jsjs | Simple JavaScript meta-loop interpreter | Frontend Utils library

 by   bramblex JavaScript Version: Current License: No License

kandi X-RAY | jsjs Summary

kandi X-RAY | jsjs Summary

jsjs is a JavaScript library typically used in User Interface, Frontend Utils, Nodejs applications. jsjs has no bugs, it has no vulnerabilities and it has medium support. You can download it from GitHub.

Simple JavaScript meta-loop interpreter
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            kandi-support Support

              jsjs has a medium active ecosystem.
              It has 1310 star(s) with 150 fork(s). There are 52 watchers for this library.
              OutlinedDot
              It had no major release in the last 6 months.
              There are 9 open issues and 3 have been closed. On average issues are closed in 11 days. There are no pull requests.
              It has a neutral sentiment in the developer community.
              The latest version of jsjs is current.

            kandi-Quality Quality

              jsjs has 0 bugs and 0 code smells.

            kandi-Security Security

              jsjs has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
              jsjs code analysis shows 0 unresolved vulnerabilities.
              There are 0 security hotspots that need review.

            kandi-License License

              jsjs does not have a standard license declared.
              Check the repository for any license declaration and review the terms closely.
              OutlinedDot
              Without a license, all rights are reserved, and you cannot use the library in your applications.

            kandi-Reuse Reuse

              jsjs releases are not available. You will need to build from source code and install.

            Top functions reviewed by kandi - BETA

            kandi has reviewed jsjs and discovered the below as its top functions. This is intended to give you an instant insight into jsjs implemented functionality, and help decide if they suit your requirements.
            • 15 . 2 . 5
            • The context .
            • the main function
            • this is a safe loop
            • the main function
            • this is equivalent
            • a bit implementation
            • incoming error is thrown
            • the input handler
            • inbound tests
            Get all kandi verified functions for this library.

            jsjs Key Features

            No Key Features are available at this moment for jsjs.

            jsjs Examples and Code Snippets

            No Code Snippets are available at this moment for jsjs.

            Community Discussions

            QUESTION

            How to solve Cannot invoke "String.compareTo(String)" because "[]" is null?
            Asked 2021-Mar-24 at 02:27
            import java.util.Scanner;
            
            
            public class Search {
                static Scanner scanner = new Scanner(System.in);
                static Scanner kb = new Scanner(System.in);
                static Scanner kb2 = new Scanner(System.in);
                public static void main (String[] args)
                {
            
                    int choice;
                    System.out.print("Choose a number of students: ");
                    int n = scanner.nextInt();  
                    String name[] = new String[n+1];
                    String course[] = new String[n+1];
                    int ID[] = new int[n+1];
            
                    for(int i=1;i <= n; i++)
                    {
                        System.out.print("Enter ID number " + i + ": ");
                        ID[i] = scanner.nextInt();
                        System.out.print("Enter Student name " + i + ": ");
                        name[i] = kb.nextLine();
                        System.out.print("Enter Student course " + i + ": ");
                        course[i] = kb2.nextLine();
                        System.out.println("----------------------------------------");
                    }
            
                   
                  
                    do
                    {
                        choice = menu();
                        if(choice == 1)
                        {
                            sortID(ID);
                            printValues(ID);
            
                        }else if(choice == 2)
                        {
                            nameSort(name,n);
                            printName(name,n);
                        }else if(choice == 3)
                        {
            
                        }
                    }while(choice !=0);
                }
            
                public static int menu()
                {
                    System.out.print("\n1. Sort by ID\n2. Sort by Name\n3. Search by ID\n4. Search by Name\n5. Search by Course\n6. Display Records In table Form.\nYour Choice: ");
                    return scanner.nextInt();
                }
            
                public static void sortID(int []id)
                {
                    int temp;
                    int index, counter;
                    for (counter=0; counter < id.length -1; counter++) {
                        for (index=0; index < id.length - 1 - counter; index++) {
                            if (id[index] > id[index+1]) {
                                temp = id[index];
                                id[index]=id[index+1];
                                id[index+1]=temp;
                            }
                        }
                    }
                }
            
                public static void printValues (int[]array) {
                    
                    System.out.println ("\nSorted Id Number: ");
                    for(int i = 1; i < array.length; i++){
                        System.out.print ("\n" + array[i]);            
                    }
                        
                }
            
                public static void printName (String[]array,int a) {
                    
                    for (int i = 0; i <= a - 1; i++) 
                    {
                        System.out.print(array[i] + ", ");
                    }
                        
                }
            
               public static void nameSort(String[] name,int a)
                {
                    String temp;
                    for (int i = 0; i < a; i++) 
                    {
                        for (int j = i + 1; j < a; j++) { 
                            if  (name[i].compareTo(name[j])>0) 
                            {
                                temp = name[i];
                                name[i] = name[j];
                                name[j] = temp;
                            }
                        }
                    }
                }
            
            
            
            
            
            
            }
            
            ...

            ANSWER

            Answered 2021-Mar-24 at 02:27
            String name[] = new String[n+1];
            

            Source https://stackoverflow.com/questions/66773616

            QUESTION

            Regular Expression for a pattern
            Asked 2020-Jan-20 at 09:38

            I was looking for regular expression to find a string having a particular pattern (first occurrence )

            Consider this as my strings

            ...

            ANSWER

            Answered 2020-Jan-20 at 09:32

            You could take a positive look ahead for a colon and take the first element of the match.

            Source https://stackoverflow.com/questions/59820312

            QUESTION

            How do I select json from my table Postgres
            Asked 2018-Dec-08 at 05:14

            I have a table named loans with a lot of fields such as (id, name, description, etc.) And inside that table, I have a column which type is JSON(TermsMessage)(ARRAY OF JSON), and when I make a select query it returns me an error

            my query

            `

            ...

            ANSWER

            Answered 2018-Dec-08 at 05:14

            PostgreSQL doesn't have a built-in equality operator for the datatypes JSON or arrays of JSON. When grouping by a JSON field, the db needs to know which are the same values and can be grouped together. You can work around this by turning the JSON (or JSON arrays) into either text arrays or JSONb arrays using something like ::jsonb[] or ::text[]. I'd probably do in the innermost part of the query so that it flows through later naturally:

            Source https://stackoverflow.com/questions/53679567

            QUESTION

            MySQL, move data from rows to columns with Images pictures
            Asked 2018-Jun-06 at 09:32

            table1

            ...

            ANSWER

            Answered 2018-Jun-05 at 13:44

            You can do it with INSERT INTO ... SELECT statement:

            Source https://stackoverflow.com/questions/50701492

            QUESTION

            json file offline can read by chart.js, but online json can't read
            Asked 2017-Jul-12 at 08:41

            I have some problem with my chart.js, if i read offline json file, chart will be appear, but if I use online json from Firebase, it won't show anything, jsut white screen.

            I export json file from firebase, before I use it in online mode, but it won't show anything.

            this is my code, just one page.

            ...

            ANSWER

            Answered 2017-Jul-12 at 07:20

            Just remove window.onload = function() {} inside $.getJSON()

            Because window is already loaded before getting JSON

            Source https://stackoverflow.com/questions/45050731

            Community Discussions, Code Snippets contain sources that include Stack Exchange Network

            Vulnerabilities

            No vulnerabilities reported

            Install jsjs

            You can download it from GitHub.

            Support

            For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .
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            CLONE
          • HTTPS

            https://github.com/bramblex/jsjs.git

          • CLI

            gh repo clone bramblex/jsjs

          • sshUrl

            git@github.com:bramblex/jsjs.git

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