mbitenieks Posted February 22, 2012 Report Posted February 22, 2012 Meerkjis ir pievienot <div></div> arpus <ul> ja ir vairaak par vienu <li></li> Bet ar sho nesanaak: $j("ul").find("li").eq(1).addClass("vairak-par-vienu"); $j("li.vairak-par-vienu").parent().parent().append('<div></div>'); Paldies! Quote
0 codez Posted February 22, 2012 Report Posted February 22, 2012 http://jsfiddle.net/Gy3MC/ $('ul').each(function(){ if($(this).children().length>1){ $(this).after($('<div>hello</div>')); } }); Quote
0 mbitenieks Posted February 22, 2012 Author Report Posted February 22, 2012 (edited) Paldies par tik aatru atbildi! Google Chrome izmet Uncaught SyntaxError: Unexpected token ILLEGAL Divaini jo taa nevajadzeetu buut. bet tomeer palaizhot lapaa. nestradaa! $j(document).ready(function(){ $j('#circle-slider').each(function(){ if($j(this).children().length>1){ $j(this).after($j('<div class="single-blog-nav"> <div id="next2"></div> <div id="prev2"></div> <div class="clear"></div> </div>')); } }); }); Paldies! Edited February 22, 2012 by mbitenieks Quote
0 codez Posted February 22, 2012 Report Posted February 22, 2012 Google Chrome izmet Uncaught SyntaxError: Unexpected token ILLEGAL http://stackoverflow.com/questions/4404526/unexpected-token-illegal-in-webkit Quote
0 mbitenieks Posted February 22, 2012 Author Report Posted February 22, 2012 Iztesteeju jau visu ko tur aprakstiija. bet tikuntaa neliidzeeja. Varbuut ir kaada savaadaaka iespeeja? cits funkcijas veids? Paldies! Quote
0 mbitenieks Posted February 22, 2012 Author Report Posted February 22, 2012 Sanaaca $j(document).ready(function(){ if($j('#circle-slider').children().length>1){ $j('#circle-slider').after($j('<div class="single-blog-nav"> <div id="next2"></div> <div id="prev2"></div> <div class="clear"></div> </div>')); } }); Paldies! Quote
0 codez Posted February 22, 2012 Report Posted February 22, 2012 Problēma nav kodā, bet tajā, ka esi ielicis kodā kādu neatļautu simbolu, vai nu kopējot, vai kā savādāk, jo jsfiddlē kods darbojas. Iesaku izmantot kādu IDE (piem. Netbeans) ar sintakses pārbaudītāju, tāds uzreiz uzrādīs, kurā vietā ir neatļauts simbols. Quote
0 indoom Posted February 22, 2012 Report Posted February 22, 2012 Intereses pēc paskatījos, vai nav kāds cits variants $('ul').has('li + li').after('<div>hello</div>'); lēnāks, bet ātrāks par each() tikai, ja ir vismaz 10 vai vairāk ul lapā $('ul').has('li:not(:only-child)').after('<div>hello</div>'); vispār lēns Quote
Question
mbitenieks
Meerkjis ir pievienot <div></div> arpus <ul> ja ir vairaak par vienu <li></li>
Bet ar sho nesanaak:
$j("ul").find("li").eq(1).addClass("vairak-par-vienu");
$j("li.vairak-par-vienu").parent().parent().append('<div></div>');
Paldies!
7 answers to this question
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