jquery - 如何在 jQuery 每个函数中获取这些数据?帮我解决这个问题
问题描述
function check_businessHours() {
for (var i = 1; i <= 7; i++) {
var weekday = document.getElementById('weekday'+i);
var opening_time = document.getElementById('opening_time'+i);
var closing_time = document.getElementById('closing_time'+i);
var close = document.getElementById('close'+i);
var always = document.getElementById('always'+i);
if (opening_time.value == "" && closing_time.value == "") {
if (close.checked == true || always.checked == true) {
$('#check_business').html('');
$('#check_business').removeClass('bg-danger');
} else {
var value = weekday.value+' business hours required';
console.log(value)
}
} else {
$('#check_business').html('');
$('#check_business').removeClass('bg-danger');
}
}
}
它像这样在控制台日志中返回
Saturday business hours required
Sunday business hours required
Monday business hours required
Wednesday business hours required
Thursday business hours required
我想返回下面给出的 div,但我不能这样做。我无法解决这个问题
<div id="check_business" class="check-input"></div>
谁能帮我解决这个问题?
解决方案
DIV 清除需要在循环开始之前进行,否则它将只包含最后一个元素。
function check_businessHours()
{
$("check_business").empty().removeClass("bg-danger");
for (var i = 1; i <= 7; i++)
{
var weekday = document.getElementById('weekday'+i);
var opening_time = document.getElementById('opening_time'+i);
var closing_time = document.getElementById('closing_time'+i);
var close = document.getElementById('close'+i);
var always = document.getElementById('always'+i);
if (opening_time.value == "" && closing_time.value == "")
{
if (!(close.checked || always.checked))
{
$("#check_business").append(`<div>${weekday.value} business hours required</div>`).addClass("bg-danger");
console.log(value)
}
}
}
}
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