刪除查詢參數使用node.js不工作

Question

我有這段代碼從url中刪除查詢參數，但它不工作。你能看看嗎？

const url = require('url') 
 

 
const obj = url.parse('http://www.example.com/path?query1=val1&query2=val2', true) 
 
delete obj.query.query2 
 
const link = url.format(obj) 
 
console.log(link) // I was expecting the removal of query2 but it didn't happen

它登錄的網址與上面過去了，爲什麼query2不會被刪除？由於

Answer 1

你需要從對象

const obj = url.parse('http://www.example.com/path?query1=val1&query2=val2', true) 
delete obj.query.query2 
delete obj.search 

const link = url.format(obj) 
console.log(link) 

刪除搜索節點這將返回URL http://www.example.com/path?query1=val1

Answer 2

即使你從查詢對象中刪除query2，query2仍然存在於搜索領域。

const url = require('url'); 

const obj = url.parse('http://www.example.com/path?query1=val1&query2=val2', true) 
console.log(obj); 
delete obj.query.query2 
delete obj.search 
console.log(obj); 
const link = url.format(obj) 
console.log(link) 

Answer 3

如果通過源URL模塊（https://github.com/defunctzombie/node-url/blob/master/url.js）看看。你可以看到 它會首先查找搜索節點（第413行）。除去這個，以便查詢對象被評估。

delete obj.search;

Answer 4

const url = require("url") 
 

 
const urlObj = url.parse('http://www.example.com/path?query1=val1&query2=val2', true) 
 

 
delete urlObj.query.query2 
 
delete urlObj.search 
 

 
const newUrl = url.format(urlObj) 
 

 
console.log(newUrl) // print >> http://www.example.com/path?query1=val1