I want to read only one line from the code so by using node.js I: import fs = require('fs'); then: var output = fs.readFileSync('/direction/projectname/server.js','utf8'); console.log(output);
I output this way whole file from this direction... BUT I WANT TO PRINT ONLY ONE SPECIFIC LINE OF THAT SOURCE CODE, for example 51st line.
Any idea how to do it?
via Zack Zilic
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