Parse error: syntax error, unexpected $end in

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Lid geworden
11 jul 2007
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138
ik weet eigenlijk niet wat deze error betekend... ik heb toch nergens $end gedeclareerd?

Parse error: syntax error, unexpected $end in script.php on line 94

PHP: 
<body>
<?php 
	if ($_POST['submit_check']){
	$username = $_POST['username'];
	$password = $_POST['password'];
	$database = $_POST['database'];
	Proc_Form_login ($username,$password,$database);
		if ($_POST['table']){
		$table      = $_POST['table'];
		$ref_levels = $_POST['ref_levels'];
		$ref_comm   = $_POST['ref_comm'];
		$ref_url	= $_POST['ref_url'];
		$name		= $_POST['name'];
		$info_url   = $_POST['info_url'];
		$paymeth 	= $_POST['paymeth'];
		$paymin		= $_POST['paymin'];
		
		Proc_Form_mysql ($table,$ref_levels,$ref_comm,$ref_url,$name,$info_url,$paymeth,$paymin);
		echo "proccesed and added data <br>";
		}
		else { 
		Show_Form_mysql ();
		}
		
	}
	else {
	Show_Form_login ();
	}
function Proc_Form_login ($username,$password,$database)
{
	mysql_connect("$database", "$username", "$password") or die(mysql_error());
	echo "Connected to MySQL<br />";
		
	mysql_select_db("$database") or die(mysql_error());
	echo "Connected to Database<br>";
}
function Proc_Form_mysql ($table,$ref_levels,$ref_comm,$ref_url,$name,$info_url,$paymeth,$paymin)
{
	echo "ready to insert data into: $table<br>";
	
    mysql_query("INSERT INTO $table
    (name, url, refcom, reflev,paymin,paymeth,webinf )
    VALUES
    ('$name','$url','$refcom','$reflev','$paymin','$paymeth','$webinf')") or die(mysql_error()); 
} 
function Show_Form_login ()
{
	print<<<_HTML_
	<form method="post" action="<?php echo $PHP_SELF;?>">
    Username : <input type="text" size="12" maxlength="30" name="username"> <br >
    Password : <input type="password" size="12" maxlength="30" name="password"> <br >
    Database : <input type="text" size="12" maxlength="30" name="database"><br>
	<input type="hidden" name="submit_check" value="1">
    <input type="submit" value="submit" name="submit"><br />
    </form> 
	
	_HTML_;
}
function Show_Form_mysql ()
{
	print<<<_HTML_
	
	<form method="post" action="<?php echo $PHP_SELF;?>">
     table : <input type="text" size="30" maxlength="30" name="table"> <br >
     website name : <input type="text" size="30" maxlength="30" name="name"> <br >
     website url : <input type="text" size="30" maxlength="60" name="ref_url"><br>
     referral levels : <input type="text" size="30" maxlength="3" name="ref_levels"> <br >
     referral commission : <input type="text" size="30" maxlength="30" name="ref_comm"> <br >
     payout minimum : <input type="text" size="30" maxlength="3" name="paymin"><br>
     payout methods : <input type="text" size="30" maxlength="30" name="paymeth"> <br >
     website info url : <input type="text" size="30" maxlength="45" name="info_url"><br>
     <input type="submit" value="submit" name="submit"><br />
     </form> 
	 
	 _HTML_;
	 
}
	

	
?>
	

	
</body>
</html>
 
PHP: 
function Show_Form_login ()
{
    print <<<_HTML_
    <form method="post" action="<?php echo $PHP_SELF;?>">
    Username : <input type="text" size="12" maxlength="30" name="username"> <br >
    Password : <input type="password" size="12" maxlength="30" name="password"> <br >
    Database : <input type="text" size="12" maxlength="30" name="database"><br>
    <input type="hidden" name="submit_check" value="1">
    <input type="submit" value="submit" name="submit"><br />
    </form>
    
    _HTML_;
}

ik weet niet wat je met dit stukje wil doen wat hierboven staat. maar in ieder geval sluit hij de tekst hier niet af... dus neemt hij vervolgens alles tot eind van je pagina mee in deze print. De fout zit dus in dit stukje.

ik gok dat je dit wil hebben:
PHP: 
function Show_Form_login ()
{
    print '
    <form method="post" action="<?php echo $PHP_SELF;?>">
    Username : <input type="text" size="12" maxlength="30" name="username"> <br >
    Password : <input type="password" size="12" maxlength="30" name="password"> <br >
    Database : <input type="text" size="12" maxlength="30" name="database"><br>
    <input type="hidden" name="submit_check" value="1">
    <input type="submit" value="submit" name="submit"><br />
    </form>';
    
}
 
Laatst bewerkt:
bedankt! er ging wel nog iets mis met de php server[self] maar ik heb het opgelost, ik heb nu een nieuw probleem, de functie ad_data word overgeslagen ik weet wel hoe het komt, maar wat kan ik eraan doen??

