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uniqueideaman

Member Since 12 Oct 2016
Offline Last Active Jun 23 2017 04:13 PM
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Posts I've Made

In Topic: Numerically Indexed Arrays

20 May 2017 - 04:28 AM

Guys,
 
I'm referring to this tutorial:
 
It shows 2 methods of creating an array:
 
 
<?php      
/* First method to create array. */
$numbers = array( one, two, three, four, five);
         
foreach( $numbers as $value ) {
    echo "Value is $value <br />";
    }
         
    /* Second method to create array. */
    $numbers[0] = "one";
    $numbers[1] = "two";
    $numbers[2] = "three";
    $numbers[3] = "four";
    $numbers[4] = "five";
         
foreach( $numbers as $value ) {
    echo "Value is $value <br />";
    }
?>
      
   </body>
</html>
 
 
 
On the first method, I just switched the numerical numbers to numbers in words but I get error! It should've worked no matter what the values of the arrays are on (note the first method below).
I changed it to:
 
 
<?php
//2 examples On How To Create Numeric Arrays:
?>
<html>
   <body>
   
<?php      
/* First method to create array. */
$numbers = array( one, two, three, four, five);
         
foreach( $numbers as $value ) {
    echo "Value is $value <br />";
    }
         
    /* Second method to create array. */
    $numbers[0] = "one";
    $numbers[1] = "two";
    $numbers[2] = "three";
    $numbers[3] = "four";
    $numbers[4] = "five";
         
foreach( $numbers as $value ) {
    echo "Value is $value <br />";
    }
?>
      
   </body>
</html>
 
 
This is the result I get:
 
Notice: Use of undefined constant one - assumed 'one' in C:\xampp\htdocs\test\test.php on line 41
 
Notice: Use of undefined constant two - assumed 'two' in C:\xampp\htdocs\test\test.php on line 41
 
Notice: Use of undefined constant three - assumed 'three' in C:\xampp\htdocs\test\test.php on line 41
 
Notice: Use of undefined constant four - assumed 'four' in C:\xampp\htdocs\test\test.php on line 41
 
Notice: Use of undefined constant five - assumed 'five' in C:\xampp\htdocs\test\test.php on line 41
Value is one 
Value is two 
Value is three 
Value is four 
Value is five 
Value is one 
Value is two 
Value is three 
Value is four 
Value is five 
 
 
My point:
 
If following works without any errors then the further following too should work:
 
/* First method to create array. Example, according to tutorial. This sows no errors. */
         $numbers = array( 1, 2, 3, 4, 5);
 
/* First method to create array. Example, according to my editing. This shows errors. */
         $numbers = array( one, two, three, four, five);

In Topic: How To Display In iFrame A Url From Mysql ?

06 May 2017 - 05:35 AM

Man,

Atleast one url should load properly in my inframe since not every website in the world would not be putting measures to foil iframes. 
How-about you try loading a page and when it works then give me the url to check out on my end.

I've now taken out all the session code altogether.
Code now looks like this:

<html>
<head>
<title>
Home Page
</title>
</head>
<body>
<body background=".png">

<?php
include 'config.php';

    $sql = "SELECT * FROM users WHERE usernames = 'USERNAME WENT HERE'";
    $result = mysqli_query($conn,$sql);
    $numrows = mysqli_num_rows($result);
    if($numrows >0)
    {    
        while ($row = mysqli_fetch_assoc($result))
        {
            $db_id = $row["ids"];
            $db_username = $row["usernames"];
            $db_first_name = $row["first_names"];
            $db_surname = $row["surnames"];
            $db_email = $row["emails"];
            $db_blog_url = $row["blogs_urls"];
    
            //Welcome user by name.
            echo "Welcome <b><h2>$db_first_name $db_surname!"?></h2></b>|

            <?php 
            //Display User's own blog Page in iframe.?>
            <iframe src='<?php echo $row["blog_urls"];?>'></iframe>
            <br>

            <?php 
            //Display 1st User's blog Page (regardless of who the user is) in iframe.?>
            <iframe src="<?php echo '.$blogs_urls[0].';?>"></iframe>
                
            <?php 
            //Display 1st User's blog Page (regardless of who the user is) in iframe.?>
            <iframe src='<?php echo ".$blogs_urls[0].";?>'></iframe>
            <?php 
            
        }
    }
    else
    {
    echo "<p>No Results!</p>\n" ;
    }    
?>

</body>
</html>

I have changed the url from my db to:

http://www.goviralno...iques-for-2017/

But I don't see any pages loading. I picked a page most likely to be unknown. However, now see no blank page but 3 iframes with 3 error messages:

1st error message:

**Access forbidden!**

**You don't have permission to access the requested object. It is either read-protected or not readable by the server.**

**If you think this is a server error, please contact the webmaster.**

**Error 403**

**localhost**
**Apache/2.4.23 (Win32) OpenSSL/1.0.2h PHP/7.0.9**


2nd error message:

**Object not found!**

**The requested URL was not found on this server. The link on the referring page seems to be wrong or outdated. Please inform the author of that page about the error.**

**If you think this is a server error, please contact the webmaster.**

**Error 404**

**localhost**
**Apache/2.4.23 (Win32) OpenSSL/1.0.2h PHP/7.0.9**


3rd error message:

Same as 1st error message.

