2016-11-27 92 views
0

我已經使用jquery創建了一個錶行,並且我想向它填充一些數據。當我使用console.log進行檢查時,數據是合適的,但未填寫該列。插入數據到jquery表

我該如何填寫欄目?

function getCorporateService(id){ 
//create new row 
    $br = "<tr id='item'>"; 
    $br += "<td> <input class='input-small' type='text' id='order_identifier' name='order_identifier' readonly></td>"; 
    $br += "<td><input class='input-small' type='text' id='service["+id+"]' name='service["+id+"]' readonly></td>"; 
    $br += "<td><select id='order_type["+id+"]' name='order_type["+id+"]'> <option> - </option> <option value='add'>Add</option> <option value='change'>Change</option> <option value='cancel'>Cancel</option> </select></td>"; 
    $br += "<td><input class='input-small' type='text' id='select_plan["+id+"]' name='select_plan["+id+"]'></td>"; 
    $br += "<td><input class='input-mini' type='text' id='qty["+id+"]' name='qty["+id+"]'></td>"; 
    $br += "<td><input class='input-small' type='text' id='unit_price["+id+"]' name='unit_price["+id+"]'></td>"; 
    $br += "<td><input class='input-small' type='text' id='total_price["+id+"]' name='total_price["+id+"]'></td>"; 
    $br += "<td><input class='input-small' type='text' id='notes["+id+"]' name='notes["+id+"]'></td>"; 
    $br += "</tr>"; 

    $(".corporatesvc").append($br); 

    //get data and parsing to column 
    $.get("{{ url('salesorder/service')}}/"+id, function(data){ 
     console.log(id); 
     console.log(data); 
     $.each(data, function (index, element){ 
      $('#service['+id+']').val(element.service_name); 
     }); 
    }); 
} 

回答

1

我發現下列編碼工作正常。 如果javascript更改爲

$('#service\\['+id+'\\]').first().val(element.service_name);