Monday, May 20, 2024

Ubuntu javascript-general

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1 Answer
114 Score
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Asked mireyag 16 Years ago, Friday, February 6, 2009 / rated 0 times / (+116) (-2)
1 Answer
57 Score
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Asked josefn 16 Years ago, Monday, February 9, 2009 / rated 0 times / (+62) (-5)
1 Answer
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Asked karladaijahf 16 Years ago, Monday, February 9, 2009 / rated 0 times / (+32) (-4)
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166 Score
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Asked colby 16 Years ago, Monday, February 9, 2009 / rated 0 times / (+167) (-1)
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Asked daijac 16 Years ago, Tuesday, February 10, 2009 / rated 0 times / (+25) (-5)
1 Answer
103 Score
77.5k Views
Asked judydestiniem 16 Years ago, Tuesday, February 10, 2009 / rated 0 times / (+108) (-5)
1 Answer
21 Score
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Asked daijab 16 Years ago, Thursday, February 12, 2009 / rated 0 times / (+22) (-1)
1 Answer
95 Score
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Asked albert 16 Years ago, Thursday, February 12, 2009 / rated 0 times / (+97) (-2)
1 Answer
98 Score
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Asked marvinm 16 Years ago, Friday, February 13, 2009 / rated 0 times / (+99) (-1)
1 Answer
168 Score
17.5k Views
Asked maxinec 16 Years ago, Tuesday, February 17, 2009 / rated 0 times / (+171) (-3)
1 Answer
119 Score
43k Views
Asked zackeryzainv 16 Years ago, Tuesday, February 17, 2009 / rated 0 times / (+123) (-4)
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62 Score
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Asked blair 16 Years ago, Tuesday, February 17, 2009 / rated 0 times / (+64) (-2)
Items 1-12 out of 19652 displayed
New Answers

Since popup.onload does not seem to work, here is a workaround: http://jsfiddle.net/WJdbk/.

var win =...
Tuesday, September 20, 2011, 12:00:00

This could be a great opportunity to use a Pipe instead of a function. This helps to avoid calling a function within an *ngFor which can have detrimental affects to performance....

Tuesday, October 9, 2018, 12:00:00

Clicking on a link means changing window.location, so how about

window.location = mailto:[email protected];
Sunday, June 26, 2011, 12:00:00

You can use .scroll() event in this way on your window:

$(window).scroll(function() {
   if($(window).scrollTop() + $(window).height() == $(document).height()) {...
Tuesday, March 13, 2012, 12:00:00
 if (n%2 === 1) {
    return 0;
 }

That will stop the recursive chain at every second number. For every odd number it will stop directly (and return 0) for every even...

Friday, February 2, 2018, 12:00:00