e.g 'F'
    $highestColumnIndex = PHPExcel_Cell::columnIndexFromString($highestColumn);
    $nrColumns = ord($highestColumn) - 64;
    echo '<select name= ' . $worksheetTitle . '>';
    for ($row = 1; $row <= $highestRow; ++ $row) {

        for ($col = 0; $col < $highestColumnIndex; ++ $col) {
            $cell = $worksheet->getCellByColumnAndRow($col, $row);
            if ($col == 0) {
                $id = $cell->getValue();
            } else if ($col == 1) {
                $val = $cell->getValue();
            }
        }
        echo '<option value=' . $id . '>';
        echo $val;
        echo '</option>';
    }
    echo '</select>';
}
return $worksheetTitle;
}

At the end of this function, I return the title and it is stored into a variable on the form page.

<?php $selectBox1 = populateDropDown("workbook1.xlsx"); ?>

When the form is submitted, how can I get the data of this variable? I've tried the following without it working:

$var = $_POST[$selectBox1];
$var = $_POST['$selectBox1'];
$var = $selectBox1;

Hardcoding the sheet name works, however:

$var = $_POST['Sheet1'];

So, I'm positive it's a problem with the variable. I'm sorry if this has been asked before, I searched for quite some time but didn't find anything that helped. Thank you for any help!

Again, just to clarify my question, how can I use $_POST to get the data from the select box, when the name of the select box was named with a variable?

Edit: added entirety of populateDropDown function

What are the <option> elements of that select? maybe you could transmit the name of the worksheet via <input type=hidden" name="worksheet" value="The Name" />? — VolkerK
– VolkerK, Commented Jul 9, 2015 at 22:45
populateDropDown obviously some framework we are not seeing here — user557846
– user557846, Commented Jul 9, 2015 at 22:47
From what I gather populateDropDown is a custom function from which the first 3 lines of code are taken. I think a more through description of what you're doing is necessary. Could you post the complete form code? I think you may have forgotten to make the form submission method POST. — jeteon
– jeteon, Commented Jul 9, 2015 at 23:30

DJC · Accepted Answer · 2015-07-09 22:52:21Z

1

Why don't you also send a hidden value with the worksheet title, giving the hidden input a name that you DO know, then you can grab in from the $_POST array.

<input type="hidden" name="my-select-name" value="<?php echo $worksheetTitle;?>">

Then after the form is posted, in your processing script you can get the value and use it to identify your select box.

$value = $_POST['my-select-name'];

$selectdata = $_POST[$value];

edited Jul 9, 2015 at 22:52

answered Jul 9, 2015 at 22:47

DJC

1,1661 gold badge15 silver badges46 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Kyle Over a year ago

Thank you, I think this is the simplest way to do it.

Mm-Art-In · Accepted Answer · 2015-07-09 22:52:06Z

What you want to do is reshape your naming of your select box, in HTML try making the select box value into an array:

<select name='[select][".$worksheetTitle."]'>
<option>...</option>
</select>

Then when you get the $_POST data, you can run a foreach through the array of [select] which will be an array of 1 value:

if (count($_POST['select']) > 0 ){
    foreach ($_POST['select'] as $key=>$value){
        $this_is_the_worksheet_name = $key;
        $this_is_the_worksheet_select_value = $value;
    }
}

This will give you access to the values, this is a little bit long winded but your question seems to be abridged, so I've worked a bit of a long way round. The reason to put the select choices into an array is that with foreach accessing the array key (thus the worksheet name) is trivial.

Collectives™ on Stack Overflow

PHP: $_POST of input that has a variable name

2 Answers 2

1 Comment

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related