ajx使用总结,
1.使用ajx提交参数,如何改变data中的参数名
1).正常使用时如下:
$.ajax({
url: store.get("path")+"/offdoc/phone_"+s,
type:"POST",
dataType: "json",
data:{
loginName : loginName,
key : key,
pid : pid,
penetrate : 'cas',
devicetype : 'phone',
processInsId : $("#processInsId").val(),
workItemId : $("#workItemId").val(),
approveNotion : $("#spyj").val().replace(/\n/g,"<br />"),
'BPMPara:participantNames' : myTag.getPeopleListName()
},
success:function(data){
if(data.success){
alert("提交成功");
}
}
});
2).如下方式是可以修改data参数名的
var subJson = '"userId":"'+user.id
+'","processInsId":'+processInsId
+',"workItemId":'+workItemId
+',"opinion":"'+newopinion
+'","approvename":"'+user.name
+'","loginName":"'+loginName
+'","key":"'+key
+'","pid":"'+pid
+'","formurl":"'+formurl
+'","penetrate":"cas","devicetype":"phone"';
$.ajax({
url: store.get("path")+"/review/processreview.do",
type:"POST",
dataType: "json",
data:eval("({"+subJson+"})"),
success:function(data){
if(data.success){
console.log("成功");
}
},
error:function(data) {
mui.toast("提交审批失败");
}
});
这种方式中data的参数名是不能动态变化获取的,