$(function () {
console.log('jQuery is working.')
updateBillingSection();
$('select').on('change', function() {
updateBillingSection();
});
function updateBillingSection() {
// (1) If the plan is not of type=BUSINESS or INDIVIDUAL, hide the billing section
var INDIVIDUAL_PLAN = "INDIVIDUAL";
var BUSINESS_PLAN = 'BUSINESS'
var paidPlans = [INDIVIDUAL_PLAN, BUSINESS_PLAN]
var planType = $('select[name="plan_type"] option:selected').val();
console.log('Plan type: ' + planType);
if(paidPlans.includes(planType)) {
$('.paid-plan-details').show("slow");
} else {
$('.paid-plan-details').hide("slow");
}
}
}
的输出是什么?
答案 0 :(得分:0)
INNER JOIN
将与指示的列的值匹配,并且仅返回找到匹配项的行。
在上面的示例中,结果表将包含值1,1,1,2,2,3-这是因为INNER JOIN
已找到表A的每个值的匹配项。 / p>