Linux, Apache, MySQL, PHP

23-Mar-11 3:19

How can I send variable to another form without include Confused | :confused:

cjoki23-Mar-11 4:32

23-Mar-11 4:32

Depends...is this second form on the same page or a different page?

Chris J
www.redash.org

Elham M24-Mar-11 3:14

24-Mar-11 3:14

Different page

cjoki24-Mar-11 4:55

24-Mar-11 4:55

This is done with either a $_POST or a $_GET variable. Your can read more info and see some code sample here http://php.net/manual/en/language.variables.external.php[^]

Chris J
www.redash.org

Elham M24-Mar-11 22:17

24-Mar-11 22:17

$_POST and $_GET just for input type but my variable isn't that.this is my code I want when user click on button insert data to database but I can't access to variable
$row[lsname],$row[teachername],...even this page and other page?another my question is why variable like$_POST[t] isn't recognaize in function?!!!even I write global
$_POST[t] but it's error

<?php

 session_start();
 echo ($_SESSION[users]);
echo "$_POST[t]";
?>
<html>


 <form method="post" action="reqire.php">
<?php
 require_once ('db.php');
$db = new db("root","","regstudent","localhost");
 function aa()
 {

     require_once ('db.php');
$db = new db("root","","regstudent","localhost");
$db->query("INSERT INTO register(`user`, `lsname`, `teachername`, `classday`, `starttime`, `endtime`, `examtime`)VALUES('$_SESSION[user]','$row[lsname]','$row[teachername]','$row[classday]','$row[starttime]','$row[endtime]','$row[examtime]') ");

echo "one record added";
 }
  echo "<table border='4' bordercolor='CC0099' width=100% height=20% >";
  echo "<tr><th>Name</th><th>Professor</th><th>Capacity</th><th>Exam</th><th     colspan='3'>classtime</th></tr>";
   $result = mysql_query("SELECT * FROM class");
    while($row = mysql_fetch_assoc($result))
    {
    if($_POST[t]==$row[lsname])
  {
        foreach($row as $key=>$var2)
     {
       echo "<td align='center'>$var2 </td>";
     }
    echo "<td align='center'><input type='submit'  value='Register' onclick='aa()'></td> ";


    }

      echo "</tr>";
    }
   echo"</table>";

?>
</form>
</html>

please help me I mystify in my project :(

effayqueue25-Mar-11 1:25

effayqueue

25-Mar-11 1:25

if you mean what I think you mean:

first page:

<?php

$sSomething='378923';

// start your form here... 

echo '<input type="hidden" name="something" value="'.htmlentities($sSomething).'">';

// end your form here...


?>

second page:

<?php

// assumes your form is post not get.. if it's GET then use $_GET instead

$sSomething=(isset($_POST['something']) ? $_POST['something'] : '');

?>

cjoki25-Mar-11 2:52

25-Mar-11 2:52

When I must debug post, get or session variables I use this at the top of the target page.

<?php
session_start();
echo "<pre>";
print_r($_POST);
print_r($_SESSION);
echo "</pre>";
/// rest of page below here
?>

This will allow you to see what is being passed to your target page so you will know if the data is making it or not.

I see that you defined your function aa()...but where is it called from?

Chris J
www.redash.org

Elham M25-Mar-11 3:15

25-Mar-11 3:15

I called it when I define submit button onclick attributes

echo "<td align='center'><input type='submit'  value='Register' onclick='aa()'></td> ";

cjoki25-Mar-11 3:32

25-Mar-11 3:32

no you are calling a javascript function called aa() and that I'm guessing does not exist?

remeber php is executed on the server. The onclick is a javascript event that is triggered on the users browser. So you are most likely getting a undefined from the browser if the form submits to itself (i.e. - the scripts name is "reqire.php") But the error I think is wiped as the page reloads.

The php function aa() is never called.

On a side note, its really bad to:
1) have a database connection based on root, to much access for what is needed.
2) display a password in a forum, big security hole.

Chris J
www.redash.org

Elham M25-Mar-11 7:05

25-Mar-11 7:05

The function aa() work but I can't access the variable my big problem is I don't know user click in which button to insert data I agree with you but I can't insert in javascript!I think I should solve this problem in another way what's your suggestion?

cjoki25-Mar-11 10:02

25-Mar-11 10:02

This is what I have issues with in your code..

