jquery - Make an ajax request based on the output from another ajax request -


what doing is, making jquery ajax call whenever user clicks button , if particular output call, 12345, need make ajax call. can making second ajax call inside success callback, believe not nice way things done. tried different methods implement this. thing is, 2 ajax calls doesn't depend on each other. mean is, don't need pass data ajax 1 ajax2. need make second ajax call if receive output, 12345 first call. did is

var first_call= ajax_call_1; ajax_call_1.done(function(data) {     if(data==='1234')     {          // call function perform second ajax call     }     else     {          // nothing     } });

so question is, there other way implement this? mean better way this? new jquery deferred object , promise. going through documentation now. appreciate get. 

    $(document).ready(function(){          $.ajax({             url: '/path/to/file',             type: 'default (other values: post)',             datatype: 'default: intelligent guess (other values: xml, json, script, or html)',             data: {param1: 'value1'},         })         .done(function(data) {             if (data==='1234') {                 nextajax(data);             }         })                function nextajax(datafromfirstajax) {             console.log(datafromfirstajax);             $.ajax({                 url: '/path/to/file',                 type: 'default (other values: post)',                 datatype: 'default: intelligent guess (other values: xml, json, script, or html)',                 data: {param1: 'value1'},             })             .done(function(data) {                 console.log('done!');             })         }      });

try this. when first ajax done function nextajax(datafromfirstajax); next ajax , data


Comments

Post a Comment

Popular posts from this blog

php - failed to open stream: HTTP request failed! HTTP/1.0 400 Bad Request -

this code i'm using find latitude , longitude of place: function coords($address){ echo $url="http://maps.googleapis.com/maps/api/geocode/json?address=".$address; $json = file_get_contents(urlencode($url)); $data = json_decode($json, true); print_r($data['results'][0]['geometry']['location']['lat']); print_r($data['results'][0]['geometry']['location']['lng']); } however returns warning: failed open stream: http request failed! http/1.0 400 bad request if use above code in simple procedure works fine not in function. note: have tried curl_init()... method produced same result? instead of encoding whole url encode address part. following code work fine $add='jamshoro phase 2'; $url="http://maps.googleapis.com/maps/api/geocode/json?address=".urlencode($add); $json = file_get_contents($url); $data = json_decode($json, true); print_r($data['r
Read more

java - How to filter a backspace keyboard input -

i have code gives message when key number pressed: txtvalueofclause.addeventfilter(keyevent.key_typed, new eventhandler<keyevent>() { @override public void handle(keyevent t) { char ar[] = t.getcharacter().tochararray(); char ch = ar[t.getcharacter().tochararray().length - 1]; if (!(ch >= '0' && ch <= '9')) { system.out.println("the char entered not number"); t.consume(); } } }); now if press on wrong button , press backspace remove it, error message. how can add backspace input if statement? i have found answer: txtvalueofclause.addeventfilter(keyevent.key_typed, new eventhandler<keyevent>() { @override public void handle(keyevent t) { char ar[] = t.getcharacter().tochararray(); char ch = ar[t.getcharacter().tochar
Read more

java - Show Soft Keyboard when EditText Appears -

i have code in android app, alert dialog , keyboard, , when dialog shown want keyboard appear. not , not sure why? here code: final alertdialog.builder alert = new alertdialog.builder(this); alert.settitle("title"); alert.setmessage("message"); final edittext input = new edittext(this); input.setinputtype(inputtype.type_class_number); alert.setview(input); input.setonfocuschangelistener(new view.onfocuschangelistener() { @override public void onfocuschange(view v, boolean hasfocus) { if (hasfocus) { log.i(tag, "in focus"); inputmethodmanager inputmethodmanager = (inputmethodmanager) getsystemservice(context.input_method_service); inputmethodmanager.showsoftinput(input, inputmethodmanager.show_implicit); } } }); but keyboard not appear though edit text in focus, , tell appear why, doesn't come up? thanks in advance.
Read more