如何从php中的视图页面传递值到控制器？

Question

I want to upload several files to the system. For that I used several checkboxes to select relevant upload and I want to pass that value from

form_open_multipart('controller function') 

to controller page. When selecting upload type I set value to label in view page.Then using

form_open_multipart()

function I direct it to controller function. I want to get the value of label. how can I do this?

I set value from js to view element (label) using

$('#set').text(send_id);

<?php 
echo form_open_multipart('uploadFile/doupload');   
?>

<table width="50%" border="0" align="right" cellpadding="0" cellspacing="0" bordercolor="#e65C00" style="border-radius: 5px;">

    <tr id="sliding" style="display:none;">
        <td>
            <label id="set" name="set"></label>
        </td>
        <td>

            <h4>Upload File</h4>
            <div  class="panel panel-warning" id="color" style="background: #ffffff; width:100%; height: auto;">
                Select File To Upload:<br />
                <input type="file" id="userfile" name="userfile" size="20" />

                <br /><br />

                <input type="submit" name="upload_file" value="Upload" class="btn btn-success">

            </div>
        </td>
    </tr>
</table> 
<?php echo form_close(); ?>

Answer 1

You can do any of the following 3

• Post the values directly from form.
• You can keep the value in a hidden field and then post it.
• You can pas the values as the parameter for the controller function.