Hi! I’m Starx

Question 1

Ghostmancer’s Question:

I am using jQuery and Twitter Bootstrap, and want to have a tooltip appear when hovering over a link. However, the tooltip doesn’t appear the first time I hover over a link, but if I move the mouse away and back on, it works every time for that link.

I have several of these on the page:

<a href="#" onmouseover="$(this).tooltip();" class="postAuthor" data-original-title="@username">Full Name</a>

I’ve created a jsFiddle to demonstrate: jsFiddle. Any help would be appreciated – thanks in advance!

Answer

Starx

.tooltip() functions initiates the tooltip effect on the link, not show the tooltip

Explanation:

When the $(this).tooltip() is triggered on the first hover, it instantiate the plugin first. Then finally on the second hover you get the tooltip.

Solution:

Add this on your code:

$(function() {
    $("a").tooltip();
});

Solution

Question 2

Sam Peterson’s Question:

Sam Peterson

I still don’t fully understand where I’m required to use the “public static void main(String[] args)” header within the .java files of a project. Do you need to put that header in every .java file of a package?

I have been following along with chapter 3 in my book, dragging and dropping downloaded stand alone book source files into my project’s package, but some of the .java files in my package don’t like that “public static void main(String[] args)” statement, even though my opening and closing curly braces are in the correct place. Here’s an example of one of those files (the ERROR(S) are described in the code’s comments):

    public class Rectangle
{
   public static void main(String[] args){
   private double length;//ERROR: illegal start of expression
   private double width;

   /**
    * Constructor
    */

   public Rectangle(double len, double w)
   {
      length = len;
      width = w;
   }

   /**
    * The setLength method accepts an argument
    * that is stored in the length field. 
    */

   public void setLength(double len)
   {
      length = len;
   }

   /**
    * The setWidth method accepts an argument
    * that is stored in the width field.
    */

   public void setWidth(double w)
   {
      width = w;
   }

   /**
    * The set method accepts two arguments
    * that are stored in the length and width
    * fields.
    */

   public void set(double len, double w)
   {
      length = len;
      width = w;
   }

   /**
    * The getLength method returns the value
    * stored in the length field.
    */

   public double getLength()
   {
      return length;
   }

   /**
    * The getWidth method returns the value
    * stored in the width field.
    */

   public double getWidth()
   {
      return width;
   }

   /**
    * The getArea method returns the value of the
    * length field times the width field.
    */

   public double getArea()
   {
      return length * width;
   }

}//end of: public static void main(String[] args)

}//end of: public class Rectangle  ERROR: class, interface, or enum expected

The ERROR(S) came up after I added the “public static void main(String[] args)” to the existing Rectangle.java file. Any idea of why this occurs?

Answer

Starx

This occurs because You cannot have more than one main() method in a package. However overriding the main method is definitely allowed. What I mean is as, long the parameter number is differed you can override main() method.

public static void main(String[] args) {
   .....
}

And somewhere else in other or same class, you can add such class

public static void main(String arg1, String arg2) {
    .....
}

The ERROR: illegal start of expression is because you are using access modifier inside the method. Use all the access mofifier and variable declaration inside the class

public class Rectangle {

    private double length;//ERROR: illegal start of expression   
    private double width;

    public static void main(String[] args) {
    ....
    }
}

The ERROR: class, interface, or enum expected is because the class Rectangle is only housing a static method, your all the methods and parameters are inside the static method called main()

Here is your code which will compile without an error.

Question 3

Aaron’s Question:

Aaron

So what I am reading is that when adding a class to an element without any class selectors, use the addClass() method over attr().

For example, using:

.addClass('myclass')

is more efficient than

.attr('class','myclass')

I cannot find anything equivalent to addId(‘myid’) for adding id selectors, does it exist or is attr() the best way to add an id selector?

I get that id is unique, the logic behind my question is addClass() is faster than attr().

So my question is if there’s a more efficient method for creating an id than attr().

Answer

Starx

Answering directly

No, addId() does not exist and addClass() is not as same as attr()

In HTML, a class attribute can hold more than one value like <div class="tall fat brown">Someone</div> so jQuery’s function addClass() helps to add a value to the class attribute not change it. Similarly the functions toggleClass() and ‘removeClass()` help manipulate them.

