Javascript Jquery映射方法
以下是Jquery映射方法:Javascript Jquery映射方法,javascript,jquery,map,Javascript,Jquery,Map,以下是Jquery映射方法: success: function (data) { var p = data.d var person = JSON.parse(p).Person[0] var company = JSON.parse(p).Company[0] alert(p) response
success: function (data) {
var p = data.d
var person = JSON.parse(p).Person[0]
var company = JSON.parse(p).Company[0]
alert(p)
response($.map(person , function (item) {
var companyNames = item.company.join(", ");
var val = "<div class='searchedPersonItem'>" + item.pName + "<br><span class='darkestGrey'>" + companyNames + "</span></div>";
// var val = "<table><tr><td><a><div class='searchedPersonItem'>" + item.pName + "<br><span class='darkestGrey'>" + companyNames + "</span></div></td><td><div class='searchedPersonItem'>" + item.pName + "<br><span class='darkestGrey'>" + companyNames + "</span></div></a></td></tr></table>";
//var val = "<table><tr><td><div class='searchedPersonItem' id='" + item.id + "'>" + item.pName + "<br><span class='darkestGrey'>" + companyNames + "</span></div></td><td><div class='searchedPersonItem1'>" + item.pName + "<br><span class='darkestGrey'>" + companyNames + "</span></div></td></tr></table>";
return {
label: val,
value: item.pName,
hidden: JSON.stringify(item)
};
}));`
成功:函数(数据){
var p=data.d
var person=JSON.parse(p).person[0]
var company=JSON.parse(p.company)[0]
警报(p)
响应($.map)(人员、职能(项目){
var companyNames=item.company.join(“,”);
var val=”“+item.pName+”
“+companyNames+”;
//var val=“+item.pName+”
“+companyNames+”+item.pName+”
“+companyNames+”;
//var val=“+item.pName+”
“+companyNames+”+item.pName+”
“+companyNames+”;
返回{
标签:val,
值:item.pName,
隐藏:JSON.stringify(项)
};
}));`
我能够成功显示人员详细信息。但我的要求是连接/合并人员和公司详细信息。请帮助我实现这一点。只需使用jQuery将这些对象合并为一个:
var merged_data = $.extend({}, person, company);
或使用此功能:
/**
* Overwrites obj1's values with obj2's and adds obj2's if non existent in obj1
* @param obj1
* @param obj2
* @returns obj3 a new object based on obj1 and obj2
*/
function merge_options(obj1,obj2){
var obj3 = {};
for (var attrname in obj1) { obj3[attrname] = obj1[attrname]; }
for (var attrname in obj2) { obj3[attrname] = obj2[attrname]; }
return obj3;
}
(摘自此处:)“帮助您实现这一目标”?您的问题到底是什么?您被困在哪里了?@user3074683您应该有类似的运行示例