Having a problem with my first PHP script

Freedom-n-Democrazy · September 7, 2011

Previously, the PHP codebase I was using was that written by others, and I used their code to tinker with so I could get an understanding how PHP works.

Now that I am comfortable using PHP, I have written my own short PHP script from scratch with the aim of it seeing my newsletter system functioning at minimal, for the first time.

The newsletters system is made with HTML/PHP/MySQL.

I am having a problem with the PHP side of things where I am getting an error:

PHP Parse error: syntax error, unexpected T_IF on line 50

... which is:

if $_POST['action'] == 'Register' {

I am hoping someone that knows allot about PHP could take a look at my code and see whats wrong with it?

PHP code within confirm.html:

<?php
$link = mysql_connect('localhost', 'testusr', 'testpw');
mysql_select_db('testdb', $link);
$email = $_POST['e-mail'];
$query = 
	if $_POST['action'] == 'Register' {
		if $_POST['newsletter'] == 'Mens' {
			"INSERT INTO newsletters(mens) VALUES('$email')";
		}
		elseif $_POST['newsletter'] == 'Mens & Womens' {
			"INSERT INTO newsletters(mensandwomens) VALUES('$email')";
		}
		elseif $_POST['newsletter'] == 'Womens' {
				"INSERT INTO newsletters(womens) VALUES('$email')";
		}
	;}
mysql_query ($link, $query);
mysql_close($link);
?>

HTML FORM code within index.html:

<FORM action="confirm.html" method="post">
<DIV>
	<SPAN class="input">
		Action:	
			<SELECT name="action">
				<OPTION>Register</OPTION>
				<OPTION>Unregister</OPTION>
			</SELECT>&nbsp&nbsp&nbsp
		E-mail:	<INPUT name="e-mail" type="text"></INPUT>&nbsp&nbsp&nbsp
		Newsletter: 
			<SELECT name="newsletter">
				<OPTION>Mens</OPTION>
				<OPTION>Mens & Womans</OPTION>
				<OPTION>Womens</OPTION>
			</SELECT>&nbsp&nbsp&nbsp
		<INPUT class="submit" type="submit" value="Submit">
	</SPAN>
</DIV>
</FORM>

Psycho · September 7, 2011

A quick look at the manual for the "if" control structure should show you the problem: http://php.net/manual/en/control-structures.if.php

Just look at the examples and see what is different about your code.

Freedom-n-Democrazy · September 7, 2011

I see what you mean.

Is this correct?:

if $_POST['action'] == 'Register' {

... should actually be:

if ($_POST['action']) == 'Register' {

Freedom-n-Democrazy · September 7, 2011

I tried the following, but I still got the same error:

if ($_POST['action']) == 'Register' {

if ($_POST['action'] == 'Register') {

nilansanjaya · September 7, 2011

you cant just use "$query =" and put a if statement in the middle like that

voip03 · September 7, 2011

Remove ;

;}

nilansanjaya · September 7, 2011

try this


<?php
$link = mysql_connect('localhost', 'testusr', 'testpw');
mysql_select_db('testdb', $link);
$email = $_POST['e-mail'];
$query = "";
	if ($_POST['action'] == 'Register') {
		if ($_POST['newsletter'] == 'Mens') {
			$query = "INSERT INTO newsletters(mens) VALUES('$email')";
		}
		elseif ($_POST['newsletter'] == 'Mens & Womens') {
			$query = "INSERT INTO newsletters(mensandwomens) VALUES('$email')";
		}
		elseif ($_POST['newsletter'] == 'Womens') {
			$query = "INSERT INTO newsletters(womens) VALUES('$email')";
		}
	;}
mysql_query ($link, $query);
mysql_close($link);
?>

voip03 · September 7, 2011

Try to learn Switch()

Freedom-n-Democrazy · September 7, 2011

Some problems are now fixed.

I now get only one error.

The code sits at this:

<?php
$link = mysql_connect('localhost', 'testusr', 'testpw');
mysql_select_db('testdb', $link);
$email = $_POST['e-mail'];
if ($_POST['action'] == 'Register') {
	if ($_POST['newsletter'] == 'Mens') {
		"INSERT INTO newsletters(mens) VALUES('$email')";
	}
	elseif ($_POST['newsletter'] == 'Mens & Womens') {
		"INSERT INTO newsletters(mensandwomens) VALUES('$email')";
	}
	elseif ($_POST['newsletter'] == 'Womens') {
		"INSERT INTO newsletters(womens) VALUES('$email')";
	}
}
mysql_query ($link);
mysql_close($link);
?>

The error:

PHP Warning: mysql_query() expects parameter 1 to be string on line 57:

mysql_query ($link);

One problem I can see here is that PHP does not know its suppose to execute that code as a MySQL query. As can be seen in my initial post, I tried to set the "if" command to $query, and then inserted $query into "mysql_query ()", but PHP was not liking me using $query with the "if" statement. Why is that?

nilansanjaya · September 7, 2011

this ,

 mysql_query ($link);

should be

 mysql_query ($query);

and use $query = "sql query here"; between the if statement , like i mentioned before

voip03 · September 7, 2011

Variable $link is empty

please update the code

$link =	"INSERT INTO newsletters(mens) VALUES('$email')";

nilansanjaya · September 7, 2011

Variable $link is empty

please update the code
$link =	"INSERT INTO newsletters(mens) VALUES('$email')";	

link isn't empty. its the database link

 mysql_connect('localhost', 'testusr', 'testpw');

better to use a different variable for sql , ex: $query

voip03 · September 7, 2011

I am sorry nilansanjaya. And your code is working fine.

nilansanjaya · September 7, 2011

I am sorry nilansanjaya. And your code is working fine.

