当用$.ajax()向后台提交参数时,如果参数中数组的话一般在后台会用List<T>接收;但老是不成功如下面代码
var arr1=[{ "aa": "1", "bb": "2" }, { "aa": "3", "bb": "4"}];var arr2=[{ "aa": "1", "bb": "2" }, { "aa": "3", "bb": "4"}];function addUser(){ $.ajax({ url:'UserAdd', data:{list1:arr1,list2:arr2}, type:'post', success:function(msg){ if(msg=='1'){ console.log('添加成功'); }else{ console.log('添加失败') } } }); }
用Fiddler 监测之后发觉数据变成啦
list1[0][aa]=1&list1[0][bb]=2&list1[1][aa]=3&list1[1][bb]=4&list2[0][aa]=1&list2[0][bb]=2&list2[1][aa]=3&list2[1][bb]=4
C#中能识别的数组应该是这样的格式
list1[0].aa=1&list1[0].bb=2&list1[1].aa=3&list1[1].bb=4&list2[0].aa=1&list2[0].bb=2&list2[1].aa=3&list2[1].bb=4
在网上查找资料之后了解到ajax post之前会用因为jQuery需要调用jQuery.param序列化参数,我们来看下jquery源码
//在ajax()方法中,对json类型的数据进行了$.param()处理if ( s.data && s.PRocessData && typeof s.data !== "string" ) { s.data = jQuery.param( s.data, s.traditional );}//param方法中if ( jQuery.isArray( a ) || ( a.jquery && !jQuery.isPlainObject( a ) ) ) { // Serialize the form elements jQuery.each( a, function() { add( this.name, this.value ); }); } else { // If traditional, encode the "old" way (the way 1.3.2 or older // did it), otherwise encode params recursively. for ( prefix in a ) { buildParams( prefix, a[ prefix ], traditional, add ); } }
找到原因之后就好办啦
首先,traditional为false,我们可以通过设置traditional 为true阻止深度序列化
先写一个数组转为对象的方法:
Array.prototype.serializeObject = function (lName) { var o = {}; $t = this; for (var i = 0; i < $t.length; i++) { for (var item in $t[i]) { o[lName+'[' + i + '].' + item.toString()] = $t[i][item].toString(); } } return o; };
var arr1=[{ "aa": "1", "bb": "2" }, { "aa": "3", "bb": "4"}];var arr2=[{ "aa": "1", "bb": "2" }, { "aa": "3", "bb": "4"}];function addUser(){ $.ajax({ url:'UserAdd', data:$.param(arr1.serializeObject("list1"))+"&"+$.param(arr2.serializeObject("list2"), //手动把数据转换拼接 type:'post', traditional:true, //这里必须设置 success:function(msg){ if(msg=='1'){ console.log('添加成功'); }else{ console.log('添加失败') } } }); }
C#后台接收代码
public class Test { public int aa{ get; set; } public int bb{ get; set; } } public ActionResult UserAdd( List<Test> list1, List<Test> list2) { return Json(amm); }
这样一来问题就解决啦!
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