当我在函数中放入$ .ajax并且它显示未定义时，我无法从回调函数获取数据

Question

function testGetData(action = '') {
    $.ajax({
        url: options.urlTo,
        type: 'POST',
        success: function() {
            if (action === 5) {
                return 'result data here';
            }
        }
    });
}

function getDataPagi(action, calBack) {
    if (typeof calBack === 'function') {
        return calBack(action);
    }
}
var testPageResult = getDataPagi(5, testGetData);
console.log(testPageResult);