But the function attr() will just manipulate the attribute and change it directly.

For example:

<div id="person">someone</div>

And the following jQuery Statements

$("#div").addClass("tall"); // This will create the class attribute and add tall 
$("#div").addClass("fat");  // This will append fat to the existing class and make "tall fat"
$("#div").addClass("brown"); // likewise

.addClass() will just append the class name. If same needs to done by using attr() you have to do

$("#div").attr('class', 'tall');
$("#div").attr('class', 'tall fat');
$("#div").attr('class', 'tall fat brown');

Or, you can modify the attribute using

$("#div").attr('class', function(i, className) {
    return className + " brown";
});

Where as ids have one value that needs to be modified or altered so a function like addId() would do exactly which attr(‘id’, ‘idvalue’) would do.

Question 4

Ted’s Question:

Ted

I am getting some data from a page in JSON format
JSON:
{
‘name’ : ‘maverick’,
Image : ‘Jason’
}

I now need to take these values and add them to the #main div.

        <div id="main">
        <div class="a"> name: Eithan <img src="img/Eithan.jpg" /> <div>
        <div class="a"> name: Emma <img src="img/Emma.jpg" /> <div>
       </div>

How do I do that in jQuery for the case there are several objects in the JSON?

Answer

Starx

You can do this by using $.getJSON() function.

$.getJSON("link/to/json", function(data) {
    $.each(data, function(i, item) {
        $('#main').append('<div class="a"> name: ' + item.name + ' <img src="img/' + item.image + '.jpg" /></div>');
    });
});

Question 5

Robert Rocha’s Question:

robert rocha

While reading a book about php I cam across a piece of code that logically doesn’t make sense to me. The line of code is part of a class function:

private function replaceTags( $pp = false ) {
    //get the tags in the page
    if( $pp == false ) {
        $tags = $this->page->getTags();
    } else {
        $tags = $this->page->getPPTags();
    }
    //go through them all
    foreach( $tags as $tag => $data ) {
        //if the tag is an array, then we need to do more than a simple find and replace!
        if( is_array( $data ) ) {
            if( $data[0] == 'SQL' ) {
                //it is a cached query...replace tags from the database
                $this->replaceDBTags( $tag, $data[1] );
            } elseif( $data[0] == 'DATA' ) {
                //it is some cahched data...replace tags from cached data
                $this->replaceTags( $tag, $data[1] );
            }
        } else {
            //replace the content
            $newContent = str_replace( '{' . $tag . '}', $data, $this->page->setContent( $newContent ) );
            $this->page->setContent( $newContent );
        }
    }
}

The specific line that doesn’t make sense to me is:

$newContent = str_replace( '{' . $tag . '}', $data, $this->page->setContent( $newContent ) );

How can you pass the variable “$newContent” to “setContent( $newContent )” when it doesn’t have a value yet?

Any explanations?

Answer

Starx

That statement is executing in a for loop so, $newContent will hold the value in another loop to use.

In the first execution, $newContent will be empty but on the next iteration it will have a value to replace.

foreach( $tags as $tag => $data ) {
    if ....
    } else {
        //replace the content
        $newContent = str_replace( '{' . $tag . '}', $data, $this->page->setContent( $newContent ) );
     // ^
     // Now next time when the loop executes again it will have a 
     // $newContent to process.

        $this->page->setContent( $newContent );
    }
}

Question 6

Pulz’s Question:

Pulz

I’m coding the Dijkstra algorithm in Java. My first method

public void populateDijkstraFrom(Node startNode)

creates a Linked List of nodes with its respective distances and predecessors. My second method

public List<Node> getShortestPath(Node startNode, Node targetNode)

is supposed to create a list of nodes with the shortest path from the startNode to the targetNode using the list of nodes from my populateDijkstraFrom method.

However, I don’t know how to access the list of nodes from my Dijkstra method in the getShortestPath method. I could change the return type from void to LinkedList but I was told that it works using void.

How do I do it?

Thanks

Answer

Starx

There are basically two ways of solving this.