PHP: 
<body>
<?php 
function Proc_Form_login ($username,$password,$database)
{
	mysql_connect("localhost", "$username", "$password") or die(mysql_error());
	echo "Connected to MySQL<br />";
		
	mysql_select_db("$database") or die(mysql_error());
	echo "Connected to Database<br>";
}
function Proc_Form_mysql ($table,$ref_levels,$ref_comm,$ref_url,$name,$info_url,$paymeth,$paymin)
{
	echo "ready to insert data into: $table<br>";
	
    mysql_query("INSERT INTO $table
    (name, url, refcom, reflev,paymin,paymeth,webinf )
    VALUES
    ('$name','$url','$refcom','$reflev','$paymin','$paymeth','$webinf')") or die(mysql_error()); 
}
?>
<?php

function Show_Form_login ()
{
	?>
    <form method="post" action="<?php echo $PHP_SELF;?>">
    Username : <input type="text" size="12" maxlength="30" name="username"> <br >
    Password : <input type="password" size="12" maxlength="30" name="password"> <br >
    Database : <input type="text" size="12" maxlength="30" name="database"><br>
    <input type="hidden" name="submit_check" value="1">
    <input type="submit" value="submit" name="submit"><br />
    </form>
<?php    
}
function Show_Form_mysql ()
{
	?>
	
	<form method="post" action="<?php echo $PHP_SELF;?>">
     table : <input type="text" size="30" maxlength="30" name="table"> <br >
     website name : <input type="text" size="30" maxlength="30" name="name"> <br >
     website url : <input type="text" size="30" maxlength="60" name="ref_url"><br>
     referral levels : <input type="text" size="30" maxlength="3" name="ref_levels"> <br >
     referral commission : <input type="text" size="30" maxlength="30" name="ref_comm"> <br >
     payout minimum : <input type="text" size="30" maxlength="3" name="paymin"><br>
     payout methods : <input type="text" size="30" maxlength="30" name="paymeth"> <br >
     website info url : <input type="text" size="30" maxlength="45" name="info_url"><br>
     <input type="submit" value="submit" name="submit"><br />
     </form>
	 
	 
<?php	 
}
	

	if ($_POST['submit_check']){
	$username = $_POST['username'];
	$password = $_POST['password'];
	$database = $_POST['database'];
	echo "$username , $password , $database <br>";
	Proc_Form_login ($username,$password,$database);
		if ($_POST['table']){
		$table      = $_POST['table'];
		$ref_levels = $_POST['ref_levels'];
		$ref_comm   = $_POST['ref_comm'];
		$ref_url	= $_POST['ref_url'];
		$name		= $_POST['name'];
		$info_url   = $_POST['info_url'];
		$paymeth 	= $_POST['paymeth'];
		$paymin		= $_POST['paymin'];
		
		Proc_Form_mysql ($table,$ref_levels,$ref_comm,$ref_url,$name,$info_url,$paymeth,$paymin);
		echo "proccesed and added data <br>";
		}
		else { 
		Show_Form_mysql ();
		}
		
	}
	else {
	Show_Form_login ();
	}


	
?>
	

	
</body>
</html>
 
wat bedoel je met de functie ad_data ??? hehe heb geprobeerd te vinden wat je bedoelde
 
ja ****, ik had me vergist :) t moet deze zijn:

Proc_Form_mysql ($table,$ref_levels,$ref_comm,$ref_url,$name,$info_url,$paymeth,$paymin);

die word overgeslagen!
 
je bent vergeten om in de 2e post form nog een veld te zetten met submit_check hierdoor gaat hij bij de 2e post altijd naar de else, omdat hij geen submit_check gepost krijgt. zou je dit veld erbij inzetten of je geeft iets anders mee dan werkt hij.

2 mogelijkheden:

PHP: 
function Show_Form_mysql () 
{ 
    ?> 
     
    <form method="post" action="<?php echo $PHP_SELF;?>"> 
     table : <input type="text" size="30" maxlength="30" name="table"> <br > 
     website name : <input type="text" size="30" maxlength="30" name="name"> <br > 
     website url : <input type="text" size="30" maxlength="60" name="ref_url"><br> 
     referral levels : <input type="text" size="30" maxlength="3" name="ref_levels"> <br > 
     referral commission : <input type="text" size="30" maxlength="30" name="ref_comm"> <br > 
     payout minimum : <input type="text" size="30" maxlength="3" name="paymin"><br> 
     payout methods : <input type="text" size="30" maxlength="30" name="paymeth"> <br > 
     website info url : <input type="text" size="30" maxlength="45" name="info_url"><br> 
     <input type="hidden" name="submit_check" value="1"> 
     <input type="submit" value="submit" name="submit"><br /> 
     </form> 
     
     
<?php     
} 


//2e mogelijkheid
 if ($_POST['submit_check'] OR andere waarde){
 
ja, maar dan is het die submit check van de eerste 'post' , en dan krijg ik het form voor de data op te slaan in de database niet te zien!

ik heb het dus "submit_check2" van gemaakt.. maar dan springt ie na mijn form te hebben ingevult weer naar Show form Login, inplaats van eerst die else uit te voeren.. ik weet wel waarom het fout is, maar wat kan ik eraan doen? ik heb al geprobeerd om een while loop rond dat stukje te bouwen, maar het wil maar niet lukken :(

en een Goto command heb je precies niet in php ! :(


zo zie je het makkelijker:
PHP: 
als ( waarde_1  gepost )
{
	voer uit data;
	als (waarde_2 gepost)
	{
		voer uit data2;
	}
	anders 
	{
		print form2; // nadat dit gebeurt is springt ie trug naar de eerste IF! 

