如何迭代第三個單元格具有rowspan屬性的每一行

Question

for (var i = 0; row = tableAppointment.rows[i]; i++) { 
    for (var j = 0; col = row.cells[j]; j++) { 
    //iterate through columns 
    //columns would be accessed using the "col" variable assigned in the for loop. 
    } 
} 

如何遍歷第三個單元格具有rowspan屬性的每一行。

Answer 1

這會給你所有對他們有這樣的元素<tr>的：

$('td:nth-child(3)[rowspan]').parent() 

演示在http://jsfiddle.net/52aR2/1/

Answer 2

像這樣的東西應該工作

$('tr').filter(function() { 
    return $('td:eq(2)',this).attr('rowspan') !== undefined; 
}).css('border', '1px solid red')​​​​​​​​​​​​​​​​​​​​;​ 

http://jsfiddle.net/9gMRk/

Answer 3

你可以做到這一點Y：

if (cell[2].hasAttribute('rowspan')) { 
    // do something here 
} 

用途element.hasAttribute()

Answer 4

$("tr").filter(function() { 
    return this.cells[2].hasAttribute("rowspan"); 
}); 

http://jsfiddle.net/52aR2/2/

或者

for (var i = 0; row = tableAppointment.rows[i]; i++) { 
    if(row.cells[2].hasAttribute("rowspan") { 
     //This is a row that matches 
    } 
} 

Answer 5

Array.prototype.slice.call(tableAppointment.rows). 
    filter(function(row) { return row.cells[2].rowSpan > 1; }). 
    forEach(function(row) { 
     // do something with row 
    }); 

Whoaaa，看媽媽，沒有jQuery的！

Answer 6

$('td:nth-child(3)[rowspan]').parent() 

Answer 7

這實際上是正確的答案。

$('td:nth-child(3)[rowspan]').parent()