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PHP Syntax Error
I'm kind of new to PHP and I'm executing a program for my class, but it
KWilson



 
Msg#: 4507682 posted 3:17 am on Oct 13, 2012 (gmt 0)

I'm kind of new to PHP and I'm executing a program for my class, but it keeps saying { [syntax error, unexpected $end, expecting '`' on line 79 } which is the last line of the program that contains no code.. help! i've included the entire code below



<html>
<head>
<title> Automobile Repair Cost </title>
<link rel="stylesheet" type="text/css" href="mypage.css"/>
</head>
<body>
<?php

// receive input from lab5a-1.html

$custName = $_POST['custName'];

$hoursLabor = $_POST['hoursLabor'];

$partsCost = $_POST['partsCost'];

$selectGarage = $_POST['selectGarage'];

//compute $laborCost,$partsCostTax, $partsCostTotal,$totalCost
if ($selectGarage == 1)
{
print ("<p> You selected a Dealer to repair your car. </p>");
$laborCost = 35 * $hoursLabor;
$partsCostTax = $partsCost * .05;
$partsCostTotal = $partsCost + $partsCostTax;
$totalCost = $laborCost + $partsCostTotal;
}
else if ($selectGarage == 2)
{
print ("<p> You selected a Local garage to repair your car.</p>");
$laborCost = 30 * $hoursLabor;
$partsCostTax = $partsCost * .085;
$partsCostTotal = $partsCost + $partsCostTax;
` $totalCost = $laborCost + $partsCostTotal;
}
else
{
print("<p> $selectGarage is an invalid garage selection! Cannot process</p>");
$laborCost = 0;
$partsCostTax = 0;
$totalCost = 0;
}
//display $custName,$selectGarage,$hoursLabor, $laborCost,$partsCostTotal,$totalCost

print ("<p>The customerís name is: $custName</p>");
print ("<p>The garage selected is: $selectGarage</p>");
print ("<p>The hours of labor required to fix your car: $hoursLabor</p>");
print ("<p>The cost for labor to repair your car: $laborCost</p>");
print ("<p>The cost for parts and supplies including tax: $partsCostTotal</p>");
print ("<p>The total cost to repair your car is: $totalCost<p>");
?>
<style type="text/css">

body { padding-left: 11em;font-family:cursive, "Segoe Print", Times, serif; color: blue; background-color: 66FF99 }
ul.navbar { list-style-type: none;
padding: 0;
margin: 0;
position: absolute;
top: 2em;
left: 1em;
width: 9em }
h1{
font-family: Helvetica, Geneva, Arial, SunSans-Regular, sans-ser
ul.navbar li {
background: white;
margin: 0.5em 0;
padding: 0.3em;
border-right: 1em solid black }
ul.navbar a { text-decoration: none }
a:link {color: red }
a:visited { color: purple }
address {
margin-top: 1em;
padding-top: 1em;
border-top: thin dotted }
</style>
</body>
</html>

 

JohnNZ



 
Msg#: 4507682 posted 3:28 am on Oct 13, 2012 (gmt 0)

` $totalCost = $laborCost + $partsCostTotal;


Is that a spurious character at the beginning of the line?

Austin80ss



 
Msg#: 4507682 posted 11:51 am on Oct 17, 2012 (gmt 0)

Unexpected $end is a common error if you didn't close block structure in your code. (like: if(){..;} , for(){..;})

lucy24

WebmasterWorld Senior Member lucy24 us a WebmasterWorld Top Contributor of All Time Top Contributors Of The Month



 
Msg#: 4507682 posted 10:56 pm on Oct 17, 2012 (gmt 0)

Is that a spurious character at the beginning of the line?

... and if yes, does that mean the program is now holding its breath-- possibly forever-- waiting for a matching ` to close it all up?

ericlewis107



 
Msg#: 4507682 posted 11:17 am on Dec 3, 2012 (gmt 0)

Although, since you're still learning, I suggest you move straight away from the mysql_* library. It's very old, slow, and insecure. Instead, look at the improved MySQLi version or PDO. It does look more complicated at the beginning, but you won't regret it.

swa66

WebmasterWorld Senior Member swa66 us a WebmasterWorld Top Contributor of All Time 10+ Year Member



 
Msg#: 4507682 posted 10:34 pm on Dec 3, 2012 (gmt 0)

As far as security goes, this is an open door for XSS (Cross Site Scripting)

Ref: [owasp.org...]

You must encode output ... and/or validate input.

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