PHP4 to PHP5 Migration

PHP4 to PHP5 Migration

What are some good steps to follow for a smooth migration from PHP4 to PHP5. What are some types of code that are likely to break?

Solution :

I also once worked on an app which used PHP4's XML support quite heavily, and would have required quite a bit of work to move to PHP5.

One of the other significant changes I was looking at at the time was the change of the default handling of function parameters. In PHP4 if I remember, they were pass-by-copy unless you specified otherwise, but in PHP5 is changed to pass-by-reference by default. In well written code, that probably won't make a big difference to you, but it could certainly cause problems.

I think one other thing I found changed is that objects are no longer allowed to overwrite their 'this' field. I would say that was a really bad idea to begin with (and I think it may have not been an intentional feature in PHP4), but I certainly found a few parts of our system that relied on it.

Hope some of that helps.

PHP array indexing: $array[$index] vs $array[“$index”] vs $array[“{$index}”]

by Blogger 4:11:00 AM array PHP

PHP array indexing: $array[$index] vs $array[“$index”] vs $array[“{$index}”]

What is the difference, if any, between these methods of indexing into a PHP array:

$array[$index]
$array["$index"]
$array["{$index}"]

I'm interested in both the performance and functional differences.

Update:

I'm not sure that's right. I ran this code:

  $array = array(100, 200, 300);
  print_r($array);
  $idx = 0;
  $array[$idx] = 123;
  print_r($array);
  $array["$idx"] = 456;
  print_r($array);
  $array["{$idx}"] = 789;
  print_r($array);

And got this output:

Array
(
    [0] => 100
    [1] => 200
    [2] => 300
)

Array
(
    [0] => 123
    [1] => 200
    [2] => 300
)

Array
(
    [0] => 456
    [1] => 200
    [2] => 300
)

Array
(
    [0] => 789
    [1] => 200
    [2] => 300
)

Solution :

All array indices are of type 'int' first, then type 'string', and will be cast to that as PHP sees fit.

Performance wise, $index should be faster than "$index" and "{$index}" (which are the same).

Once you start a double-quote string, PHP will go into interpolation mode and treat it as a string first, but looking for variable markers ($, {}, etc) to replace from the local scope. This is why in most discussions, true 'static' strings should always be single quotes unless you need the escape-shortcuts like "\n" or "\t", because PHP will not need to try to interpolate the string at runtime and the full string can be compiled statically.

In this case, doublequoting will first copy the $index into that string, then return the string, where directly using $index will just return the string.

source: http://stackoverflow.com/questions/6628/php-array-indexing-arrayindex-vs-arrayindex-vs-arrayindex

Login Script with hidden buttons

by Blogger 4:07:00 AM Login PHP Script

Login Script with hidden buttons

I have been using PHP and Javascript for building my dads website. He wants to incorporate a login system into his website. I have the design for the login system using PHP, my problem is how do I show buttons if the person is logged in.

For Example - You have Home, Products, About Us, and Contact. Well I want to have buttons for Dealer, Distributor, and maybe other information if the user is logged in. So I will have Home, Products, About Us, Contacts, Dealer (if dealer login), Distributor (if distributor login), and so forth.

Would Javascript be a good way to do this or would php, maybe even both. Using Javascript to show and hide buttons, and php to check to see which buttons to show.

Solution:

Regarding security, you cannot trust what comes from the client:

The visitor can see all your code (HTML and Javascript, not PHP) and try stuff
The visitor may not even use a browser; it's trivially easy to send a request with a script

This means hiding the buttons is good User Interface design (because you can't use them if you are not logged in). But it's not a security feature. The security feature is checking, on the server, that the visitor is logged in before each action that requires it.

If you don't intend to show the buttons, it's not useful to send the HTML and images to the browser and then hide them with Javascript. I would check with PHP.

source : http://stackoverflow.com/questions/7031/login-script-with-hidden-buttons

MySQL/Apache Error in PHP MySQL query

by Blogger 3:48:00 AM Apache MYSQL PHP Sql Query

MySQL/Apache Error in PHP MySQL query

I am getting the following error:

Access denied for user 'apache'@'localhost' (using password: NO)

When using the following code:

<?php

include("../includes/connect.php");

$query = "SELECT * from story";

$result = mysql_query($query) or die(mysql_error());

echo "<h1>Delete Story</h1>";

if (mysql_num_rows($result) > 0) {
    while($row = mysql_fetch_row($result)){
          echo '<b>'.$row[1].'</b><span align="right"><a href="../process/delete_story.php?id='.$row[0].'">Delete</a></span>';
      echo '<br /><i>'.$row[2].'</i>';
    }
}
else {
   echo "No stories available.";
}
?>

The connect.php file contains my MySQL connect calls that are working fine with my INSERT queries in another portion of the software. If I comment out the $result = mysql_query line, then it goes through to the else statement. So, it is that line or the content in the if.

