forms - jQuery check if input empty (0) on Submit - 


<input type="text" readonly="" value="0" class="input-mini1 sum1">  <input type="text" readonly="" value="0" class="input-mini1 sum2"> <input type="text" readonly="" value="0" class="input-mini1 sum3"> <button type="button" id="btn_submit" class="btn">submit</button>

solution tried:

$("#btn_submit").click(function(){     var isformvalid = true;      $(".input-mini1 input").each(function(){       if ($.trim($(this).val()).length == 0){          alertify.error("please check hours");          isformvalid = false;       }else{       alertify.error("valid hours");       }      });       if (!isformvalid) alert("please fill in required fields (indicated *)");     return isformvalid; });

for reason, when submit form empty/zero filled in, not validate. user must not submit form if input value 0 or 0.0 (either case) , should validated.

also when click on submit button, not loop through inputs.

any ideas how make work ?

your selector .input-mini1 input wrong, looking input elements descendants of element class input-mini1.

it should input.input-mini1 since input-mini1 class of input element

so

$("input.input-mini1").each(function () {     if ($.trim($(this).val()).length == 0) {         alertify.error("please check timesheet hours");         isformvalid = false;     } else {         alertify.error("valid timesheet hours");     } });

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