Easiest one would be return the list of nodes on your method

public List<Node> populateDijkstraFrom(Node startNode) {
       // ^ Configure to return list instead of void

    // ......
    return nodeList;
}

public List<Node> getShortestPath(Node startNode, Node targetNode) {
    List<Node> pdList = populateDijkstraFrom(startNode);
    // ^ --- Get the list by simply passing the same parameter to the method
}

Another is stated by Gian

Question 7

K20GH’s Question:

K20GH

I’m trying to do something fairly simple as per below. Basically, I want to check to see if a user exists using their ‘gid’. If they don’t exist, add them to the database, if they do, echo “User Already Exists”

For whatever reason, it will always add a new user, even if their gid exists.

$checkuser = "SELECT gid FROM user WHERE gid = '$id'";
$adduser = "INSERT INTO user (gid, name, pic) VALUES ('$id','$name','$img')";

if (!mysql_num_rows($checkuser))
{
    if (!mysqli_query($con,$adduser))
      {
      die('Error:  ' . mysqli_error($con));
      }
    echo "1 record added";
}
else {
    echo "User Already Exists";
    }

Answer

Starx

There is many things wrong with your code. First and the most important is you have a query a statement, if you do not query, you will not receive any result-set and the resultant is new registration every time.

Function is_resource() will help you decide whether or not the query
returns valid result.

Add here is the correct way to solve your problem.

$checkuser = "SELECT gid FROM user WHERE gid = '$id'";
$adduser = "INSERT INTO user (gid, name, pic) VALUES ('$id','$name','$img')";

$result = mysqli_query($con, $checkuser)); 
          // ^-- You have to query first to find out if there are any users or not

if(is_resource($result) && mysqli_num_rows($result) > 0) {
   // ^ This will confirm the query has returned a result and it contains rows

     echo "User Exists";
} else {
   // If it does not you can finally add it

     $result = mysqli_query($con, $adduser) or die(mysqli_error());
     echo "User added"; 
}

Question 8

Mr Me’s Question:

Mr Me

I’ve already checked the issue disable access to included files , but I am wondering if that’s the best solution.

Context: I’m building a bootstrap class for my PHP Framework, and realized that there are multiple solutions to this security issue.

After researching and reading posts like the one I mentioned at first and others related to htaccess, I think that there are basically three types of solutions:

1 – Checking a constant (like in the post I linked)

if(!defined('THIS_IS_NOT_A_DIRECT_ACCESS')) {
    header('HTTP/1.1 404 Not Found');
    include('./../error/404.php');
    die;
}

Or

require('../error/include_file.php');
//
if(!defined('THIS_IS_NOT_A_DIRECT_ACCESS'))
{
    header('HTTP/1.1 404 Not Found');
    include('404.php');  
    die; 
}

2 – Redirecting all calls to the Bootstrap and making a clever filtering.

//Something like
// if $urlRequested it's a file
    // Go to Error
// else if $urlRequested it's not a controller
    // Go to Error
// else 
    // Execute Controller Logic.

3 – Setting htaccess.

# Redirecting calls for non-directories, files or links
RewriteCond %{REQUEST_FILENAME} !-d
RewriteCond %{REQUEST_FILENAME} !-f
RewriteCond %{REQUEST_FILENAME} !-l
RewriteRule ^(.+)$ index.php?url=$1 [QSA]

# Redirecting the rest of the calls to the error page.
RewriteRule ^(.+)$ index.php?url=error/404 [QSA]

Personally I think the solution 3 is the most interesting, but I am pretty new in the Htaccess control so I wonder if this is a safe solution.

For the purists and minimalists, the Question in here would be:
Are these (the three examples) nice direct access control systems for Apache-PHP applications? Else, which would be the safest approach? And the simplest?

Answer

Starx

This is a debatable topic but .htaccess rules applies to all the document on that particular directory. While applying 1 or 2 you may have to include that portion on every file.

Question 9

User2386255’s Question:

user2386255

I have the following code working pretty well, but I want to be able to select more than one option if need be: http://jsfiddle.net/exlondoner/MKa3n/

Can anyone help?

JS:

$('.media-select-option').on('click', function() {
    $('.media-select-option').removeClass('selected');
    $(this).addClass('selected');     
    $('.m-overlay-close-btn').addClass('checked');       
});

Answer

Palash Mondal

Try this:

$('.media-select-option').on('click', function () {
    $(this).toggleClass('selected');
    $('.m-overlay-close-btn').addClass('checked');
});

FIDDLE

Answer

Starx

It is not working because there is no such element with class m-overlay-close-btn

$('.overlay-close-btn').addClass('checked');

Working fiddle

Question 10

Wizzme’s Question:

wizzme

I cannot figure out what is wrong with my dropdown menu. When I over over the main level link, the drop down appear but at the left of my screen instead of underneath the main link.
I have been on this for a couple of hours and any help would be greatly appreciated.