I have been searching the net for any solutions, and most seem to be related to too many MySQL connections or that the user I am logging into MySQL as does not have permission. I have checked both. I can still perform my other queries elsewhere in the software, and I have verified that the account has the correct permissions.

Answer :

And if it matters at all, apache@localhost is not the name of the user account that I use to get into the database. I don't have any user accounts with the name apache in them at all for that matter.

If it is saying 'apache@localhost' the username is not getting passed correctly to the MySQL connection. 'apache' is normally the user that runs the httpd process (at least on Redhat-based systems) and if no username is passed during the connection MySQL uses whomever is calling for the connection.

If you do the connection right in your script, not in a called file, do you get the same error?

source : http://stackoverflow.com/questions/2900/mysql-apache-error-in-php-mysql-query

How to include PHP files that require an absolute path?

by Blogger 3:41:00 AM Files PHP PHP Include

How to include PHP files that require an absolute path?

I have a directory structure like the following;

script.php

inc/include1.php
inc/include2.php

objects/object1.php
objects/object2.php

soap/soap.php

Now, I use those objects in both script.php and /soap/soap.php, I could move them, but I want the directory structure like that for a specific reason. When executing script.php the include path is inc/include.php and when executing /soap/soap.php it's ../inc, absolute paths work, /mnt/webdev/[project name]/inc/include1.php... But it's an ugly solution if I ever want to move the directory to a different location.

So is there a way to use relative paths, or a way to programically generate the "/mnt/webdev/[project name]/"?

Answer :

This should work

$root = realpath($_SERVER["DOCUMENT_ROOT"]);

include "$root/inc/include1.php";

Edit: added imporvement by aussieviking

source :http://stackoverflow.com/questions/4369/how-to-include-php-files-that-require-an-absolute-path

PHP: show a number to 2 decimal places

by Blogger 7:37:00 AM 2 decimal places decimal float number number_format() PHP

PHP: show a number to 2 decimal places

What's the correct way to round a PHP string to 2 decimal places?

$number = "520"; // It's a string from a DB

$formatted_number = round_to_2dp($number);

echo $formatted_number;

Output is "520.00";

How function round_to_2dp() definition should be?

Best Answer :

can use number_format():

return number_format((float)$number, 2, '.', '');

Example:

$foo = "105";
echo number_format((float)$foo, 2, '.', '');  // Outputs -> 105.00

This function returns a string.

PHP file_get_contents() returns “failed to open stream: HTTP request failed!”

by Blogger 2:05:00 AM cURL curl_init() file_get_contents() PHP

PHP file_get_contents() returns “failed to open stream: HTTP request failed!”

Problem :

I am having problems calling a url from PHP code. I need to call a service using a query string from my PHP code. If I type the url into a browser, it works ok, but if I use file-get-contents() to make the call, I get:

Warning: file-get-contents(http://.... ) failed to open stream: HTTP request failed! HTTP/1.1 202 Accepted in ...

The code I am using is:

$query=file_get_contents('http://###.##.##.##/mp/get?mpsrc=http://mybucket.s3.amazonaws.com/11111.mpg&mpaction=convert format=flv');
echo($query);

Like I said - call from the browser and it works fine. Any suggestions?

I have also tried with another url such as:

$query=file_get_contents('http://www.youtube.com/watch?v=XiFrfeJ8dKM');

This works fine... could it be that the url I need to call has a second http:// in it?

Best answer

using cURL. if cURL not install , insatll first cURL

<?php

$curl_handle=curl_init();
curl_setopt($curl_handle, CURLOPT_URL,'http://###.##.##.##/mp/get?mpsrc=http://mybucket.s3.amazonaws.com/11111.mpg&mpaction=convert format=flv');
curl_setopt($curl_handle, CURLOPT_CONNECTTIMEOUT, 2);
curl_setopt($curl_handle, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($curl_handle, CURLOPT_USERAGENT, 'Your application name');
$query = curl_exec($curl_handle);
curl_close($curl_handle);

?>

All In One Script

PHP4 to PHP5 Migration

PHP4 to PHP5 Migration

Solution :

PHP array indexing: $array[$index] vs $array[“$index”] vs $array[“{$index}”]

PHP array indexing: $array[$index] vs $array[“$index”] vs $array[“{$index}”]

Update:

Solution :

Login Script with hidden buttons

Login Script with hidden buttons

Solution:

MySQL/Apache Error in PHP MySQL query

MySQL/Apache Error in PHP MySQL query

Answer :

How to include PHP files that require an absolute path?

How to include PHP files that require an absolute path?

Answer :

PHP: show a number to 2 decimal places

PHP: show a number to 2 decimal places

Best Answer :

PHP file_get_contents() returns “failed to open stream: HTTP request failed!”

PHP file_get_contents() returns “failed to open stream: HTTP request failed!”

Problem :

Best answer

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