我正在使用一種表單,我無法更改標記&無法使用jQuery。 當前表單將結果發佈到新窗口。是否有可能將其更改爲ajax表單,以便在提交時顯示結果而不改變任何標記? 將結果(標記)從結果頁面拉回到表單頁面。簡單的Ajax形式使用JavaScript沒有jQuery
以下是窗體的標記。
<form class="form-poll" id="poll-1225962377536" action="/cs/Satellite" target="_blank">
<div class="form-item">
<fieldset class="form-radio-group">
<legend><span class="legend-text">What mobile phone is the best?</span></legend>
<div class="form-radio-item">
<input type="radio" class="radio" value="1225962377541" name="option" id="form-item-1225962377541">
<label class="radio" for="form-item-1225962377541">
<span class="label-text">iPhone</span>
</label>
</div><!-- // .form-radio-item -->
<div class="form-radio-item">
<input type="radio" class="radio" value="1225962377542" name="option" id="form-item-1225962377542">
<label class="radio" for="form-item-1225962377542">
<span class="label-text">Android</span>
</label>
</div><!-- // .form-radio-item -->
<div class="form-radio-item">
<input type="radio" class="radio" value="1225962377543" name="option" id="form-item-1225962377543">
<label class="radio" for="form-item-1225962377543">
<span class="label-text">Symbian</span>
</label>
</div><!-- // .form-radio-item -->
<div class="form-radio-item">
<input type="radio" class="radio" value="1225962377544" name="option" id="form-item-1225962377544">
<label class="radio" for="form-item-1225962377544">
<span class="label-text">Other</span>
</label>
</div><!-- // .form-radio-item -->
</fieldset>
</div><!-- // .form-item -->
<div class="form-item form-item-submit">
<button class="button-submit" type="submit"><span>Vote now</span></button>
</div><!-- // .form-item -->
<input type="hidden" name="c" value="News_Poll">
<input type="hidden" class="pollId" name="cid" value="1225962377536">
<input type="hidden" name="pagename" value="Foundation/News_Poll/saveResult">
<input type="hidden" name="site" value="themouth">
任何提示/教程十分讚賞。 :)
'.elements'爲'只返回輸入節點FORM'? –
是的。表單對象的.elements只返回輸入節點 –
該解決方案將發送未經檢查的單選按鈕和複選框。 –