Here is the html part :

<div class="nav">

        <ul id="menu">
           <li><a href="#" class="current">Home</a></li>
              <li><a href="#">Apetiziers</a>
              <ul>
               <li><a href="#">Sub-Link 1</a></li>
               <li><a href="#">Sub-Link 2</a></li>
               <li><a href="#">Sub-Link 3</a></li>
               <li><a href="#">Sub-Link 4</a></li>
              </ul>
              </li>

              <li><a href="#">Entree</a>
             <ul>
             <li><a href="#">Sub-Link 1</a></li>
             <li><a href="#">Sub-Link 2</a></li>
             <li><a href="#">Sub-Link 3</a></li>
             <li><a href="#">Sub-Link 4</a></li>
             </ul>
              </li>

              <li><a href="#">Main Course</a>
             <ul>
             <li><a href="#">Sub-Link 1</a></li>
             <li><a href="#">Sub-Link 2</a></li>
             <li><a href="#">Sub-Link 3</a></li>
             <li><a href="#">Sub-Link 4</a></li>
             </ul>      
              </li>

              <li><a href="#">Dessert</a>
             <ul>
              <li><a href="#">Sub-Link 1</a></li>
              <li><a href="#">Sub-Link 2</a></li>
              <li><a href="#">Sub-Link 3</a></li>
              <li><a href="#">Sub-Link 4</a></li>
             </ul>      
              </li>


              <li><a href="#">Contact Us</a></li>
        </ul>

 </div>

And the .css :

ul#menu {
   float: left;
  margin: 0;
  width: auto;
    padding: 0px 40px 0px;
    background: #333; color: #fff;
    line-height: 100%;
}

ul#menu li {
  display: inline; 
}


/* top level link */
ul#menu a {
  float: left;
  padding: 10px 16px;
  margin-right: 0px;
  background: #789; color: #fff;
  text-decoration: none;
  border-right: 1px solid #e2e2e2;
}

/* main level link hover */
ul#menu a.current {
  background: #f60; color: #fff;
}


ul#menu li:hover > a {
  color: #fff; background: #ff4500;
  text-decoration: underline;
}

/* dropdown */
ul#menu li:hover > ul {
    display: block; /* shows the sub-menu (child ul of li) on hover */
}

/* sub level list */
ul#menu ul {
    display: none; /* hides the sub-menu until you hover over it */
    margin: 0;
    padding: 0;
    width: 140px;
    position: absolute;
    top: 35px;
    left: 0;
    background: #000;
    border: solid 1px #ccc;
}

ul#menu ul li {
    float: none;
    margin: 0;
    padding: 0;
}

ul#menu ul a {
    font-weight: normal;
    background: #9BB3BF; color: #036;
}

/* sub levels link hover */
ul#menu ul li a:hover {
color: #036; background: #DDDF99;
}

Answer

Starx

It is probably jumping towards the closest relative container. So configure your list to act as relative container:

ul#menu > li {
  position: relative;
}

Also there was unnecessary float in your anchor tags, your li are already set to display as inline there is no point in float them

ul#menu a {
    display: inline-block;
    padding: 10px 16px;
    /* ... */
}

Here your fixed code

Hi! I’m Starx

experienced Software Developer. And this is my blog

Blog Page

Javascript tooltip only appears on second hover

Ghostmancer’s Question:

Converting a stand alone java file to a file that's part of a project package?

Sam Peterson’s Question:

addClass vs attr and creating an ID

Aaron’s Question:

Adding div from JSON in jQuery?

Ted’s Question:

php variable assignment confusion

Robert Rocha’s Question:

How to check if there aren't any result after querying and execute another statement?

K20GH’s Question:

Disable direct access to files in PHP

Mr Me’s Question:

Selecting more than one option

User2386255’s Question:

Dropdown Menu CSS

Wizzme’s Question:

Please fill the form - I will response as fast as I can!