<?php 
	session_start(); 
	echo ($_SESSION[users]);
	echo "$_POST[t]"; // "$_POST[t]" should be $_POST['t']
?>
<html> 
	<form method="post" action="reqire.php">
	<?php 
		require_once ('db.php');
		// never ever show log in data in code to others. I will make a 
		// seperate script just for the connection function and include the page and call the function for the page that will run queries
		$db = new db("root","","regstudent","localhost"); 
		
		function aa()  // function is defined... but not called, this is useless
		{
			require_once ('db.php');
			// calling this again when you could just use the one defined outside of the function like this
			// global $db;....or you could pass the $db as a parameter of the function like aa($db)
			$db = new db("root","","regstudent","localhost");
			
			$db->query("INSERT INTO register(`user`, `lsname`, `teachername`, `classday`, `starttime`, `endtime`, `examtime`)VALUES('$_SESSION[user]','$row[lsname]','$row[teachername]','$row[classday]','$row[starttime]','$row[endtime]','$row[examtime]') ");
			
			echo "one record added"; 
		}
		
		echo "<table border='4' bordercolor='CC0099' width=100% height=20% >";  
		echo "<tr><th>Name</th><th>Professor</th><th>Capacity</th><th>Exam</th><th colspan='3'>classtime</th></tr>";   
		
		// add a die condition in case of failure. In production code it should write to a log, but is ok to display for development
		$result = mysql_query("SELECT * FROM class");   
		 // btw - the * will make a additional query on the db to look up the field names, best practice is to include field names yourself.
		 
		// this is how I would write the die....test it by forcing an error, like an incorrect table name in the query.
		$result = mysql_query("SELECT * FROM class") or die("<br>...LINE ".__LINE__."<br>...SQL: ".$sql."<br>...ERROR: ".mysql_error()."<br>");
		
		while($row = mysql_fetch_assoc($result)) 
		{    
			if($_POST[t]==$row[lsname])  // $_POST[t] should be $_POST['t'], but why do you want to show either 1 record or 0 records?
			{
				foreach($row as $key=>$var2)     
				{       
					// this is risky, the table will look like hell if the db schema changes with more or less
					//	fields than the seven columns expected in the header of this table
					echo "<td align='center'>$var2 </td>"; 
				}    
				echo "<td align='center'><input type='submit'  value='Register'  onclick='aa()'></td> ";    // <-- onclick is to javascript not php...it will not call the php function of defined above.
			}      
			echo "</tr>";    // where is the opening <tr>...?
		}   
		echo"</table>";
	?>
	</form>

This is a framework of how I would structure the same thing...

<?php
	session_start();
	include("resources.php"); // this would contain a list of includes for what was needed within my scripts.
	$dbh=dbconn(); // this function would be found in db.php referenced from resources.php
	if(isset($_POST['action'])) // I name my submits action
	{
		//....process logic in here. You could call your function from here too or take the logic out of the function
	}
	echo "<form action='reqire.php' method='post'>\n";
	//.. your form structure is here....
	echo "</form>\n";
	?>

Chris J
www.redash.org

Elham M25-Mar-11 21:17

Re: Remove a File Using PHP

25-Mar-11 21:17

Thanks alot for your help Wink | ;)

Remove a File Using PHP

Haroon Mughal23-Mar-11 2:35

Haroon Mughal

23-Mar-11 2:35

To remove a file from php use the unlink() function. You can enter either the full path or the relative path to the file. You can see the examples below. All of them are valid.

unlink('deleteme.txt');
unlink('/home/arman/public_html/deleteme.txt');
unlink('./deleteme.txt');
unlink('../deleteme.txt');

PHP will give you a warning if the file you want to remove doesn't exist so you may want to force php to be quiet using the '@' character like this :

@unlink('./deleteme.txt');
Or you can first test if the file exist or not like thigs

$fileToRemove = '/home/arman/public_html/deleteme.txt';
if (file_exists($fileToRemove))) {
// yes the file does exist
unlink($fileToRemove);
} else {
// the file is not found, do something about it???
}

PHP will also give you a warning if you don't have enough permission to delete the file. It would be better if your code check if the file is removed successfully or not like show below

$fileToRemove = '/home/arman/public_html/deleteme.txt';
if (file_exists($fileToRemove)) {
// yes the file does exist

if (@unlink($fileToRemove) === true) {
// the file successfully removed
} else {
// something is wrong, we may not have enough permission
// to delete this file
}
} else {
// the file is not found, do something about it???
}

Joan M30-Mar-11 7:48

PHP Tips By Google Master

30-Mar-11 7:48

You should post those things in the tips/tricks forum...

