[Solved] Setting default positions?

[microbe]

No doubt this has been answered before but I can't find it.

I have multiple tips on a page. They all have the same position info, only the content changes.

Currently I have

JS Code

$("#tip1").qtip({
	  		 content: 'Tip 1',
	   			style:{
				tip:"topLeft"
			},
	   		position: {
		  		corner: {
			 		target: 'bottomRight',
			 		tooltip: 'topLeft'
		 		 }
	  		}
		});
		$("#tip2").qtip({
	  		 content: 'Tip 2',
	   			style:{
				tip:"topLeft"
			},
	   		position: {
		  		corner: {
			 		target: 'bottomRight',
			 		tooltip: 'topLeft'
		 		 }
	  		}
		});

I have tried

JS Code

$().qtip({   ***(and $qtip({)***
	   		position: {
		  		corner: {
			 		target: 'bottomRight',
			 		tooltip: 'topLeft'
		 		 }
	  		}
		});
	$("#tip1").qtip({
	  		 content: 'Tip 1',
	   			style:{
				tip:"topLeft"
			}
	   	});
		$("#tip2").qtip({
	  		 content: 'Tip 2',
	   		style:{
				tip:"topLeft"
			     }
	  		}
		});

I am sure there is a way to do it, but I am too thick to work it out.

Thanks in advance.

[kiddailey]

You can setup a default configuration object and then use the jQuery extend method to combine it with your other configuration option. Like so:

JS Code

tipOptions = 
	{
		style: {
			tip:"topLeft"
		},
		position: {
			corner: {
				target: 'bottomRight',
				tooltip: 'topLeft'
			}
		}
	}
 
$("#tip1").qtip($.extend(tipOptions, 
	{
		content: 'Tip 1'
	})
);
 
$("#tip2").qtip($.extend(tipOptions, 
	{
		content: 'Tip 2'
	})
);

Working jsFiddle example:

http://jsfiddle.net/kiddailey/RMmhY/