javascript - 无法在给定的文件扩展名中有效地找到文件的文件扩展名

Question

这里我需要在给定的文件名中找到文件扩展名，即我有2个变量，一个是fileData，它包含文件扩展名，filename它包含文件名，所以在这里我们必须检查文件名是否具有该特定扩展名。然后我们必须检索文件扩展名

fileData = ['exe', 'obj', 'file', 'data'];
filename = ['one.exe', 'two.obj', 'three.p', null, undefined];

constructor() {
  var fileSplit;
  for (var i of this.filename) {
    fileSplit = i.substring(i.lastIndexOf('.') + 1, i.length) || i;

    if (this.fileData.includes(fileSplit)) {

      console.log('File name::', i, 'and file extension::', fileSplit);
    } else {
      console.log('File name not there::', i)
    }
  }
}

出现此错误:错误

Error: Cannot read property 'substring' of null

stackblitz 链接 https://stackblitz.com/edit/angular-ott3vh

Answer 1

您在进行拆分时没有检查未定义的情况。这样做:

fileData =['exe','obj','file','data'];
filename =['one.exe','two.obj','three.p',null,undefined];

constructor(){
   var fileSplit = "";
   for(var i of this.filename){
        fileSplit = "";
        if(i){ 
            fileSplit = i.substring(i.lastIndexOf('.')+1, i.length) || i;
            if(this.fileData.includes(fileSplit)){
                console.log('File name::',i,'and file extension::',fileSplit);
            } else{
                 console.log('File name not there::',i);
            }
        } // Do all the above if(i)
   }
}