Javascript using window.location seems to lose state


Posted on 16th Feb 2014 07:03 pm by admin

Not sure what forum so let me know if I'm in the wrong place. I have a main window with a ajax grid that has a column of links that uses javascript and window.location to go to another page to show check stub information. When I use the backbutton on the browser everything persists on main window. Ok so far. If I navigate to the window with the show check stub information, there is a 'Print' button that again thru javascripts prints the grid with the check info on it. (Code I got of the internet with some minor changes I made).view plaincopy to clipboardprint?function Print() { var curVisible = true; var grid_obj = document.getElementById('<%= grdVoucherDetail.ClientID %>'); for (var i = 1; i < grid_obj.rows.length-1; i++) { grid_obj.rows[i].cells[8].style.display = curVisible ? "none" : "inline"; grid_obj.rows[i].cells[9].style.display = curVisible ? "none" : "inline"; } if (grid_obj != null) { var new_window = window.open('print.html'); //print.html is just a dummy page with no content in it. new_window.document.write(grid_obj.outerHTML); new_window.document.close(); new_window.focus(); new_window.print(); new_window.close(); } curVisible = false; for (var i = 1; i < grid_obj.rows.length-1; i++) { grid_obj.rows[i].cells[8].style.display = curVisible ? "none" : "inline"; grid_obj.rows[i].cells[9].style.display = curVisible ? "none" : "inline"; } } function Print() { var curVisible = true; var grid_obj = document.getElementById('<%= grdVoucherDetail.ClientID %>'); for (var i = 1; i < grid_obj.rows.length-1; i++) { grid_obj.rows[i].cells[8].style.display = curVisible ? "none" : "inline"; grid_obj.rows[i].cells[9].style.display = curVisible ? "none" : "inline"; } if (grid_obj != null) { var new_window = window.open('print.html'); //print.html is just a dummy page with no content in it. new_window.document.write(grid_obj.outerHTML); new_window.document.close(); new_window.focus(); new_window.print(); new_window.close(); } curVisible = false; for (var i = 1; i < grid_obj.rows.length-1; i++) { grid_obj.rows[i].cells[8].style.display = curVisible ? "none" : "inline"; grid_obj.rows[i].cells[9].style.display = curVisible ? "none" : "inline"; } }
Did you know?Explore Trending and Topic pages for more stories like this.
Now after the print and closing that popup I click the backbutton again and things don't seem to persist. I did notice that the pageload of the main window (one I'm going back to) gets triggered after the printing of the grid where if I don't click the print button and go back the pageload does not get triggered. Any ideas would be appreciated. I have searched and searched for an answer and am getting nowhere. When this happens I usually feel that I've done something stupid. Oh well Thanks for any help that you can give me
No comments posted yet

Your Answer:

Login to answer
87 Like 21 Dislike
Previous forums Next forums
Other forums

getting most records by count
Code: [Select]<?php

$connect = mysql_connect("localhost","dam

Xml parsing
I need a suggestion about parsing xml with multiply parts like pervious...
i.e. different device

Code error with Index.php
Error: Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/runevid/public_

question about stripslashes and real_escape_string
im cleaning up an old app that I wrote fixing some of the vulernabilities from attacks.

I hav

first id from db not showing
I have a php script which displays the content of a mysql table as a html table with sorting, delete

php mail form text wont appear / javascript included
Guys/gals...

I am running into a problem whereby I have a great piece of javascript code that

How to kill asynchronous postback / current postback?
Hi,here is my problem:I have a web site with many pages of which some may take time to process resul

MySQL Does not UPDATE- SQLString Problem
vb Syntax (Toggle Plain Text) 1. SQL = "UPDATE sampletable SET column1 = 'C1sample1'"

need help in php variable
i have a php variable on one page
Code: [Select]$lastId = mysql_insert_id($db);
echo $lastId;<

mysql query with single quotes in a variable
$sitedetails = "INSERT INTO vars (address, sitename, description, ownername, theme) VALUES ('$u

Sign up to write
Sign up now if you have flare of writing..
Login   |   Register
Follow Us
Indyaspeak @ Facebook Indyaspeak @ Twitter Indyaspeak @ Pinterest RSS



Play Free Quiz and Win Cash