我正在创建一个基于下面动态表单值的JSON,当用户提交时,我在#results中显示json feed
生成JSON时是否也可以获取表单中的所有值,我想得到
输入值等
然后按以下顺序创建/显示JSON?
可在此查看工作版本:
http://jsfiddle.net/dev1212/GP2Y6/25/
目前它没有重新调整任何值并得到一些未定义的.. 我尝试过的代码是
<script>
x = function(selector){
var attrs = [];
$(selector + " input").each(function(){
var attrObject = {};
$(this.attributes).each(function(index, attr){
attrObject[attr.name] = attr.value;
attrObject[attr.va] = attr.value;
//console.log(attrObject)
});
attrs.push(attrObject);
attrObject = {};
});
return attrs;
}
$(document).ready(function(){
alert(JSON.stringify(x("#myform")));
});
</script>
答案 0 :(得分:0)
<script type="text/javascript" src="/www/include/js/jquery.min.js"></script>
<script>
x = function(selector){
var attrs = [];
$(selector + " input").each(function(){
var attrObject = {};
$(this.attributes).each(function(index, attr){
attrObject[attr.name] = attr.value;
//console.log(attrObject)
});
attrs.push(attrObject);
attrObject = {};
});
return attrs;
}
$(document).ready(function(){
alert(JSON.stringify(x("#myForm")));
});
</script>
<form id="myform" class="form-wd">
<input class="ui-dform-text" type="text" title="data for network.node1.eth0.ipaddr" name="network.node1.eth0.ipaddr">
</form>