无效的情况：

  function ManHourCheck() {var StartDate = $("#StartDate").val();//日报日期var EndDate = $("#EndDate").val();//完成日期var UserID = $("#UserID").val();//员工ID$.ajax({async: false,//加了这一行也没用!!!!!!!!!!!!!!!type: "POST",url: "/Daily/ManHourCheck",data: {"StartDate": StartDate,"EndDate": EndDate,"UserID": UserID,},success: function (result) {if (result.success) {return true;//取不到！！！！！！！！}else {layer.alert(result.message, { area: ['500px', ''] });return false;//也取不到！！！！！！！！}}});}

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修改为：

  function ManHourCheck() {var StartDate = $("#StartDate").val();//日报日期var EndDate = $("#EndDate").val();//完成日期var UserID = $("#UserID").val();//员工IDvar check = false;//看这里！！！！！！！！！！！！！$.ajax({async: false,type: "POST",url: "/Daily/ManHourCheck",data: {"StartDate": StartDate,"EndDate": EndDate,"UserID": UserID,},success: function (result) {if (result.success) {          check = true;//看这里！！！！！！！！！！！！！}else {  layer.alert(result.message, { area: ['500px', ''] });     check = false;//看这里！！！！！！！！！！！！！}}});//看这里！！！！！！！！！！！！！if (check == true) {return true;}else {return false;}}

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说明：

①不要在ajax的success里面做 return 值 ！！！取不到的！！！

②用个变量去接 ajax的return 值, 然后 return 这个变量！！！就行了

③再不行的话，把ajax的return之后的操作的function写在success里.... 不推荐

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其他参考一下这个： https://blog.csdn.net/qq_28938475/article/details/82800656