使用eval()JS动态创建对象

Question

我试图在JavaScript中动态创建一个对象.这是我写的JS代码:

var table = $("#eidtPersonalInfoTbl");
var trs = table.find('tr');

var obj = { };

$(trs).each(function(index, row){
     var field = $(row).find('td').eq(0).html();
     var value = $(row).find('td').eq(1).html();
     eval('obj.' + field + ' = ' + value );
});

这是表格的HTML标记:

<table class="table" border="1" id="eidtPersonalInfoTbl">
<tr> 
    <td class="span3 hidden-phone" > Name </td>
    <td class="span5"> Name </td>       
</tr>
<tr>
    <td class="span3 hidden-phone"> Address</td>
    <td class="span5"> Address </td>
</tr>
<tr>
    <td class="span3 hidden-phone">Area</td>
    <td class="span5"> Area</td>                                        
<tr>
<tr>
    <td class="span3 hidden-phone">Gender</td>
    <td>Male</td>                                                       </tr>           
<tr> 
    <td class="span3 hidden-phone" > Salutation </td>
    <td class="span5"> Dr</td>                                      
</tr>   
<tr>
    <td class="span3 hidden-phone">State</td>
    <td class="span5"> State </td>                                          
<tr>
    <tr>
    <td class="span3 hidden-phone">City</td>
    <td class="span5"> City </td>                                       
</tr>
<tr> 
    <td class="span3 hidden-phone" > Postel Code </td>
    <td class="span5"> Postel Code </td>                                        
</tr>
<tr> 
    <td class="span3 hidden-phone" > Phone# </td>
    <td class="span5"> 04128741 </td>                                       
</tr>
<tr> 
    <td class="span3 hidden-phone" > Mobile# </td>
    <td class="span5"> 03218741525</td>                                     
</tr>
<tr> 
    <td class="span3 hidden-phone" > Cover Letter </td> 
    <td>Cover letter goes here</td>
</tr>                       
<tr>
    <td> <input type="submit" name="per-det" class="btn btn-success span5" value="Update and Cont."></td>
</tr>                                       

每当我尝试执行此操作时,它都会给我这个错误

未定义的变量名称

Answer 1

使用它更简单,更安全,更快捷:

obj[field] = value;

......而不是eval('obj.' + field + ' = "' + value + '"'),显然有相同的目的.

你看到你现在看到的,因为value应该用引号括起来.例如,如果两个field和value等于'Name'(字符串),因为它现在站在evalled表达的样子......

obj.Name = Name

...显然导致'Undefined variable Name'错误.

这里有两个旁注.首先,trs在这一行中再次包装jQuery对象是没有意义的......

$(trs).each(function(index, row)

...因为它已经是一个jQuery对象(结果table.find('tr')).如果遵循一个简单的约定,则更容易看到这种冗余:用于存储jQuery对象的所有变量的前面名称$:

var $table = $("#eidtPersonalInfoTbl");
var $trs   = $table.find('tr');
// ...

// $($trs) - obviously redundant

其次,在这些行中两次浏览DOM有点浪费:

 var field = $(row).find('td').eq(0).html();
 var value = $(row).find('td').eq(1).html();

我宁愿把它改写成这样:

 var $tds = $(row).find('td');
 var field = $tds.eq(0).html(); // or just $tds[0].innerHTML;
 var value = $tds.eq(1).html(); // or just $tds[1].innerHTML;