JQuery得到了形式和形式方法

Question

我喜欢这个

<form id="popisgolubova_form">
    <input name="pregledaj" type="button" formaction="uredigoluba.php" formmethod="post" formtarget="_self" value="pregledaj" class="button" onclick="popisgolubova_radiobutton(this)">
    <input name="rodovnik" type="button" formaction="rodovnik.php" formmethod="post" formtarget="_blank" value="rodovnik" class="button" onclick="popisgolubova_radiobutton()">
    <input name="podaci" type="button" value="poodaci" formaction="podaciogolubu.php" formmethod="post" formtarget="_blank" class="button" onclick="popisgolubova_radiobutton()">
</form>

和javascript

function popisgolubova_radiobutton(element)
{
    alert($(element).find("[formaction]").val());
    var popisgolubova_radiobutton=$("input[name=RadioGroup1]").is(":checked");
    if(popisgolubova_radiobutton==false)
    {
        alert("nop");
    }
    else
    {
        $("form#popisgolubova_form").submit();
    }   
}

首先,我正在检查是否检查了任何复选框,如果是,我可以提交表格.但问题是形成,形式方法和形式目标.如何获取并提交它们

Answer 1

要获取表单的操作或方法属性,您可以尝试以下操作:

$(function() { 
    var action = $("#formid").attr('action'),
        method = $("#formid").attr('method');
});