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0

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Ajax response looks undefined

I'm trying to get the ajax response and attribute it to a variable (let's say _has_weekend_hollidays) as follows:


JS Call:

_has_weekend_hollidays = checkWeekendHollidays( _i );
console.log ( checkWeekendHollidays( _i ) );

AJAX Call

$.ajax({
    url: "./directory/ajax_check_weekend_hollidays.php",
    type: "POST",
    data: { 
            start_date: _date1,
            final_date: _date2 
          }
}).done(function (_result) {
    return ( _first_weekday == 0 || _last_weekday_id >= 6 || _result!="0" );
}).fail(function (_result) {
    console.log("ERROR:" + _resultado);
});

The AJAX _result is OK but the return is not working.

So console.log shows undefined.

Any ideas?

Thanks in advance!

about 4 years ago · Juan Pablo Isaza
1 answers
Answer question

0

If you need the result to continue, do your work after the Ajax callback.

It's undefined because there are no variable return by Ajax:

{
    url: "./directory/ajax_check_weekend_hollidays.php",
    type: "POST",
    data: { 
            start_date: _date1,
            final_date: _date2 
          }
}

This line:

return ( _first_weekday == 0 || _last_weekday_id >= 6 || _result!="0" );

is not executed until Ajax is "Done".

It you wish to log the return data, use console.log in the return function.

Example:

$.ajax({
    url: "./directory/ajax_check_weekend_hollidays.php",
    type: "POST",
    data: { 
            start_date: _date1,
            final_date: _date2 
          }
}).done(function (_result) {
    console.log(_first_weekday == 0 || _last_weekday_id >= 6 || _result!="0");
    //do something else here too
}).fail(function (_result) {
    console.log("ERROR:" + _resultado);
});
about 4 years ago · Juan Pablo Isaza Report
Answer question
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