Where is mistake?

Question about the task

Request parser

Java Core, Level 5, Lesson 12

Under discussion

In what is problem?

Read a URL from the console.
Display a space-delimited list of all the parameters (The parameters follow the "?" and are separated by "&", e.g. "lvl=15").
The URL contains at least 1 parameter.
The parameters must be displayed in the same order in which they are present in the URL.
If the obj parameter is present, pass its value to the relevant alert method.
alert(double value) - for numbers (fractional numbers have a decimal point)
alert(String value) - for strings
Note that the alert method must be called AFTER the list of all parameters is displayed.

Example 1

Input:
http://codegym.cc/alpha/index.html?lvl=15&view&name=Amigo

Output:
lvl view name

Example 2

Input:
http://codegym.cc/alpha/index.html?obj=3.14&name=Amigo

Output:
obj name double: 3.14

Requirements:

The program should read only one line from the keyboard.
The Solution class must not have static fields.
The program must display data on the screen in accordance with the task conditions.
The program should call the alert method with the double parameter if the obj parameter can be correctly converted to a double.
The program should call the alert method with the String parameter if the obj parameter CANNOT be correctly converted to a double.

package com.codegym.task.task15.task1527; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.ArrayList; import java.util.List; /* Request parser */ public class Solution { public static void main(String[] args) { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); List<String> list; String x = "", y = ""; try { String s = br.readLine(); list = findString(s); for (String l : list) { if(!(l.charAt(0) > 47 && l.charAt(0) < 58)) x += l + " "; } alert(x); for (String l : list) { if(l.contains(".") && doubl_e(l)) { double d = Double.parseDouble(l); alert(d); } } } catch (Exception e) { } } public static void alert(double value) { System.out.println("double: " + value); } public static void alert(String value) { System.out.println("String: " + value); } public static boolean doubl_e(String s) { try { double d = Double.parseDouble(s); return true; } catch (Exception e) { return false; } } public static List<String> findString(String s) { List<String> list = new ArrayList<>(); String a = ""; for(int i = 0; i < s.length(); i++) { if(s.charAt(i) == '&' || s.charAt(i) == '?' || s.charAt(i) == '=') { for(int j = i+1; j < s.length(); j++) { if(s.charAt(j) > 96 && s.charAt(j) < 123) a += s.charAt(j); else if(s.charAt(j) > 47 && s.charAt(j) < 58 || s.charAt(j) == 46) { a += s.charAt(j); } else { if(!a.isEmpty()) list.add(a); a = ""; break; } } } } return list; } }

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Jon Hjerting

Level 20 , Oslo, Norway

10 September 2019, 17:04

You make up a list of both parameter names and values in your findString function. You are only meant to print a list of the paramter names. After printing out the parameter names, you only need to send the value of the obj parameter to the alert function. (And if it is a number, convert it to a double first.)

Jon Hjerting

Level 20 , Oslo, Norway

10 September 2019, 16:59

I would recommend you drop the use of 'magic numbers' in your code as it makes it very hard to read. If you change the code from if(s.charAt(j) > 96 && s.charAt(j) < 123) to if (s.charAt(j) >= 'a' && s.charAt(j) <= 'z') This would save both the writer and reader of the code to look up ascii values.

Jon Hjerting

Level 20 , Oslo, Norway

6 September 2019, 18:00

I also found the specs confusing on this one. I wasn't aware of the URL class, so I used the String split method to split the url first at '?', then at '&' and then at '='. However you do it you need to make a list of the parameter names and print out those with a space between them. (Check out the array join method.) Then you have to see if the parameter 'obj' is present. If it is, try to convert its value to a number. If this works send the number to alert. Otherwise send the string value to alert.

Xubres

Level 16 , Uranus, Space

28 December 2018, 13:12

I will try to explain my solution. I don't know if it's the best but it works. I used URL class, so I can get parameters into an array without much trouble. Then check if parameter contains an =, and I print only the part before the =; Then iterate through parameters again and check if they contain "obj". If they do, try parsing to double. If it fails keep as String. (this time you only print the part after the '='). The methods I used are string methods, string contains and substring. Also a lot of try catch. Hope this helps anyone who is stuck and is not too much information to be taken down :)

Michael Martin

Level 19 , Arlington, United States

26 November 2018, 13:28

I never could get this to work. The requirements don't make sense. I recommend skipping it :)

Mack

Level 20 , Osiedle-Nowiny, Poland

26 November 2018, 15:58

so I will ;)

Robert P

Level 20 , Nashville, United States

27 November 2018, 05:30

The wording is weird in the conditions page, and the examples took a bit of staring at to figure out what they wanted exactly. But solution is not terribly difficult, just time consuming, and complicated. I have a working solution to this problem, but it's kinda complicated, and I don't want to paste it, to have it taken down by the admins. Please message me, if you're interested in me explaining it. But you are very close, Mack. when I pass: http://codegym.cc/alpha/index.html?obj=3.14&name=Amigo I get: String: obj name ^^^^^^ ------------------ this and only this needs to go. double: 3.14 so that it returns: obj name double: 3.14 Similarly http://codegym.cc/alpha/index.html?lvl=15&view&name=Amigo String: lvl view name ^^^^^^ --------------------this shouldn't be included I hope this helps.