Using value of array element in javascript function does not work

Question

This batch does not work, although I put in all possible checks to know wthere a certain assignment does not come through:

alert("valorsumo = "+valorsumo);  // okay
for (j = 0; j <= "<?php echo $fVN ?>"; j++) {
  horp = valornbroj[j];
  alert("horp = "+horp);  // okay
  elcentoj[j] = Math.round(horp/valorsumo * 1000) / 10;  // percentage with one decimal
  alert("percentage = "+elcentoj[i]);  //  undefined !

The queer thing is that the same kind of code, met in a separate file produces the desired answer:

function het()  {     
  var nbroj = [],
  sumo = 3894,
  elcentoj = [];
  nbroj[0] = "290046";   
  elcentoj[0] = Math.round(nbroj[0]/sumo * 1000) / 10;
  alert("percentage = "+elcentoj[0]);  // okay
}

Am I blind? Thanks in advance !

Try changing elcentoj[i] to elcentoj[j] :) Semi-blind :)

Hardy
– Hardy

2013-12-28 08:28:22 +00:00
Commented Dec 28, 2013 at 8:28 — Hardy
– Hardy, Commented Dec 28, 2013 at 8:28

Tareq Salah · Accepted Answer · 2013-12-28 08:34:19Z

2

In this line

alert("percentage = "+elcentoj[i]);

I is not defined replace i with j

alert("percentage = "+elcentoj[j]);

So your loop will be like this

for (j = 0; j <= "<?php echo $fVN ?>"; j++) {
  horp = valornbroj[j];
  alert("horp = "+horp);  // okay
  elcentoj[j] = Math.round(horp/valorsumo * 1000) / 10;  // percentage with one decimal
  alert("percentage = "+elcentoj[j]);  
}

edited Dec 28, 2013 at 8:34

answered Dec 28, 2013 at 8:29

Tareq Salah

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1 Comment

user2772060 Over a year ago

Shame on me! The wrong index hid the real mistake, which happily I found already: a variable was defined locally, within function het(), while it needed to be passed to another function. Thank you !

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