[RESOLVED] Unexpected T_IF error ?

chrisguk

Hi,

From the code below, hopefully you can see what I am trying to achieve.

Basically I want the result printed where I choose to put the result anywhere on the page.

I keep getting the unexpected T_IF error on line 39, where am I going wrong here?

<html>
<head></head>
<body>
<?php 

session_start();

$self = $_SERVER['PHP_SELF'];


$localhost = 'localhost';
$dbuser = 'root';
$dbpass = 'password';
$dbname = 'north';

$connect = mysql_connect($localhost, $dbuser, $dbpass);

mysql_select_db($dbname, $connect);
if(isset($_POST['submit']))
{
$id = $_POST['id'];
$title = $_POST['title'];
$description = $_POST['description'];
$keywords = $_POST['keywords'];
$_SESSION['id']=$id;
$_SESSION['title']=$title;
$_SESSION['description']=$description;
$_SESSION['keywords']=$keywords;

$query = "INSERT INTO URL VALUES ('','$id','$title','$description','$keywords')";
mysql_query($query);

$success = 'You have successfully updated your data';

} 

$form.="<div style=\"border: 1px solid;padding: 10px 0 25px 0;\">";
$form.="<p>Database result:" . if(isset($success)) { print $success; }."</p>";
$form.="</div>";
$form.="<form action=\"$self\" method=\"post\" style=\"border: 1px solid;\" >";
$form.="<legend>";
$form.="<label>Page Name:</label> <input type=\"text\" name=\"id\"><br>";
$form.="<label>Title of Page:</label> <input type=\"text\ name=\"title\"><br>";
$form.="<label>Description: </label><input type=\"text\" name=\"description\"><br>";
$form.="<label>Keywords:</label> <input type=\"text\" name=\"keywords\"><br>";
$form.="<input type=\"submit\" name=\"submit\">";
$form.="</legend>";
$form.="</form>";

print $form;
?>


</body>
</html>

Derokorian

This line:

$form.="<p>Database result:" . if(isset($success)) { print $success; }."</p>";

Is wrong. If you want to do inline like that use ternary, other wise make it its own line:

$form.="<p>Database result:";
if(isset($success)) {
   $form .= $success; 
}
$form .= "</p>"; 

// OR
$form.="<p>Database result:" . ( isset($success) ? $success : '' ) ."</p>";

chrisguk

Now I am getting:

Notice: Undefined variable: form in /var/www/test/index.php on line 37

$query = "INSERT INTO URL VALUES ('','$id','$title','$description','$keywords')";
mysql_query($query);

$success = 'You have successfully updated your data';

}

LINE 37 = $form.="<div style=\"border: 1px solid;padding: 10px 0 25px 0;\">";
$form.="<p>Database result:";
if(isset($success)) {
   $form .= $success;
}
$form.= "</p>"; 
$form.="</div>";
$form.="<form action=\"$self\" method=\"post\" style=\"border: 1px solid;\" >";
$form.="<legend>";
$form.="<label>Page Name:</label> <input type=\"text\" name=\"id\"><br>";
$form.="<label>Title of Page:</label> <input type=\"text\ name=\"title\"><br>";
$form.="<label>Description: </label><input type=\"text\" name=\"description\"><br>";
$form.="<label>Keywords:</label> <input type=\"text\" name=\"keywords\"><br>";
$form.="<input type=\"submit\" name=\"submit\">";
$form.="</legend>";
$form.="</form>";

Derokorian

Where do you ever initialize $form before you try to concatenate onto it? (Hint: you don't, so before that line, it is undefined)

chrisguk

Derokorian;11002548 wrote:
Where do you ever initialize $form before you try to concatenate onto it? (Hint: you don't, so before that line, it is undefined)

Sorry Derokorian, I dont understand what you mean?

vivek151189

you have not define $form variable before line 37 and you are using
$form.="<div style=\"border: 1px solid;padding: 10px 0 25px 0;\">";

instead use

$form="<div style=\"border: 1px solid;padding: 10px 0 25px 0;\">";

Ashley_Sheridan

You're trying to do short string concatenation with .= but $form isn't declared the first time you use it. The first time you use $form, you should just use = and not .=

chrisguk

Ashley Sheridan;11002552 wrote:
You're trying to do short string concatenation with .= but $form isn't declared the first time you use it. The first time you use $form, you should just use = and not .=

Yes of course, my bad.

Thanks Ashley

chrisguk

One last thing.

The script seems to be looking at this line again once the form is posted:

$title = $_POST['title'];

and stating it is undefined. I assume this is because the POST has already processed and now it doesnt have a string value?

Ashley_Sheridan

How do you mean already processed? What your script is trying to tell you is that $POST['title'] doesn't exist. Do a var_dump($POST); to see what actually exists in your $_POST array, because what you think is in there may not be at all.

chrisguk

Ashley Sheridan;11002557 wrote:
How do you mean already processed? What your script is trying to tell you is that $POST['title'] doesn't exist. Do a var_dump($POST); to see what actually exists in your $_POST array, because what you think is in there may not be at all.

Ye I can see its missing the title altogether, trying to figure out why 😕

array(4) { ["id"]=> string(6) "3erere" ["description"]=> string(4) "Test" ["keywords"]=> string(4) "erer" ["submit"]=> string(12) "Submit Query" }

Derokorian

Can you show us your code as it stands now after the changes? Also have you made sure you're refreshing the form and sending a new request (aka not refreshing an old request?)

chrisguk

All fixed thanks for your assistance guys!!!🙂🙂🙂

It was a type on the form input box

Ashley_Sheridan

What does the form HTML look like once it's been output by your PHP script?

One aside, I notice in your above code that your <label> tags aren't attached to your actual <input> tags in any way. You should either wrap the labels around the inputs, or associated them by id with a for attribute on the label.