<html>
<head>
<title>bd</title>
</head>
<form id="ceshi" name="ceshi" action="#" METHOD="POST" onsubmit = "return confirm('ceshi');" >
<input type="button" onclick="this.form.submit()" value="click" />
</form>
</html>
上面的代码,点击click按钮,直接提交表单,并不会触发onsubmit事件
如果是this.form.onsumbit()的话,则只是触发时间,但并不提交表单...
那我现在怎么既触发事件又提交表单呢??
<head>
<title>bd</title>
</head>
<form id="ceshi" name="ceshi" action="#" METHOD="POST" onsubmit = "return confirm('ceshi');" >
<input type="button" onclick="this.form.submit()" value="click" />
</form>
</html>
上面的代码,点击click按钮,直接提交表单,并不会触发onsubmit事件
如果是this.form.onsumbit()的话,则只是触发时间,但并不提交表单...
那我现在怎么既触发事件又提交表单呢??
需要调用 fireEvent方法。尝试一下:this.form.fireEvent('onsubmit');
变成<input type="submit" value="click" />
<html>
<head>
<title>bd</title>
</head>
<form id="ceshi" name="ceshi" action="#" METHOD="POST" >
<input type="button" onclick="if(confirm('ceshi'))this.form.submit()" value="click" />
</form>
</html><html>
<head>
<title>bd</title>
</head>
<form id="ceshi" name="ceshi" action="#" METHOD="POST" onsubmit = "return confirm('ceshi');" >
<input type="submit" value="click" />
</form>
</html>
因为不能使用
<input type="submit" />
所以只能用onsubmit来解决
自己想到一个方法...
<html>
<head>
<title>bd</title>
</head>
<body onload="alert('flaus')">
<form id="ceshi" name="ceshi" action="#" METHOD="POST" onsubmit = "if(confirm('ceshi')) this.submit();" >
<input type="button" onclick="this.form.fireEvent('onsubmit')" value="click" />
</form>
</body>
</html>
测试通过
<html>
<head>
<title>bd</title>
</head>
<body onload="alert('flaus')">
<form id="ceshi" name="ceshi" action="#" METHOD="POST" onsubmit = "if(confirm('ceshi')) this.submit();" >
<input type="button" onclick="this.form.onsubmit()" value="click" />
</form>
</body>
</html>