每個孩子的循環jQuery

Question

我如何編寫使用for（i = 1; ...）的短代碼？函數，允許我遍歷每個孩子並添加一個不同的類。

$('#WebPartWPQ3 > table:first-child').addClass('blogpost1'); 
$('#WebPartWPQ3 > table:nth-child(2)').addClass('blogpost2'); 
$('#WebPartWPQ3 > table:nth-child(3)').addClass('blogpost3'); 
$('#WebPartWPQ3 > table:nth-child(4)').addClass('blogpost4'); 
$('#WebPartWPQ3 > table:nth-child(5)').addClass('blogpost5'); 
$('#WebPartWPQ3 > table:nth-child(6)').addClass('blogpost6'); 
$('#WebPartWPQ3 > table:nth-child(7)').addClass('blogpost7'); 
$('#WebPartWPQ3 > table:nth-child(8)').addClass('blogpost8'); 
$('#WebPartWPQ3 > table:nth-child(9)').addClass('blogpost9'); 

Answer 1

嘗試

$('#WebPartWPQ3 > table').each(function(i,j){ 

$(this).addClass("blogpost"+(i+1)); 
});