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What i'm doing worng in this conditional?


By : Asaru
Date : October 15 2020, 08:10 AM
it should still fix some issue You are using 'or' || instead of 'and' &&. Col is greater than or equal to left OR less than or equal to right. Col will always be greater than left in the first range. This means the first if statement is always true (if you're editing in either of the ranges).
code :
if(colIndex >= watchRange1.left || //should be &&

colIndex <= watchRange1.rigth && //right is misspelled
rowIndex >= watchRange1.top || //should be &&
rowIndex <= watchRange1.bottom && 
e.value != 0) //access key value instead of call function


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Codility worng answer on TapeEquilibrium for php


By : R.Mohan
Date : March 29 2020, 07:55 AM
wish of those help Two elements test case is probably {-1000, 1000}. This array can be splited only in one place and the distance is |-1000-1000| = 2000. Your program for this input returns 0, because in the second iteration of the for loop $arr_h is 0 and $sum_total is also 0.
Simply change condition in loop for ($i = $count_-1; $i > 0 ; $i--) to get 100%.

java easy ifs worng output


By : Dynamic Vietnam
Date : March 29 2020, 07:55 AM
will help you I am new to java, and currently Im playing around with simple if statments. I have two questions. x First one: if answer yes, the last statment "You need to answer yes or no" is printed anyway. I only want it to print if the answer is something other than yes or no. , The answer to both the questions ..
code :
public static void main(String[] args) {
    Scanner keyboard = new Scanner(System.in);
    while(true) {
        System.out.printf("Are you doing good?");
        String ord = keyboard.nextLine();

        if (ord.equalsIgnoreCase("yes")) {
           System.out.println ("nice to hear!");
           break;
        }

        else if (ord.equalsIgnoreCase("no")) {
            System.out.println ("that makes me sad!");
            break;
        }

        else {
            System.out.println("you need to  answer yes or no");
        }
    }
}

what is worng in ajaxSubmit calling?


By : harv
Date : March 29 2020, 07:55 AM
I wish this help you , You forgot id selector try this:-
code :
$scope.submitTheForm = function(htmlcode){  
    var idOfForm = "formOfCalCPDF";
    var dataPassed = $.param({'htmlcode':htmlcode,'mode':'getpdf'});
    alert("coming here");
    $('#'+idOfForm).ajaxSubmit({
        url: 'ggs.erm.payrollJava.Taxsummary',
        type: "POST",
         data: dataPassed,
        error:function (){},
        success: function (data){}
    });
};

PHP POST method adding the worng value


By : Hafiz Jahid Ahmed
Date : March 29 2020, 07:55 AM
I wish this helpful for you The way you are printing this is actually repeating the same hidden value called apprentice_name. The they are going to be see by the form as the last value printer.
The closest thing that would work without modifying your code too much is to print different forms for each value.
code :
while($row = mysqli_fetch_array($result)){ ?>
<form action="some-script.php">
  <input type="hidden" value="<?php echo $row['apprentice_name'] ?>" name="apprentice_name"> 
  <input type="submit" value="Login" name=<?php $row['apprentice_name'] ?>>
</form>
}

'While' loop count is worng?


By : SunnyD
Date : March 29 2020, 07:55 AM
To fix this issue with this piece of code 0.95 ** rush_height_counter * rush_height_gain do you mean
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