我只想使用jQuery打开弹出窗口三次

Question

我创建了一个弹出窗口。我想打开三遍。并非一直如此

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Answer 1

// make falsy (null, undefined, empty string) = 0
var walk_in = localStorage.getItem('walk_in') || 0; 
walk_in++; // cast to number and increment
if (walk_in<=3) {
  $.getJson( .... 
    if (result.country == 'IN') {
      localStorage.setItem('walk_in', walk_in); 

测试stringVar++：

var counter = ""
console.log(counter,typeof counter)
counter = counter || 0
console.log(counter,typeof counter)

counter = "1"
console.log(counter,typeof counter)
counter = counter || 0
counter++; // casts and adds
console.log(counter,typeof counter)