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Thread: underscore

  1. #1
    Join Date
    Sep 2006
    Posts
    375

    underscore

    Hi,

    I want to display few rows per page (say 6), from a mysql table with php.
    Now if there are more than 6 I have set up a form at the bottom where you click on a number eg 1 2 3.... page no. you can view at the bottom of this forum.

    How can you display a number with an underscore using php?

    echo "<form action=\"image_nextrow.html\" action=\"post\">";
    $num="<u>2</u>" ;
    echo "<input type=\"image\" value=".$num." />";//fails

    Unless you can use a simple hyperlink and send information to that page another way?

  2. #2
    Join Date
    Aug 2004
    Location
    Ankh-Morpork
    Posts
    18,921
    You can use a link with a query string:
    HTML Code:
    <a href='target_page.php?page=2'>2</a>
    Then in the target page, the value will be accessed from the $_GET array ($_GET['page'] in the above example).
    "Please give us a simple answer, so that we don't have to think, because if we think, we might find answers that don't fit the way we want the world to be."
    ~ Terry Pratchett in Nation

    eBookworm.us

  3. #3
    Join Date
    Sep 2006
    Posts
    375
    Your solution looks easier than what I am doing.

    Can you get this to work?
    q) On 1 form I have this
    echo "<form action=\"image_nextrow.html\" action=\"post\">";
    echo "<input type=\"hidden\" name=\"data1\" value=\"ss\"> ";
    echo "<input type=\"hidden\" name=\"data2\" value=\"ww\"> ";
    echo "<input type=\"image\" value=\"submit\" src=\"2\" alt=\"2\" />";
    echo "</form>";

    On image_nextrow.html in the php section I have this and it fails to display anything, however in the url I notice the variables are being passes, so what is going on?

    $data1= $_POST['data1'];
    $data2= $_POST['data2'];

    echo $data1 ."hrrrrr";
    echo $_POST['data2'];

  4. #4
    Join Date
    Oct 2006
    Location
    Indonesia
    Posts
    70
    it because your script(image_nextrow.html) detected by server as html not as php script, so it doesn't execute your php code, so you need to change the page as image_nextrow.php.
    alternativly, if you just want your page as .html but can run php code (assume u use apache server and do php as apache module), you should add this line
    AddType application/x-httpd-php .html
    on your(file) httpd.conf, on the part of
    <IfModule mod_php5.c>
    (if u use php5, otherwise just fine the line like AddType application/x-httpd-php, and below it add AddType application/x-httpd-php .html)

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