我想将变量的值传递给ajaxForm数据。将值传递给ajaxform数据
value1 = "dynamic_value1";
value2 = "dynamic_value2";
$('form').ajaxForm({
data: {
key1: value1,
key2: value2
}
});
期待这样的:
date:{Key1:"dynamic_value1", Key2:"dynamic_value2"}
所以在PHP我可以访问像
echo $_POST['key1'];
================== ==== COMPLETE脚本
<script src="../../bin/addons/jquery-1.7.2.js"></script>
<script src="../../bin/addons/jquery.form.js"></script>
<script>
// jQuery Form-plugin
(function() {
var value1 = "dynamic_value1";
var value2 = "dynamic_value2";
$('.dummyForm1').ajaxForm({
data:{
key1: value1,
key2: value2
}
complete: function(xhr) {
txt = xhr.responseText;
alert(txt);
}
});
})();
</script>
<form class="dummyForm1" action="form-php.php" method="post">
<input type="submit" value="Hit!" />
</form>
form-php.php
<?
echo "Key1 value:". $_POST['key1'];
?>
所以,你只是想InitCase键值名的当前状态? – Chandu 2012-07-10 17:10:12
我需要在AjaxForm调用期间在JSON中传递的变量值? – Shiv 2012-07-10 17:52:26
你现在的代码应该可以正常工作。你面临的问题是什么? – Chandu 2012-07-10 17:54:13