2017-04-18 70 views
-1
<tr class="form-row form-row-odd form-row-err form-cols-2 form_element_company_id"> 
        <th>Company&nbsp;<span class="required">*</span></th> 
        <td><select name="company_id" class="selectmenu input-inf select2-initialized select2-hidden-accessible" tabindex="-1" aria-hidden="true"><option value="">— None —</option> 
<option value="1" class="sub-option-0 ">Option 1</option> 
<option value="14" class="sub-option-0 ">Option 2</option> 
<option value="45" class="sub-option-1 " data-parent-name="Option 2="14">Option 2-1</option> 
<option value="46" class="sub-option-1 " data-parent-name="Option 2" data-parent-id="14">Option 2-2</option> 
<option value="47" class="sub-option-1 " data-parent-name="Option 2" data-parent-id="14">Option 2-3</option> 
<option value="29" class="sub-option-0 ">Option 3</option> 
<option value="30" class="sub-option-0 ">Option 4</option> 
<option value="31" class="sub-option-0 ">Option 5</option> 
     </tr> 

我想安裝一些測試與硒IDE,但我很難確定如何使用jquery或js選擇一個隨機選項。Select2隨機選擇與jQuery

+3

您是否嘗試過任何東西,如果是這樣,我們可以看到JS/jQuery的你試過嗎? –

+0

你見過'Math.random'嗎? – evolutionxbox

+0

https://jsfiddle.net/dalinhuang/w8ug2af4/1/ –

回答

0
function setUserCompany(){ 
var companyField = $('select[name="company_id"]'); 
if (companyField.length == 0) { 
return; 
} 
var options = companyField.find('option'); 
if (options.length == 0) { 
return; 
} 
var randomIdx = Math.floor((Math.random() * options.length)+1); 
var randomValue = $(options.get(randomIdx)).attr('value'); 
companyField.select2().val(randomValue).trigger('change'); 
} 

我的問題是,它不承認我的$選擇作爲一個jQuery對象