NO problem

Freedom-n-Democrazy · September 7, 2011

Hey guys!

What I did was:

if ($_POST['action'] == 'Register') {

if ($_POST['newsletter'] == 'Mens') {

$query = "INSERT INTO newsletters(mens) VALUES('$email')";

... and set mysql_query to "mysql_query ($query);" and it worked!!!!!!!!

Thanks so much for your help guys. Any time you need a favor (even if its not related to PHP) just send me through a PM and ask.

voip03 · September 7, 2011

Mark it solved

Freedom-n-Democrazy · September 7, 2011

Arghhhhhhh!!!

Now I am getting another error. I can register an e-mail address for the mens newsletter, but cannot register an e-mail address for the Mens & Womens or Womens newsletters.

I get the error:

PHP Notice: Undefined variable: query on line 57

mysql_query ($query);

The code sits at this:

<?php
$link = mysql_connect('localhost', 'testusr', 'testpw');
	mysql_select_db('testdb', $link);
	$email = $_POST['e-mail'];
	if ($_POST['action'] == 'Register') {
		if ($_POST['newsletter'] == 'Mens') {
			$query = "INSERT INTO newsletters(mens) VALUES('$email')";
		}
		elseif ($_POST['newsletter'] == 'Mens & Womens') {
			$query = "INSERT INTO newsletters(mensandwomens) VALUES('$email')";
		}
		elseif ($_POST['newsletter'] == 'Womens') {
			$query = "INSERT INTO newsletters(womens) VALUES('$email')";
		}
	}
mysql_query ($query);
mysql_close($link);
?>

PFMaBiSmAd · September 7, 2011

The mysql_query() statement is outside/after the end of the if($_POST['action'] == 'Register'){} conditional statement where the $query variable is defined, so, every time the page is requested, regardless of the $_POST['action'] value, $query is referenced, resulting in a php error being generated.

The mysql_query() statement should be inside the if($_POST['action'] == 'Register'){...} conditional statement where the $query variable is defined so that it is only executed when the form was submitted and the $query variable has a value.

Also, your if/elseif logic testing the $_POST['newsletter'] value allows the query to be executed when the $_POST['newsletter'] value does not match any of the possible choices. You should make sure that value is one of the choices. If it is not, you should output a user messages alerting them that an invalid value was selected and don't execute the query.

Finally, if after making those changes to the logic, you think you are submitting one of the choices but your code did not match it, you would need to troubleshoot what the actual submitted value is. Using echo or var_dump on the $_POST['newsletter'] variable would display what it contains.

Psycho · September 7, 2011

Comments, comments, comments - Use Them! A good programmer writes just as many lines of comments as he does of code (if not more). By adding comments, it gives context to the code and many times you can "see" logical mistakes that have been done.

In addition to what PFMaBiSmAd stated I think there must be another problem if you cannot register for the other two options. But, I have to ask, why are there THREE values and not TWO? You only need a "mens" value and a "womens" value - both in your form and in the database. The database table should have a field for the email address and then two fields to specify mens and womens which would be 0 (not selected) or 1 (selected). I don't know what your form looks like but, personally, I would use two checkboxes.

Also, if something doesn't operate as you expect - add debugging code to echo variables and results to the page to see what went wrong where.

Here is some updated code - I give no warranty as to whether it fixes your problems, but it should help identify the problem.

<?php

if ($_POST['action'] == 'Register')
{
    #Only need to connect to database if there was data submitted
    //Connect to DB server
    $link = mysql_connect('localhost', 'testusr', 'testpw');
    if(!$link)
    {
        die("Unable to connect to database");
    }
    //Select database
    if(!mysql_select_db('testdb', $link))
    {
        die("Unable to select database";
    }

    //Validate email address
    $email = mysql_real_escape_string(trim($_POST['e-mail']));
    if(empty($email))
    {
        echo "Error: Email address is required";
    }
    else
    {
        //Generate query based upon newsletter selection
        $query = false;
        if ($_POST['newsletter'] == 'Mens')
        {
            $query = "INSERT INTO newsletters(mens) VALUES('$email')";
        }
        elseif ($_POST['newsletter'] == 'Mens & Womens')
        {
            $query = "INSERT INTO newsletters(mensandwomens) VALUES('$email')";
        }
        elseif ($_POST['newsletter'] == 'Womens')
        {
            $query = "INSERT INTO newsletters(womens) VALUES('$email')";
        }

        //Check that a valid selection was made
        if(!$query)
        {
            echo "Error: No valid newsletter selection made: '{$_POST['newsletter']}'";
        }
        else
        {
            //Run query
            if(!mysql_query($query))
            {
                echo "Error: Unable to add record.";
                echo "<br>Query: {$query}<br>Error: " . mysql_error(); #only enable this line for debugging
            }
            else
            {
                echo "Your submission has been registered";
            }
        }
    }
    //Close DB connection
    mysql_close($link);
}

?>

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Having a problem with my first PHP script

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