Solved

SQL Error

Posted on 2009-07-15
10
183 Views
Last Modified: 2012-05-07
I'm trying to add a line to the coding below. The one I am having issues with is commented out, put it is causing a white page to occur. I know it's a simple syntex error that I'm over looking.
The line above, and below work fine, so I copied one of them and simply changed the information it was calling. Please help!!
<?php
$connection = mysql_connect($host, $user, $pass) or die ("Unable to connect!"); 
mysql_select_db($db) or die ("Unable to select database!"); 
$query = "SELECT * 
FROM `voc2_gift` 
LIMIT 0 , 30 "; 
$result = mysql_query($query) or die ("Error in query: $query. ".mysql_error()); 
if (mysql_num_rows($result) > 0) { 
echo "<table bgcolor=#000000 cellpadding=50 width=100 border=1 bordercolorlight=#00FF00 bordercolorlight=#00FF00 bordercolordark=#00FF00>";
 
  $numItems = mysql_numrows($result);
  $itemsPerRow = 4;
  $itemsCounter = 0;
 
  for($i=0;$i<$numItems;$i++) {
    if($itemsCounter==0) { echo "<tr>"; }
    echo "<td><center><img src='".mysql_result($result,$i,"GiftURL")."'>";
echo "<input type=button value='Send' onclick='sendImage(".mysql_result($result,$i,"GiftURL").")'>";
    // echo "<p>".mysql_result($result,$i,"GiftPrice").;
    echo "<p>".mysql_result($result,$i,"GiftName")."</center></td>"; 
 
    if($itemsCounter==3) { echo "</tr>"; }
    $itemsCounter = ($itemsCounter+1)%$itemsPerRow;
  }
 
  echo "</table>"; 
 
} else { 
  echo "No rows found!"; 
} 
?>

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Question by:mimoser
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10 Comments
 
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Accepted Solution

by:
Xemorph earned 500 total points
ID: 24865192
You have a '.' at the end of your statement, that will cause a parse error
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Expert Comment

by:flob9
ID: 24865207
Please, show us the fields of the table voc2_gift.
0
 

Author Comment

by:mimoser
ID: 24865216
I left the period in the line, because the line below also has a period at the begining and end, which works.

Those table fields are

GiftID
GiftName
GiftURL
GiftPrice
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LVL 4

Expert Comment

by:Xemorph
ID: 24865254
I would try to check your PHP logs.  That will tell you if there was a fatal error that occurred, and what it was.
0
 
LVL 14

Expert Comment

by:flob9
ID: 24865275
Add this before the mysql_connect :

error_reporting(E_ALL);

then, uncomment the line

echo "<p>".mysql_result($result,$i,"GiftPrice");

and tell us what happen.
0
 

Author Comment

by:mimoser
ID: 24865324
[15-Jul-2009 18:59:13] PHP Parse error:  syntax error, unexpected ';' in public_html/gift.php on line 118

0
 
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Expert Comment

by:flob9
ID: 24865379
remove the dot before ";" :)
0
 
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Expert Comment

by:Xemorph
ID: 24865382
If line 118 is line 19 in your posted code, I believe it is the '.' like I said before.  If the code changed at all, post what was ran.

Thanks.
0
 

Author Closing Comment

by:mimoser
ID: 31604046
That worked. Thank you, it had me confused because of the . in front of
.mysql_result
so I thought one was needed at the end as well.
0
 
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Expert Comment

by:flob9
ID: 24865510
nice one :)
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