通外范围的逻辑的回调函数的JavaScript的NodeJS

Question

/file_iterator.js

var asyncFileObjectReader = require('../core/async_forEach_Array.js') 
var sampleFunction; 

function allAcitemFileIterator(filesObject, mapContainer, root_directory) {  

    asyncFileObjectReader.forEachFunction(filesObject, sampleFunction); 

    sampleFunction = function (fileName) { 
     mapContainer.set(basePathAlterColons.removeBasePathAlterColons(fileName, root_directory), 
      extractUriFromList.extractUriFromList(getReferences.getReferences(fileName))); 
    } 
} 

/async_foreach_Array.js

function forEachFunction(filesObject, filesObjectFunction) { 
    filesObject.forEach(function (fileName) { 
     if (path.extname(fileName) === acitem_file_extension) { 
      filesObjectFunction(fileName); 
     } 
    }); 
} 

当试图通过sampleFunction作为回调函数到forEachFunction，但仍然保持外部函数的状态，我得到一个错误，因为“未定义为不是函数”请帮助？

Answer 1

在定义它之前，您正在使用该功能！互换这两个语句：

sampleFunction = function (fileName) { 
    mapContainer.set(basePathAlterColons.removeBasePathAlterColons(fileName, root_directory), 
     extractUriFromList.extractUriFromList(getReferences.getReferences(fileName))); 
}; 

asyncFileObjectReader.forEachFunction(filesObject, sampleFunction); 

哦，真的没有理由让sampleFunction全球你的模块。特别是当你使用mapContainer和root_directory作为闭包变量时，在异步函数中。如果allAcitemFileIterator被称为太快，这将导致可怕的竞争条件。所以请使用var sampleFunction = …，或者只是一个函数声明。