hello all,
i m using Jquery ajax post method to edit a form on same page, but if there is some mistake then how do i send user back on that page where data were
loaded.
now i describe u what i do?
i have a page manageMovies.php there are list of movie name, now when i click on a name of any movie,
then i load editMovie.php on same page
now when i do some mistakes( i.e when validations fails) then i want to go back on same page
manageMovies.php
loaded with editunit.php regarding that movie on the page.
here is my page structure
manageMovies.php
<div id="display"></div>
<div id="movieList">
<table >
<tr><td id="mov_10">Apharan</td></tr>
<tr><td id="mov_11">Gangaajal</td></tr>
<tr><td id="mov_12">Rajniti</td></tr>
</table>
</div>
<script type="text/javascript">
jQuery('td').click(function () {
jQuery('#movieList').hide(); // hide the div 'movielist'
jQuery.post('editMovie.php', {
idForEdit: jQuery(this).attr('id')
}, function (data) {
jQuery("#display").html(data); //display the editMovie.php page on 'display' div
});
});
</script>
now when i do some mistakes on editunit.php and go further for post, then i need to go back on same page (manageMovies.php) where
editMovie.php is shown on display div and movielist div should be hidden
