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[英]How to open a specific web page once submit button is clicked in the form?
[英]Open a form action url page in a popup box when only a specific submit button is clicked
我有一個帶有兩個提交按鈕的表單(發送消息和更新狀態)。 單擊send_message按鈕時,我想打開一個彈出窗口,並顯示send_message表單。 請建議如何進行
<form name="send_messages" action="/sendMessage/" method="post" accept-charset="UTF-8">
<div class="user-add">
<span>
<select name="user-status" id="user-status" onchange="showDiv(this)">
<option value="">Select Status</option>
<option value="0">Enable</option>
<option value="1">Disable</option>
</select></span>
<button id="hidden_div" style="display:none;" name="button-status" value="user-status-send"
onchange="this.form.submit()"><i class="fa fa-pencil statusup"></i>Update Status
</button>
<span><button id="merge_button" name="button-status" value="message-send" >
<i class="fa fa-envelope"> </i>Send Message</button>
</span>
</div>
您可以在模式<iframe id='submitFormFrame' />
創建框架,然后將target屬性設置為指向該框架<form target='#submitFormFrame'>
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