<分区>
我编写了这个函数来处理传入的数据并用相应的数据填充表单。我想知道是否有更好的方法来简化这段代码:
//Handling of "regular weeks" (no "blackouts" or empty days)
//Itteration through specific day # to ensure specificity
for (var a = 0; a < data.days.length; a++) {
var default_start_time = data.days[a].default_start_time;
var default_end_time = data.days[a].default_end_time;
var select_start_time = data.days[a].start_time;
var select_end_time = data.days[a].end_time;
var time_chunk = data.days[a].time_chunk_id;
if (data.days[a].day_of_week_Number == 2) {
$("#mondayHead").html("Monday");
$("#mondayDate").html(data.days[a].date);
$("#mondayStart").html(buid(default_start_time, default_end_time, select_start_time));
$("#mondayEndTime").html(buid(default_start_time, default_end_time, select_end_time));
$("#mondayTimeChunkID").val(time_chunk);
if (data.days[a].comment != null) {
var subMonday = data.days[a].comment.substring(0, 10);
//Comment is held not diplayed **to be fixed
$("#mondayCommentLink").html(subMonday + "..." + "<input type='hidden' id='mondayCommentText' name='mondayCommentText' size='10' value=data.days[a].comment />");
}
}
if (data.days[a].day_of_week_Number == 3) {
$("#tuesdayHead").html("Tuesday");
$("#tuesdayDate").html(data.days[a].date);
$("#tuesdayStart").html(buid(default_start_time, default_end_time, select_start_time));
$("#tuesdayEndTime").html(buid(default_start_time, default_end_time, select_end_time));
$("#tuesdayTimeChunkID").val(time_chunk);
if (data.days[a].comment != null) {
var subTuesday = data.days[a].comment.substring(0, 10);
$("#tuesdayCommentLink").html(subTuesday + "...");
}
}
if (data.days[a].day_of_week_Number == 4) {
$("#wednesdayHead").html("Wednesday");
$("#wednesdayDate").html(data.days[a].date);
$("#wednesdayStart").html(buid(default_start_time, default_end_time, select_start_time));
$("#wednesdayEndTime").html(buid(default_start_time, default_end_time, select_end_time));
$("#wednesdayTimeChunkID").val(time_chunk);
if (data.days[a].comment != null) {
var subWednesday = data.days[a].comment.substring(0, 10);
$("#wednesdayCommentLink").html(subWednesday + "...");
}
}
if (data.days[a].day_of_week_Number == 5) {
$("#thursdayHead").html("Thursday");
$("#thursdayDate").html(data.days[a].date);
$("#thursdayStart").html(buid(default_start_time, default_end_time, select_start_time));
$("#thursdayEndTime").html(buid(default_start_time, default_end_time, select_end_time));
$("#thursdayTimeChunkID").val(time_chunk);
if (data.days[a].comment != null) {
var subThursday = data.days[a].comment.substring(0, 10);
$("#thursdayCommentLink").html(subThursday + "...");
}
}
if (data.days[a].day_of_week_Number == 6) {
$("#fridayHead").html("Friday");
$("#fridayDate").html(data.days[a].date);
$("#fridayStart").html(buid(default_start_time, default_end_time, select_start_time));
$("#fridayEndTime").html(buid(default_start_time, default_end_time, select_end_time));
$("#fridayTimeChunkID").val(time_chunk);
if (data.days[a].comment != null) {
var subFriday = data.days[a].comment.substring(0, 10);
$("#fridayCommentLink").html(subFriday + "...");
}
}
if (data.days[a].day_of_week_Number == 7) {
$("#saturdayHead").html("Saturday");
$("#saturdayDate").html(data.days[a].date);
$("#saturdayStart").html(buid(default_start_time, default_end_time, select_start_time));
$("#saturdayEndTime").html(buid(default_start_time, default_end_time, select_end_time));
$("#saturdayTimeChunkID").val(time_chunk);
if (data.days[a].comment != null) {
var subSaturday = data.days[a].comment.substring(0, 10);
$("#saturdayCommentLink").html(subSaturday + "...");
}
}
}
我很好奇是否有一种方法可以使用两个 for 循环并遍历此函数来更新这些特定值。我觉得有一堆 if 语句不是很好的做法，看起来也不优雅。