未捕获的TypeError：onmouseover时无法读取null错误的属性“1”

Question

DIV就是那样

<div onmouseover="showTrackInfo(<?php echo $info['ID']; ?>);" id="pl_table_div_<?php echo $info['ID']; ?>">

JS是

function showTrackInfo(a) {
Tipped.create("#pl_table_div_"+a, 'test', {
      skin: "light",
      maxWidth: 400,
      fixed: true,
      hook: "middleleft"
   });
}

我哪里错了？