Question

如果格式sAddToBasket中的广播被检查，如何提交表单radioform？

<form name="radioform" method="post" action="something1.php">
    <input type="radio" onchange="this.form.submit();" class="option--input" id="radio1" name="radio" value="">
    <input type="radio" onchange="this.form.submit();" class="option--input" id="radio2" name="radio" value="">
</form>
<form name="sAddToBasket" method="post" action="something2.php">
    <input type="hidden" name="option1" value="value1"/>
    <input type="hidden" name="option2" value="value2"/>
</form>

Answer 1

我会摆脱DOM中的JS（你的onchange=...）并做这样的事情：

$(document).ready( function() {
  $("form[name=radioform] input").on("change", function() {
    $("form[name=sAddtoBasket").submit();
  }
});

Answer 2

你可以试试这样的事情

<form name="radioform" method="post" action="something1.php"> <input type="radio" onchange="document.getElementById('sAddToBasket').submit();" class="option--input" id="radio1" name="radio" value=""> <input type="radio" onchange="document.getElementById('sAddToBasket').submit();" class="option--input" id="radio2" name="radio" value=""> </form>

 <form id='sAddToBasket' name="sAddToBasket" method="post" action="something2.php"> <input type="hidden" name="option1" value="value1"/> <input type="hidden" name="option2" value="value2"/> 
</form>

Answer 3

onchange你可以像下面这样调用js函数。

<input type="radio" onchange="formSubmit(this)" class="option--input" id="radio1" name="radio" value="">

<form id="sAddToBasket">

</form>

<script>
function formSubmit(radioObj){
  if(radioObj.checked){
    document.getElementById("sAddToBasket").submit();

  }
}
</script>

Answer 4

只需在单选按钮中添加一个类。

<form name="radioform" method="post" action="something1.php">
    <input type="radio"  class="option--input subForm" id="radio1" name="radio" value="">
    <input type="radio"  class="option--input subForm" id="radio2" name="radio" value="">
</form>

并在脚本上使用：

 $(function(){
$('.subForm').click(function(){
if($(this).is(":checked")
$('form[name=sAddtoBasket]').submit();

})
})

在单选按钮检查上提交表单

4 个答案: