javascript error with php variable passing in codeigniter

Question

function something(frm,i){
//var ch=frm.outof1.value;
for(var j=1;j<=i;j++)
{
    //var b="outof" + j;
    alert(frm.outof+j.value);

}
//alert("outof" + i);
return false;
}

 $js='onClick="something(this.form,\''. $ii .'\')"';
  echo form_button('mybutton', 'Click Me', $js);

and getting output NAN where in html this is // echo form_input('outof'.$i,''); // the form input.

i need value of form input named outof1,outof2...etc...but i am getting output NaN — user1366645
– user1366645, Commented Apr 30, 2012 at 20:43
Can you post your rendered HTML here that browser get by this echo form_button('mybutton', 'Click Me', $js); ? — The Alpha
– The Alpha, Commented Apr 30, 2012 at 20:49

Ja͢ck · Accepted Answer · 2012-04-30 20:54:10Z

1

First, you will want to make sure the passed $ii is made into the correct type (not a string) by using parseInt. Then you construct the form input name by concatenating 'outof' and the number before evaluating .value.

function something(frm, i)
{
    for(var j = 1; j <= parseInt(i); ++j) {
        alert(frm['outof' + j].value);
    }
}

answered Apr 30, 2012 at 20:54

Ja͢ck

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2 Comments

Matt Lo Over a year ago

+1, however i should expect an integer versus having to convert it within the method. Probably unquote that second arg. Agree?

Ja͢ck Over a year ago

@MattLo I normally pass verbatim variables to JavaScript using json_encode to make sure the intended type is passed without causing a parse error if the type was anything else :)

iBiryukov · Accepted Answer · 2012-04-30 20:42:19Z

0

alert(parseInt(frm.outof) + parseInt(j.value))

Most likely, you're trying to sum up strings, and not integers

answered Apr 30, 2012 at 20:42

iBiryukov

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