window.open parameters as a variable

Question

Within a function I have a line of code that opens a window whilst drawing its parameters either from the functions own parameters (u.n) or variables created within the function.

This works fine in the script.

win2 = window.open(u, n, 'width=' + w + ', height=' + h + ', ' + 'left=' + wleft + ', top=' + wtop + ', ' + tools);

As this is called several other times in the script but as win3, win4 etc. , to reduce code, I wanted to put the parameters, which are the same each time, into a variable and just use that each time.

myparameters =  u + ',' + n + ',width=' + w + ', height=' + h + ', ' + 'left=' + wleft + ', top=' + wtop + ', ' + tools;

win3 = window.open(myparameters);

I have tried playing around with this without much luck, can it be done?

Thanks.

you're trying to pass a single string whereas all those are expected as individual parameters. each parameter has a different meaning. Refer developer.mozilla.org/en-US/docs/Web/API/Window.open — Surender
– Surender, Commented Oct 29, 2013 at 14:03

Abhitalks · Accepted Answer · 2013-10-29 14:24:35Z

2

Yes you can, to some extent by wrapping it in function call. What I typically do is to have a utility function which I can call upon whenever required.

Something like this:

popOpen: function (url, target, height, width) {
        var newWindow, args = "";
        args += "height=" + height + ",width=" + width;
        args += "dependent=yes,scrollbars=yes,resizable=yes";
        newWindow = open(url, target, args);
        newWindow.focus();
        return newWindow;
}

You can further reduce the parameters by making it an object like:

popOpen: function (params) {
        var newWindow, args = "";
        args += "height=" + params.height + ",width=" + params.width;
        args += "dependent=yes,scrollbars=yes,resizable=yes";
        newWindow = open(params.url, params.target, params.args);
        newWindow.focus();
        return newWindow;
}

And you can call it like this:

var param = { url: '...', height: '...', width: '...' };
popOpen(param);

Or,

var param = new Object;
param.url = '...';
param.height = '...';
popOpen(param);

edited Oct 29, 2013 at 14:24

answered Oct 29, 2013 at 13:58

Abhitalks

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5 Comments

putvande Over a year ago

And how are you going to execute this function with only 1 variable?

Abhitalks Over a year ago

@putvande while you were downvoting, I was editing the answer to include an object variable. This makes it one variable to pass. Is that good enough?

putvande Over a year ago

Not sure.. is up to OP to decide I guess. But it still has the same problem. (I have removed the downvote though)

slebetman Over a year ago

The OP doesn't really want to call a function with "one variable". He wants one variable that he can call window.open with that changes "magically" each time he uses that variable. What the OP really wants is a function.

Abhitalks Over a year ago

Just wrapping the call. It will still be a function call for the op, but with reduced code every time he makes a call. And yes, thanks a lot @putvande for considering.

Guilherme Sehn · Accepted Answer · 2013-10-29 14:04:34Z

1

The way you are trying is not possible. You might want to do this:

var myparameters =  'width=' + w + ', height=' + h + ', ' + 'left=' + wleft + ', top=' + wtop + ', ' + tools;
win3 = window.open(u, n, myparameters);

Fiddle: http://jsfiddle.net/aBR7C/

edited Oct 29, 2013 at 14:04

answered Oct 29, 2013 at 13:59

Guilherme Sehn

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5 Comments

Naveen Kumar Alone Over a year ago

@martin what problem you are getting?

slebetman Over a year ago

He wants the variable to be "dynamic", that is, each time he uses it the w and h gets a new value. He's got a big misunderstanding of how variables work. what he really wants is a function.

Guilherme Sehn Over a year ago

Edited the answer and added an example running at jsfiddle. It seems to be working fine.

martin Over a year ago

Yes, I found a typo in my code, this does seem to work. I will just double check it works with all browsers, then mark this as my answer. Thanks.

martin Over a year ago

Works okay in FF, but with Chrome I am getting windows that aren't sized correctly, so it probably isn't wise to do it this way.

Joel · Accepted Answer · 2013-10-29 14:22:23Z

0

You are missing some additional parameters in your function call:

var myparameters =  'width=' + w + ', height=' + h + ', ' + 'left=' + wleft + ', top=' + wtop + ', ' + tools;

win3 = window.open(u, n, myparameters);
                   ^  ^   ^
                 //Here you are passing parameters

edited Oct 29, 2013 at 14:22

Joel

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answered Oct 29, 2013 at 14:00

Naveen Kumar Alone

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