	}
}
anders {
	print form;
}
 
Laatst bewerkt:
zou je misschien het stukje broncode kunnen plakken dat je bedoeld?? ik volg het niet helemaal.... ik probeer nu op mijn eigen server voor je uit. Of zet de hele broncode nogmaals neer
 
Dit heb ik ervan gemaakt tot nu toe je mag het proberen.. werkt het niet dan wil ik iets betere uitleg..... werkt het dan moet je misschien nog ergens unset($_SESSION['form1_ingevuld'] = true;). ik hoor het wel



PHP: 
<?php 

session_start();

function Proc_Form_login ($username,$password,$database) 
{ 
    mysql_connect("localhost", "$username", "$password") or die(mysql_error()); 
    echo "Connected to MySQL<br />"; 
         
    mysql_select_db("$database") or die(mysql_error()); 
    echo "Connected to Database<br>"; 
    
    //$_SESSION['username'] = $username; 
   	//$_SESSION['password'] = $password; 
    //$_SESSION['datbabase'] = $database; 
    $_SESSION['form1_ingevuld'] = true;
} 

function Proc_Form_mysql ($table,$ref_levels,$ref_comm,$ref_url,$name,$info_url,$paymeth,$paymin) 
{ 
    echo "ready to insert data into: $table<br>"; 
     
    mysql_query("INSERT INTO $table 
    (name, url, refcom, reflev,paymin,paymeth,webinf ) 
    VALUES 
    ('$name','$url','$refcom','$reflev','$paymin','$paymeth','$webinf')") or die(mysql_error()); 
} 
?> 
<?php 

function Show_Form_login () 
{ 
    ?> 
    <form method="post" action="<?php echo $PHP_SELF;?>"> 
    Username : <input type="text" size="12" maxlength="30" name="username"> <br > 
    Password : <input type="password" size="12" maxlength="30" name="password"> <br > 
    Database : <input type="text" size="12" maxlength="30" name="database"><br> 
    <input type="hidden" name="submit_check" value="1"> 
    <input type="submit" value="submit" name="submit"><br /> 
    </form> 
<?php     
} 
function Show_Form_mysql () 
{ 
    ?> 
     
    <form method="post" action="<?php echo $PHP_SELF;?>"> 
     table : <input type="text" size="30" maxlength="30" name="table"> <br > 
     website name : <input type="text" size="30" maxlength="30" name="name"> <br > 
     website url : <input type="text" size="30" maxlength="60" name="ref_url"><br> 
     referral levels : <input type="text" size="30" maxlength="3" name="ref_levels"> <br > 
     referral commission : <input type="text" size="30" maxlength="30" name="ref_comm"> <br > 
     payout minimum : <input type="text" size="30" maxlength="3" name="paymin"><br> 
     payout methods : <input type="text" size="30" maxlength="30" name="paymeth"> <br > 
     website info url : <input type="text" size="30" maxlength="45" name="info_url"><br> 
     <input type="hidden" name="submit_check" value="1"> 
     <input type="submit" value="submit" name="submit"><br /> 
     </form> 
     
     
<?php     
} 
     
if ($_POST['submit_check2']){ 
        $table      = $_POST['table']; 
        $ref_levels = $_POST['ref_levels']; 
        $ref_comm   = $_POST['ref_comm']; 
        $ref_url    = $_POST['ref_url']; 
        $name        = $_POST['name']; 
        $info_url   = $_POST['info_url']; 
        $paymeth     = $_POST['paymeth']; 
        $paymin        = $_POST['paymin']; 
         
        //Proc_Form_mysql ($table,$ref_levels,$ref_comm,$ref_url,$name,$info_url,$paymeth,$paymin); 
        echo "proccesed and added data <br>"; 
        } 

    elseif ($_POST['submit_check'] OR $_SESSION['form1_ingevuld']){ 
    
    if($_POST['submit_check'])
    {
    	$username = $_POST['username']; 
    	$password = $_POST['password']; 
    	$database = $_POST['database'];
    	
    	echo "$username , $password , $database <br>"; 
    	Proc_Form_login ($username,$password,$database); 
    }

elseif($_SESSION['form1_ingevuld'])
{
 $username=$_SESSION['username']; 
$password=$_SESSION['password'] ; 
 $database =  $_SESSION['datbabase'];

Proc_Form_login ($username,$password,$database);
}

        
        else { 
        Show_Form_mysql (); 
        } 
         
    } 
    else { 
    Show_Form_login (); 
    } 


     
?>
 
ik ga het je morgen laten weten! :P ik heb vd geen tijd meer :)
 
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