Note from my most recent code, I have 3 iframes coding 3 different ways with php. Looking at these error messages and my 3 iframe codes, what is your conclusion ? What do you make out of all this ?

In Topic: How To Display In iFrame A Url From Mysql ?

01 May 2017 - 06:16 AM

Error settings is full set to show error messages (config.php):
 
/*
* ERROR HANDLING
* ini_set('display_errors', 1);
*   ini_set('display_startup_errors', 1);
 
* For All Error, Warning and Notice
*   error_reporting(E_ALL); OR error_reporting(-1);
* For All Errors
*   error_reporting(E_ERROR);
* For All Warnings
*   error_reporting(E_WARNING);
* For All Notice
*   error_reporting(E_NOTICE);
*/
error_reporting(E_ALL);

In Topic: How To Display In iFrame A Url From Mysql ?

01 May 2017 - 05:56 AM

Ok,
 
Here's the original script.
I shortened it 2 nights ago so you guys don't have to wade through many lines of code. And while shortening it, I made mistakes. And so, here's the full script for that particular page.
config.php contains the db connection details and other files.
 
 
 
   
    <html>
    <head>
    <title>
    Home Page
    </title>
    </head>
    <body>
    
    <?php
    include 'config.php';
    
    /*Check if user is logged-in or not by checking if session is set or not. 
    If user is not logged-in then redirect to login page. Else, show user's account homepage.*/
    
    if(!isset($_SESSION["user"])) 
    {
        header("location:login.php");
    }
    else 
    {
     $user = $_SESSION["user"];
        $sql = "SELECT * FROM users WHERE usernames = '".$user."'";
     $result = mysqli_query($conn,$sql);
        $numrows = mysqli_num_rows($result);
        if($numrows >1)
     { 
     while ($row = mysqli_fetch_assoc($sql))
     {
     $db_id = $row["id"];
     $db_username = $row["usernames"];
     $db_first_name = $row["first_names"];
     $db_surname = $row["surnames"];
     $db_email = $row["emails"];
     $db_blog_url = $row["blogs_urls"];
        
         //Welcome user by name.
     echo "Welcome <b><h2>$db_first_name $db_surname!"?></h2></b>|
    
     <?php
     //Display log-out link.
     echo "<a href='logout.php'>$user Log Out</a>";?>|<br>
        
     <?php 
     //Display User's own blog Page in iframe.?>
     <iframe src="<?php echo $db_blog_url;?>"></iframe>
     <br>
    
     <?php 
     //Display 1st User's blog Page (regardless of who the user is) in iframe.?>
     <iframe src="<?php echo ".$blogs_urls[0].";?>"></iframe>
     
     <?php 
     //Display All Users' blogs Pages in iframe.?>
     <iframe src="<?php echo $row['blogs_urls'];?>"></iframe>
     <?php 
     }
        } 
    }
    ?>
    
    </body>
    </html>
   
   
   
Ignore the part where my code is not secure from sql injection. I will deal with that later. In the meanwwhile, concentrating on how to fix this frame thing. I attempted 3 different ways to code the iframe to load a webpage but no luck. I don't get any errors but a complete blank page. (Error settings is full set to show error messages).

In Topic: How To Display In iFrame A Url From Mysql ?

29 April 2017 - 04:56 PM

Hi,

Imagine there is a tbl called "users" and it has a column called "your blog url".
Now, imagine you want to display that blog's url in an iframe on one of your pages. How would you code it ?
I did the following but the iframe fails to load the webpage.
I guess this is due to me failing to get the iframe to pic one url or one entry from the "your blog url" column.

The best attempt I made was this:

<?php 
    //Display User's blog?>
    <iframe src="<?php $row["blog"];?>"></iframe>
​​​​​​​

How would you code it yourself ? Imagine, you want to display in the iframe the url that is the final entry in the "your blog url" column in users tbl in mysql db.

Imagine the tbl looks like this:

     users
id|your_blog_url
6|http://myownblog.com

And since the "http://myownblog.com" is the final entry in the "your_blog_url" column you want to load that in the iframe.

Whatever sample you provide, make sure it works by checking it in your wamp/xampp before adding your code in this thread.

Thanks!