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Johng123

By Johng123
in PHP Coding Help

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Johng123 Newbie

Johng123

Posted
Posted

Okay so I do not really know php that well and I am trying to write a simple website code in it but I am running into a problem.  I have a form and I want it to add to the end of a list.  I have a text file called codes.txt on the server and idk why this won't work.

 

<html>
<body>

Your friend code <?php echo $_POST['fcode']; 
$myFile = "codes.txt";
echo $myFile;
$fh = fopen($myfile,'a+') or die("can't open file");
fwrite($fh , $fcode);
?>
has been added to the list

</body>
</html>

 

 

site here: http://invitemasteronline.t35.com/test.html

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crabfinger Newbie

crabfinger

Posted
Posted

i believe this is along the lines of what you want. But I would suggest imposing some security restrictions on who can send the form and what data can be sent, you dont want just anybody posting huge amounts of unregulated or even malicious data to a file on your server.

 

<?php
function show_form() {
?>
<form action="" method="post">
	Friend Code: <input type="text" name="fcode" />
	<input type="submit" value="submit" />
</form>
<?php
}
$friend_file = 'friends.txt';
$friend_content = file_get_contents($friend_file);
if(isset($_POST['fcode'])) {
	$friend_code = $_POST[fcode];
	$friend_content = $friend_content . "\r\n" . $friend_code;
	if(file_put_contents($friend_file,$friend_content)) {
		print $friend_code . ' has been added to the friends file.<hr />';
	}
}
show_form();
?>

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Johng123 Newbie

Johng123

Posted
  • Author
Posted

i believe this is along the lines of what you want. But I would suggest imposing some security restrictions on who can send the form and what data can be sent, you dont want just anybody posting huge amounts of unregulated or even malicious data to a file on your server.

 

<?php
function show_form() {
?>
<form action="" method="post">
	Friend Code: <input type="text" name="fcode" />
	<input type="submit" value="submit" />
</form>
<?php
}
$friend_file = 'friends.txt';
$friend_content = file_get_contents($friend_file);
if(isset($_POST['fcode'])) {
	$friend_code = $_POST[fcode];
	$friend_content = $friend_content . "\r\n" . $friend_code;
	if(file_put_contents($friend_file,$friend_content)) {
		print $friend_code . ' has been added to the friends file.<hr />';
	}
}
show_form();
?>

thanks but I don't really wanna change the whole thing, cause then I don't understand what's happening at all.  I just wanna figure out why mine isn't working

Whats $fcode?

$fcode=$_POST['fcode']

 

 

HTH

Teamatomic

sorry, that was the input from a form.

Also its not even reading the file still I think.

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teamatomic Newbie

teamatomic

Posted
Posted

Why dont you give the full path the the file. if its in the same folder as the script then ./file.txt otherwise /path/to/file.txt. And if you have control of the server and the code is not for distribution and you have php v5+ then use

file_put_contents("$myFile", "$fcode", FILE_APPEND, LOCK_EX);

 

who owns the file? what are the perms?

 

so what if its form output. To use $fcode you need to give it a value

 

also...show us the error you get..."it's broken" is not really good enough

 

HTH

Teamatomic

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ignace Advanced Member

ignace

Posted
Posted

@crabfinger don't just write some code without thinking about it always make sure that heavy operations (or soon to be heavy) are lazy-loaded

 

$friend_file = 'friends.txt';
$friend_content = file_get_contents($friend_file);
if(isset($_POST['fcode'])) {
    $friend_code = $_POST[fcode];
    $friend_content = $friend_content . "\r\n" . $friend_code;
    if(file_put_contents($friend_file,$friend_content)) {
        print $friend_code . ' has been added to the friends file.<hr />';
    }
}

 

In the above code is the file (friends.txt) data read but not used until the form is submitted.

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crabfinger Newbie

crabfinger

Posted
Posted

@crabfinger don't just write some code without thinking about it always make sure that heavy operations (or soon to be heavy) are lazy-loaded

 

$friend_file = 'friends.txt';
$friend_content = file_get_contents($friend_file);
if(isset($_POST['fcode'])) {
    $friend_code = $_POST[fcode];
    $friend_content = $friend_content . "\r\n" . $friend_code;
    if(file_put_contents($friend_file,$friend_content)) {
        print $friend_code . ' has been added to the friends file.<hr />';
    }
}

 

In the above code is the file (friends.txt) data read but not used until the form is submitted.

 

Sorry when i wrote that i was going to add

 

print '<hr />' . $friend_content;

 

to the end of the file for debugging purposes

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crabfinger Newbie

crabfinger

Posted
Posted

How does that put 

$friend_content = file_get_contents($friend_file);

in the correct place?

 

$friend_file = 'friends.txt';
$friend_content = file_get_contents($friend_file);
if(isset($_POST['fcode'])) {
    $friend_code = $_POST[fcode];
    $friend_content = $friend_content . "\r\n" . $friend_code;
    if(file_put_contents($friend_file,$friend_content)) {
        print $friend_code . ' has been added to the friends file.<hr />';
    }
}
print '<hr />' . $friend_content;

 

means that i only have to load the file once because it is being used wether or not the form is posted.

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Johng123 Newbie

Johng123

Posted
  • Author
Posted

@crabfinger don't just write some code without thinking about it always make sure that heavy operations (or soon to be heavy) are lazy-loaded

 

$friend_file = 'friends.txt';
$friend_content = file_get_contents($friend_file);
if(isset($_POST['fcode'])) {
    $friend_code = $_POST[fcode];
    $friend_content = $friend_content . "\r\n" . $friend_code;
    if(file_put_contents($friend_file,$friend_content)) {
        print $friend_code . ' has been added to the friends file.<hr />';
    }
}

 

In the above code is the file (friends.txt) data read but not used until the form is submitted.

 

ok thank you soo much that worked great!! Now I am really confused though I tried to change the code, so it would check if the user has tried to submit a code before.  It does not work!

<html>
<body>
<?php
$friend_file = 'friends.txt';
$friend_content = file_get_contents($friend_file);
if(isset($_POST['fcode'])) {
    $pos = strpos($friend_content, ($_POST['fcode']);
        if ($pos === false) {
           $friend_code = $_POST[fcode];
           $friend_content = $friend_content . "\r\n" . $friend_code;
           if(file_put_contents($friend_file,$friend_content)) {
        print $friend_code . ' has been added to the friends file.<hr />';
        } }
        if ($pos === !false){
     print $friend_code . ' is already on the list!<hr />';
    }
}
?>

</body>
</html>

 

 

error is: "Parse error: syntax error, unexpected ';' in /home/freehost/t35.com/i/n/invitemasteronline/addcode.php on line 7"

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Johng123 Newbie

Johng123

Posted
  • Author
Posted

Change 

   $pos = strpos($friend_content, ($_POST['fcode']);

to: 

   $pos = strpos($friend_content, $_POST['fcode']);

There seems to have been an error when the code was written, that is line 7.

 

ahh thanks now it works perfectly!! thanks so much every one!!

 

I will probably have more questions down the road but for now i'm good.  I <3 u all :D

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