parsingurlterminalhtml-escape-charactersuri-scheme

How can I add a link as a parameter inside a link so I can start from terminal

I need to start the following link from the terminal

vidyo://login/?username=x&password=y&portal=https://nexi.alpha.vidyo.com&minimize=true

I get this error:

zsh: parse error near `&'

Now I'm a android programmer, so I did in a android app a console log on Uri.parse(link), but I get the same link, as "parsed" But that doesn't feel right. What do I need to do to my portal in order for the link to be parsed correctly

PS: I also tried this:

vidyo://login/?username=x&password=y&portal=https%3A%2F%2Fnexi.alpha.vidyo.com&minimize=true

But no luck PS: I figured out that if I try 

vidyo://login/?portal=https%3A%2F%2Fnexi.alpha.vidyo.com&minimize=true

It works, and the same for 

vidyo://login/?username=x&password=y

Apparently adding the 3rd parameter gives me an parse error in terminal, why?

Solution

  • open "vidyo://login/?username=test2&password=test2&6portal=https://nexi.alpha.vidyo.com&minimize=false" works, or open vidyo://login/?username=test2%26password=test2%26portal=https://nexi.alpha.vidyo.com%26minimize=false also works. The issue was the second "&"