Or even better you could make an article with a bunch of php tools...

[www.tamelectromecanica.com] Robots, CNC and PLC machines for grinding and polishing.

Haroon Mughal23-Mar-11 2:33

Haroon Mughal

23-Mar-11 2:33

Profile your code to pinpoint bottlenecks
Hoare's dictum states that Premature optimization is the root of all evil, an important thing to keep in mind when trying to make your web sites faster. Before changing your code, you'll need to determine what is causing it to be slow. You may go through this guide, and many others on optimizing PHP, when the issue might instead be database-related or network-related. By profiling your PHP code, you can try to pinpoint bottlenecks.
Upgrade your version of PHP
The team of developers who maintain the PHP engine have made a number of significant performance improvements over the years. If your web server is still running an older version, such as PHP 3 or PHP 4, you may want to investigate upgrading before you try to optimize your code.
Migrating from PHP 4 to PHP 5.0.x
Migrating from PHP 5.0.x to PHP 5.1.x
Migrating from PHP 5.1.x to PHP 5.2.x
Use caching
Making use of a caching module, such as Memcache, or a templating system which supports caching, such as Smarty, can help to improve the performance of your website by caching database results and rendered pages.
Use output buffering
PHP uses a memory buffer to store all of the data that your script tries to print. This buffer can make your pages seem slow, because your users have to wait for the buffer to fill up before it sends them any data. Fortunately, you can make some changes that will force PHP to flush the output buffers sooner, and more often, making your site feel faster to your users.
Output Buffering Control
Avoid writing naive setters and getters
When writing classes in PHP, you can save time and speed up your scripts by working with object properties directly, rather than writing naive setters and getters. In the following example, the dog class uses the setName() and getName() methods for accessing the name property.
class dog {
public $name = '';

public function setName($name) {
$this->name = $name;
}

public function getName() {
return $this->name;
}
}
Notice that setName() and getName() do nothing more than store and return the name property, respectively.
$rover = new dog();
$rover->setName('rover');
echo $rover->getName();
Setting and calling the name property directly can run up to 100% faster, as well as cutting down on development time.
$rover = new dog();
$rover->name = 'rover';
echo $rover->name;
Don't copy variables for no reason
Sometimes PHP novices attempt to make their code "cleaner" by copying predefined variables to variables with shorter names before working with them. What this actually results in is doubled memory consumption (when the variable is altered), and therefore, slow scripts. In the following example, if a user had inserted 512KB worth of characters into a textarea field. This implementation would result in nearly 1MB of memory being used.
$description = strip_tags($_POST['description']);
echo $description;
There's no reason to copy the variable above. You can simply do this operation inline and avoid the extra memory consumption:
echo strip_tags($_POST['description']);
Avoid doing SQL queries within a loop
A common mistake is placing a SQL query inside of a loop. This results in multiple round trips to the database, and significantly slower scripts. In the example below, you can change the loop to build a single SQL query and insert all of your users at once.
foreach ($userList as $user) {
$query = 'INSERT INTO users (first_name,last_name) VALUES("' . $user['first_name'] . '", "' . $user['last_name'] . '")';
mysql_query($query);
}
Produces:
INSERT INTO users (first_name,last_name) VALUES("John", "Doe")
Instead of using a loop, you can combine the data into a single database query.
$userData = array();
foreach ($userList as $user) {
$userData[] = '("' . $user['first_name'] . '", "' . $user['last_name'] . '")';
}
$query = 'INSERT INTO users (first_name,last_name) VALUES' . implode(',', $userData);
mysql_query($query);
Produces:
INSERT INTO users (first_name,last_name) VALUES("John", "Doe"),("Jane", "Doe")...

PHP performance tips

Haroon Mughal23-Mar-11 2:32

Haroon Mughal

23-Mar-11 2:32

Author: Eric Higgins, Google Webmaster
Recommended experience: Beginner to intermediate PHP knowledge
PHP is a very popular scripting language, used on many popular sites across the web. In this article, we hope to help you to improve the performance of your PHP scripts with some changes that you can make very quickly and painlessly. Please keep in mind that your own performance gains may vary greatly, depending on which version of PHP you are running, your web server environment, and the complexity of your code.

css and link styles

Joan M22-Mar-11 4:59

Why do images are not shown?

