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For my PHP framework I have a debug window which pops up when debugging is enabled

 

var dwindow = window.open('debug.php','debugwindow',',toolbar=1,resizable=1,scrollbars=yes');

 

Now at the moment this line is used every time a page is loaded (in the main window) it always opens in the same single pop up window.

 

Im wondering is there a way I can either check to see if the window is open then issue the line of code above.

OR

If I have to use window.open every time, is there a way I can only let a set of code run once the debug.php has been opened?

 

Thanks

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Im wondering is there a way I can either check to see if the window is open then issue the line of code above.

 

I don't quite understand what you are trying to achieve.

 

If you want to open the window only if you have opened it before.

 

In theory:

1: Make an object

2: Give that object a trigger property of 0.

3: Set it to one when you open a new window.

4: Make a method that checks if trigger == 0.

Will try explain again:

 

User opens index.php

If debug mode is enabled a single pop up window is opened (only one window is required, code in first post does this)

When user requests additional pages eg index.php?page=news (as an example)

I need some javascript that will check to see if the previous debug window that has been opened, is still open.

 

To put additional items into my debug window I use jQuery prepend function. So I need to check that the debug page has loaded before I can prepend items to my div table.

 

Guess the last one is the most important, as I need to make sure that the debug window is loaded first before trying to make changes in the window

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