javascript错误，PHP变量在codeigniter中传递

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.

Answer 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);
    }
}

Answer 2

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

最有可能的是，您试图求和字符串而不是整数