[SOLVED] PHP Parse Error: syntax error Line 21 unexpected T_STRING

DeepakJ · July 26, 2007

Could someone please help me figure out what is wrong with this script?

I fixed the initial error but a new one popped up .

Parse error: syntax error, unexpected $end in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\verification.php on line 23

<html><head><title>AlibreCam Verification System</title></head>

<center><h3>AlibreCam Verification System</h3></center>

<center><TABLE border=0 cellpadding=3><form name="input" action="verification.php" method="get">
<tr><td><center>Customer ID:</center></td>
<td><input type="text" name="user"></td></tr>
<tr><td colspan=2><center><input type="submit" value="Submit"></center></td></tr>
</form><br></table><br>

<?php 
include("connectioninfo.php");
$customer = $_GET['user'];
$query= sprintf("SELECT * FROM invoiceid WHERE customerid=%s,$customer);
$query= mysql_real_escape_string($query);
$cResult = mysql_query($query);
$value=0;
while($row=mysql_fetch_array($cResult)){
$value=$value+$row[numoflicenses];
}
echo 'Total number of licenses purchased: '.$value; 
?>
</html>

csplrj · July 26, 2007

Change this line

$value=$value+$row[numoflicenses];

to

$value=$value+$row['numoflicenses'];

Bye for now

CSJakharia

DeepakJ · July 26, 2007

That doesn't work. I get another error if I do that.

yarnold · July 26, 2007

<html><head><title>AlibreCam Verification System</title></head>

<center><h3>AlibreCam Verification System</h3></center>

<center><TABLE border=0 cellpadding=3><form name="input" action="verification.php" method="get">
<tr><td><center>Customer ID:</center></td>
<td><input type="text" name="user"></td></tr>
<tr><td colspan=2><center><input type="submit" value="Submit"></center></td></tr>
</form><br></table><br>

<?php 
include("connectioninfo.php");
$customer = $_GET['user'];
$query= sprintf("SELECT * FROM invoiceid WHERE customerid=%s,$customer");
$query= mysql_real_escape_string($query);
$cResult = mysql_query($query);
$value=0;
while($row=mysql_fetch_array($cResult)){
$value=$value+$row['numoflicenses'];
}
echo "Total number of licenses purchased: ".$value; 
?>
</html>

DeepakJ · July 26, 2007

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\verification.php on line 19 Thats the error I get when I change the variable to ['x']

teng84 · July 26, 2007

the select statement is missing " quote at the end of it

DeepakJ · July 26, 2007

<html><head><title>AlibreCam Verification System</title></head>

<center><h3>AlibreCam Verification System</h3></center>

<tr><td><center>Customer ID:</center></td>

</form><br></table><br>

<?php

$hostname = "localhost";

$username = "root";

$password = "xxxxxxxx";

$dbname = "licensinginformation";

mysql_connect($hostname, $username, $password) or DIE("Unable to connect to MySQL server $hostname");

$selected = mysql_select_db($dbname) or DIE("Could not select requested db $dbname");

;

$customer = $_GET['user'];

$query= sprintf("SELECT * FROM invoiceid WHERE customerid=%s",$customer);

$query= mysql_real_escape_string($query);

$cResult = mysql_query($query);

$value=0;

while($row=mysql_fetch_array($cResult)){

$value=$value+$row['numoflicenses'];

}

echo "Total number of licenses purchased: ".$value;

?>

</html>

Thanks for the help but now it is saying "

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\verification.php on line 25"

teng84 · July 26, 2007

$query= "SELECT * FROM invoiceid WHERE customerid=".$customer."";

try

DeepakJ · July 26, 2007

Same error, I don't think it works.

DeepakJ · July 26, 2007

It seems like the format of my query is off but it seems fine. The thing says it isn't able to perform that function on it.

DeepakJ · July 26, 2007

Please help

Trium918 · July 26, 2007

Are you stilll getting an error? If so, what is it?

Trium918 · July 26, 2007

Try this!

<html><head><title>AlibreCam Verification System</title></head>

<center><h3>AlibreCam Verification System</h3></center>

<center><TABLE border=0 cellpadding=3><form name="input" action="verification.php" method="get">
<tr><td><center>Customer ID:</center></td>
<td><input type="text" name="user"></td></tr>
<tr><td colspan=2><center><input type="submit" value="Submit"></center></td></tr>
</form>
</table>


<?php 
$hostname = "localhost";
$username = "root";
$password = "xxxxxxxx";
$dbname = "licensinginformation";

mysql_connect($hostname, $username, $password) or DIE("Unable to connect to MySQL server $hostname");
$selected = mysql_select_db($dbname) or DIE("Could not select requested db $dbname");

$customer = $_GET['user'];
$query= sprintf("SELECT * FROM invoiceid WHERE customerid='$customer'");
//$query= mysql_real_escape_string($query);
$cResult = mysql_query($query);
$value=0;
while($row=mysql_fetch_array($cResult)){
$value=$value+$row['numoflicenses'];
}
echo "Total number of licenses purchased: ".$value; 
?>
</html>

teng84 · July 26, 2007

do this first

$cResult = mysql_query($query) or die(mysql_error());

DeepakJ · July 27, 2007

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\verification.php on line 25"

This is the error:

I entered that, same error, no death message.

teng84 · July 27, 2007

can you paste your code now

DeepakJ · July 27, 2007

<html><head><title>AlibreCam Verification System</title></head>

<center><h3>AlibreCam Verification System</h3></center>

<tr><td><center>Customer ID:</center></td>

</form><br></table><br>

<?php

$hostname = "localhost";

$username = "root";

$password = "98989lol";

$dbname = "licensinginformation";

$value=0;

$customer = $_GET['user'];

mysql_connect($hostname, $username, $password) or DIE("Unable to connect to MySQL server $hostname");

$selected = mysql_select_db($dbname) or DIE("Could not select requested db $dbname");

$query= sprintf("SELECT * FROM invoiceid WHERE customerid=%s",$customer);

//$query= "SELECT * FROM invoiceid WHERE //customerid==".$customer.""; (doesn't work)

$query= mysql_real_escape_string($query);

$cResult= mysql_query($query);

while($row=mysql_fetch_array($cResult)){

$value=$value+$row['numoflicenses'];

}

echo "Total number of licenses purchased: ".$value;

?>

</html>

Here is my current code

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[SOLVED] PHP Parse Error: syntax error Line 21 unexpected T_STRING

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