Print photos

[Ben]

Hi Valch,

Sorry for the wait!

To output an image as an actual image instead of just the URL/filename, you'll need something like this. I made it as simple as possible. This is based on the "table format" HTML/Printer-friendly Export Type.

Code:

<h1>{$form_name} - {$view_name}</h1>

<table cellpadding="2" cellspacing="0" width="100%" class="print_table">
<tr>
  {foreach from=$display_fields item=column}
    <th>{$column.field_title}</th>
  {/foreach}
</tr>
{strip}
{foreach from=$submissions item=submission}
  {assign var=submission_id value=$submission.submission_id}
  <tr>
    {foreach from=$display_fields item=field_info}
      {assign var=col_name value=$field_info.col_name}
      {assign var=value value=$submission.$col_name}
      <td>
        {if $field_info.field_name == "field2"}
          <img src="http://www.yoursite.com/folder/{$value}" />
        {else}
          {smart_display_field form_id=$form_id view_id=$view_id
            submission_id=$submission_id field_info=$field_info
            field_types=$field_types settings=$settings value=$value}
        {/if}
      </td>
    {/foreach}
  </tr>
{/foreach}
{/strip}
</table>

The important bit is this:

Code:

{if $field_info.field_name == "field2"}
          <img src="http://www.yoursite.com/folder/{$value}" />
        {else}
          {smart_display_field form_id=$form_id view_id=$view_id
            submission_id=$submission_id field_info=$field_info
            field_types=$field_types settings=$settings value=$value}
        {/if}

That looks at the current field name. If it's "field2" (which here, contains the image field - you will need to change that to whatever field name your image field(s) have).

You'll also need to update the URL to the file as well.

Hope this helps -

Ben