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View Full Version : error in code for updating records



ciaracous
08-23-2006, 07:21 PM
I'm trying to update records that are stored in my db on products. what I am doing is I am clicking on an image of a product to update. When I click on a product it brings me to my admin_update2.php page and passes in the value of shopName and prodId. I can see from the url that these values are being passed through ok. However I am getting an error saying

Parse error: parse error, unexpected $ in /home/c/ciaracousins/public_html/admin_update2.php on line 143

line 143 is just a blank line at the very bottom of my page. Any idea what is wrong with this??



<?php

include("db.php");

$prodId = $_GET['prodId'];

$sql = "select * from product where prodId = '$prodId'";

//db
//$result = mysql_query($sql,$conn) or die(mysql_error());


if (isset($_POST['submittedUpdate']))
{
$prodId = mysql_real_escape_string(trim($_POST['prodId']));
$sName = mysql_real_escape_string(trim($_POST['shopName']));
$dept = mysql_real_escape_string(trim($_POST['dept']));
$brand = mysql_real_escape_string(trim($_POST['brand']));
$type = mysql_real_escape_string(trim($_POST['type']));
$image = mysql_real_escape_string(trim($_POST['image']));
$price = mysql_real_escape_string(trim($_POST['price']));

$query2 = "UPDATE product SET prodName='$prodName', dept='$dept', brand='$brand', type='$type', image='$image', price='$price' WHERE prodId ='$prodId' and shopName='$shopName'";

$query = "SELECT * FROM product WHERE prodId = '$prodId'";


if ($result = mysql_query($query))
{

$row = mysql_fetch_array($result, MYSQL_ASSOC)
?>
<form action="admin_update2.php" method="post">
<center>
<table align="center" cellspacing="0" cellpadding="5" bgcolor="#ffffff" border=1 bordercolor="#2696b8">

<TR><TD align="left" bgcolor="#2696b8" width="30%"><FONT color="white"><B>Product Id: </B></FONT></TD>
<TD width="70%"><?php echo $row['prodId'] ?></TD></TR>

<TR><TD align="right" bgcolor="#2696b8"><FONT color="white"><B>Shop Name:</B></FONT></TD>
<TD width="70%"><?php echo $row['shopName'] ?>"></TD></TR>

<TR><TD align="right" bgcolor="#2696b8"><FONT color="white"><B>Product Name:</B></FONT></TD>
<TD><INPUT type="text" maxLength=45 size=45 name=name value="<?php echo $row['prodName'] ?>"></TD></TR>

<TR><TD align="right" bgcolor="#2696b8"><FONT color="white"><B>Department:</B></FONT></TD>
<TD><INPUT type="text" maxLength=45 size=45 name="surname" value="<?php echo $row['dept'] ?>"></TD></TR>

<TR><TD align="right" bgcolor="#2696b8"><FONT color="white"><B>Brand:</B></FONT></TD>
<TD><INPUT type="text" maxLength=45 size=45 name="address" value="<?php echo $row['brand'] ?>"></TD></TR>

<TR><TD align="right" bgcolor="#2696b8"><FONT color="white"><B>Type:</B></FONT></TD>
<TD><INPUT type="text" maxLength=45 size=45 name="address2" value="<?php echo $row['type'] ?>"></TD></TR>

<TR><TD align="right" bgcolor="#2696b8"><FONT color="white"><B>Image:</B></FONT></TD>
<TD><INPUT type="text" maxLength=45 size=45 name="county" value="<?php echo $row['image'] ?>"></TD></TR>

<TR><TD align="right" bgcolor="#2696b8"><FONT color="white"><B>Price:</B></FONT></TD>
<TD><INPUT type="text" maxLength=45 size=45 name="county" value="<?php echo $row['price'] ?>"></TD></TR>

<TR><TD></td>
<TD><CENTER><input name=submit type=submit value="UPDATE">
<FONT color=navy></FONT></P></TD></TR>
</TBODY>


<input type="hidden" name="prodId" value='.$prodId.'>
<input type="hidden" name="submittedUpdate" value="TRUE"/></td></tr>

</TABLE>
</FORM>
<?php
}

?>

</body>
</html>

//line 143

Fumigator
08-23-2006, 07:30 PM
Nearly always this means you haven't enclosed code blocks properly with the {} brackets. In your case, you have two "if" blocks, but only one closing } bracket.



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