22-Mar-11 4:59

Hello all,

Trying to get css working here...

Let's say I have a php file.

In that file I have something like:

if ($lang == "AAA"){
  echo('<a href="'.$linkname.'" class="linkidiomaseleccionat">AAAA</a> | ');
}
else {
  echo('<a href="'.$linkname.'" class="linkidioma">AAAA</a> | ');
}

Let's say the CSS file has this inside:

A.linkidiomaseleccionat, A.linkidiomaseleccionat:VISITED, A.linkidiomaseleccionat:ACTIVE, A.linkidiomaseleccionat:FOCUS, A.linkidiomaseleccionat:LINK{
  color: #ff6d00;
  text-decoration: none;
}

A.linkidiomaseleccionat:HOVER{
  color: #ff6d00;
  text-decoration: underline;
}

A.linkidioma, A.linkidioma:VISITED, A.linkidioma:ACTIVE, A.linkidioma:FOCUS, A.linkidioma:LINK{
  color: #FFFFFF;
  text-decoration: none;
}

A.linkidioma:HOVER{
  color: #ff6d00;
  text-decoration: underline;
}

A.linkNF, A.linkNF:VISITED, A.linkNF:ACTIVE, A.linkNF:FOCUS, A.linkNF:LINK{
  color: #ff6d0N;
  text-decoration: none;
}
A.linkNF:HOVER{
  color: #ff6d0N;
  text-decoration: underline;
}

A.linkNU, A.linkNU:VISITED, A.linkNU:ACTIVE, A.linkNU:FOCUS, A.linkNU:LINK{
  color: #FFFFFF;
  text-decoration: none;
}
A.linkNU:HOVER{
  color: #ff6d0N;
  text-decoration: underline;
}

Why on earth if I change the text that assigns the class to my links in the php file I don't get a good result if I'm not using "linkidiomaseleccionat" and "linkidioma"... I've tried to change the styule names to get something more consistent (bad initial design) and now I can't see it working if I change the name for it's counterparts: "linkNF" and "linkNU"...

Any of you could help me with that?

I'm feeling really bad here... I can't get it and I don't understand why changing only the name of the style class prevents it from working properly... Dead | X|

We... thank you in advance!

[www.tamelectromecanica.com] Robots, CNC and PLC machines for grinding and polishing.

Re: css and link styles

Luc Pattyn22-Mar-11 5:25

Luc Pattyn

22-Mar-11 5:25

Hi Joan,

three comments:

1.
I don't really understand your question. It isn't clear to me what you want, what you get, and where they differ.

2.
you may suffer from cache effects; try CTRL/F5 to force the browser to reload the page and its dependencies (the CSS files).

3.
you probably could simplify by assigning more than one class to those anchor tags, as in:

echo('<a href="'.$linkname.'" class="linkidiomaseleccionat mySpecialColorScheme">AAAA</a> | ');

Luc Pattyn [Forum Guidelines] [My Articles] Nil Volentibus Arduum

Please use <PRE> tags for code snippets, they preserve indentation, improve readability, and make me actually look at the code.

Joan M21-Mar-11 5:43

Re: Why do images are not shown?

21-Mar-11 5:43

Hello all again,

Given that code:

<div id="IndexHeader">
  <?php 
    echo('<div id="logoIndex"><img src="'.$path_images.'/logo.png" alt="alt text"></div>');
    include($path_base.'/menu_idioma.php'); 
  ?>
</div>

If I'm echoing the path, copying it to the clipboard and then paste it into the explorer address bar, it seems right enough in order to allow my windows system to load the right image...

So... any idea on why is this happening? I'm really confused... those things should be extremely easy and they don't seem so... Dead | X|

As always thank you in advance...

[www.tamelectromecanica.com] Robots, CNC and PLC machines for grinding and polishing.

Luc Pattyn21-Mar-11 5:50

Luc Pattyn

21-Mar-11 5:50

would the image show if you hadn't the include statement?
it smells like file-not-found, possibly caused by your earlier include-changes-directory issue.

ADDED: does any of your code contain chdir()?

Smile | :)

Luc Pattyn [Forum Guidelines] [My Articles] Nil Volentibus Arduum

Please use <PRE> tags for code snippets, they preserve indentation, improve readability, and make me actually look at the code.

Re: Why do images are not shown?

Joan M21-Mar-11 6:10