Send data to form using link

Question

I don't know to do the script, when click link, it will send data to form. What im trying to do is . When user click link data will appear at the form. below is my link. generated from database.

<div id="menu_bar"  region="west" split="true" title="Pages listing" style="width:200px;padding:10px;">
<?php 
$parent = mysql_query("select * from pages where parent = 0");
            echo "<ul id='sitemap'>";

        while($row = mysql_fetch_array($parent)){
            $parent_id = $row['id'];
            $parent_name = $row['name'];
        echo "<li><a href='#' onclick='editPage()'>$parent_name ($parent_id)</a>";
        echo "<ul>";
                $child = mysql_query("select * from pages where parent = '$parent_id'");
                while($row = mysql_fetch_array($child)){
                $child_id = $row['id'];
                $child_name = $row['name'];
                    echo "<li><a href='list2.php?id=$child_id' onclick='editPage()'>$child_name ($child_id)</a></li>";
                }
        echo "</ul>";
        }

        echo "</li>";
        echo "</ul>";
?>
</div>

And the form.

Basic Information
Name:

Parent:

Order:

Body:

Special:

Please help me, im still in programming field. im using PHP, JSON,JQUERY, EASY-UI.

Thanks

Could you please post the sendId funcion? Also the line says sendId('$id') but I don't see the $id variable in your PHP code. — Carlos Vergara, Jul 27 '12 at 01:09
i still not create the sendId function. iam not sure what i want to put inside the function — hemiz, Jul 27 '12 at 01:16
I just saw the $id variable. How about you catch the data and display it in a message box for now, only to see if it's actually getting the value? — Carlos Vergara, Jul 27 '12 at 01:32
Is it like when a user clicks a name all his details should be displayed on a new form — Sibu, Jul 27 '12 at 04:23
@Sibu : Yes exactly.. can u give me example of javascript or jquery function? — hemiz, Jul 27 '12 at 06:45
do you want to display the content on a new page or same page.. — Sibu, Jul 27 '12 at 07:31

score 0 · Answer 1 · answered Jul 27 '12 at 09:29

If you want to edit page detail in new page then no need to call onclick='editPage()' instead of that pass the page url in href.

<?php 
    $parent = mysql_query("select * from pages where parent = 0");
                echo "<ul id='sitemap'>";

            while($row = mysql_fetch_array($parent)){
                $parent_id = $row['id'];
                $parent_name = $row['name'];
            echo "<li><a href='edit_page.php?id=".$row['id']."' >$parent_name ($parent_id)</a>";
            echo "<ul>";
                    $child = mysql_query("select * from pages where parent = '$parent_id'");
                    while($row = mysql_fetch_array($child)){
                    $child_id = $row['id'];
                    $child_name = $row['name'];
                        echo "<li><a href='edit_page.php?id=".$row['id']."' >$child_name ($child_id)</a></li>";
                    }
            echo "</ul>";
            }

            echo "</li>";
            echo "</ul>";
    ?>

edit page code should be like this

<?php 
    $id = $_REQUEST['id'];
    //use ur table name here
    //I am considering that ur table contain all the required information
    $result = mysql_query("select * from table where id='".$id."'");
    $row = mysql_fetch_assoc($result);
    ?>
    <form>
    Name <input type="text" name="name" id="name" value="<?php echo $row['name']; ?>" />
    Parent <input type="text" name="name" id="name" value="<?php echo $row['parent']; ?>" />
    Order <input type="text" name="name" id="name" value="<?php echo $row['order']; ?>" />
    Body <input type="text" name="name" id="name" value="<?php echo $row['body_text']; ?>" />
    Special <input type="text" name="name" id="name" value="<?php echo $row['special']; ?>" />
    </form>

hope this will help you.

Send data to form using link

1 Answers1