自己看吧,超级简单,就不用挨个获取表单名称和值对装在Json里往php传了,直接传个form就可以。
【HTML】
<form method="post" id="form1">
Order: <input type="text" name="order" value="">
Name: <input type="text" name="name" value="">
Old: <input type="text" name="old" value="">
Gender: <input type="radio" name="gender" value="female">男
<input type="radio" name="gender" value="male">女
Choose: <select name="choose">
<option value="c1">c1</option>
<option value="c2">c2</option>
<option value="c3">c3</option>
</select>
<!-- 什么复选框,hidden的表单啊,我都没往上写,都是可以的。-->
</form>
<button>触发</button>
result: <p></p>
【jQuery】
$("button").click(function(){
$.ajax({
url:'2.php',
type:'POST',
data:$("#form1").serialize(),
success:function(response){
$("p").text(response);
}
});
});
【PHP】
$order=$_POST['order']; //可以用$_POST获取到input表单,name所对应的值
$name=$_POST['name'];
$old=$_POST['old'];
$gender=$_POST['gender'];
$choose=$_POST['choose']; echo "Order: ".$order." ";
echo "Name: ".$name." ";
echo "Old: ".$old." ";
echo "Gender: ".$gender." ";
echo "Choose: ".$choose." ";
输出结果:Order: 1 Name: 1 Old: 1 Gender: female Choose: c3