kals
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Posts posted by kals
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try this it just may work:
change:
[code]name="ismember" value=" <?php if($ismember == "1"){echo "1";} else{echo "0";}?> [/code]
to:
[code]name="ismember" value=" <?php if($ismember == "1"){echo "on";} else{echo "off";}?> [/code]
change:
[code]$ismember = $_POST['ismember'];[/code]
to:
[code]if($_POST['ismember']=="on"){
$ismember=1;}
else{$ismember=0;}[/code]
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I encountered this problem many times and its ussually because a qoute " or a } is missing what I normally do is move blocks of code to a temprory file until the error dissapears and then check the last block I moved.
such a problem can't be found without seeing entire code (also try looking at the code in some php editing software that shows colour coded php tags )
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session_start() should actually work ok where it is as it is bieng called before outputting anything to the browser
what exactly do you mean by "doesn't work at all"?
Do you get invalid password error, warnings, a blank page or something else?
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Hi I think this is the problem
change:
[code] Is Member: <input type="checkbox" name="ismember" <?php if($ismember == "1"){echo " CHECKED";}?>>
</tr>[/code]
to:
[code] Is Member: <input type="checkbox" name="ismember" value=" <?php if($ismember == "1"){echo "1";}
else{echo "0";}?> ">
</tr> [/code]
the way it currently is although it is checked ismember has no value assigned to it and so is reverted to default value of 0.
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where do you get $LOGIN_METHOD from?
make sure it is defined and value is correct
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anyone else?
there has to be a way so many people claim to do it. -
Hi,
change this:
$id=$_POST['id'];
to:
$id=$_GET['id'];
I havent gone through your entire code but that was the first thing I noticed
POST works when submitting values from forms while GET works using urls.
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remove the following from the form:
<input name="id" type="hidden">
remove the following from the added page:
$id=$_POST['id'];
change the following:
$query = "INSERT INTO users (id, firstname, surname, email, password) VALUES('$id', '$firstname', '$surname' , '$email' , '$password' )";
to:
$query = "INSERT INTO users (firstname, surname, email, password) VALUES('$firstname', '$surname' , '$email' , '$password' )";
echo $query. "<br/>";
$result = mysql_query($query) or die("Error: ". mysql_error());
to view the id:
$id=mysql_insert_id($result);
echo $id;
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Is there some way to find out if an email has been read or not when sending emails
I mean other than the "Disposition-Notification-To:" header
I have seen many sites/programs offering over 90% accuraccy in finding if email has been read or not also many sites state that they can do so without alerting the reciepient of the email. Does anyone know how this can be done?
Thanks in advance
need help urgent cURL php
in PHP Coding Help
Posted
the error can be seen here:
http://sourcefreelance.com/godaddy/replace.php
the code is below:
[code]<?
$url="https://idp.godaddy.com/login.aspx?se=%2B&spkey=GDSWEB91&login=&target=secure%5Ftransfer%2Easp";
$vars="PasswordTextBox=pass&UsernameTextBox=user&__EVENTTARGET&__EVENTARGUMENT&__VIEWSTATE=dDw3MDI1OTkxOTI7dDw7bDxpPDA+Oz47bDx0PDtsPGk8MT47PjtsPHQ8O2w8aTw1PjtpPDg+Oz47bDx0PHA8cDxsPEltYWdlVXJsOz47bDxodHRwczovL2ltYWdlcy5nb2RhZGR5LmNvbS9zc28vYnV0dG9ucy8xL2J1dF9zZWN1cmVsb2dpbi5naWY7Pj47Pjs7Pjt0PHA8cDxsPFRleHQ7PjtsPFByb3RlY3RlZCBieSBhIEdvIERhZGR5IFNTTCBDZXJ0aWZpY2F0ZS4gU2VjdXJlIGFuZCAxMjgtYml0IGVuY3J5cHRlZCE7Pj47Pjs7Pjs+Pjs+Pjs+PjtsPExvZ2luSW1hZ2VCdXR0b247Pj6MHjUFoXuRslkGzQ7erW+h3AK/hQ==";
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_HEADER, 1);
curl_setopt($ch, CURLOPT_USERAGENT, $_SERVER['HTTP_USER_AGENT']);
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_COOKIEJAR, 'cookie.txt');
curl_setopt($ch, CURLOPT_COOKIEFILE, 'cookie.txt');
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS, $vars);
$data = curl_exec($ch);
curl_close($ch);
print $data;
?>
[/code]
Any help highly appreciated, can also be paid (first person to solve problem)