由于正在执行的脚本大概知道"了它的名称，因此加载以下脚本应该起作用:

Since the executing script presumably "knows" the name it was loaded with the following should work:

for (var s, scripts = document.getElementsbyTagName ('script'), i = scripts.length; i--;)
  if ((s = scripts[i]).src.indexOf ('scriptname.js') < 0 && s.id && !test (s.id)) {
     alert ("Script " + s.src + " id " + s.id + " not yet processed");
  }