php - mysql fetch array (LIKE%)

Question

I have a php file which fetch and print data from my mysql table. here is the code..

<?php
{
$username = 'root';
$bookid = "GCK";
$con = mysql_connect("localhost","root","password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("library", $con);

$query3 = "SELECT * FROM GCKcatalogue WHERE BookID LIKE'%$bookid%'";
$numresults=mysql_query($query3);
$numrows=mysql_num_rows($numresults);

$result3 = mysql_query($query3);
($info3 = mysql_fetch_array($result3));

}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
<head>
<meta content="en-us" http-equiv="Content-Language">
<meta content="text/html; charset=utf-8" http-equiv="Content-Type">
<title>Results</title>

</head>
<body>

<table align="center" style="width: 80%" class="style4" cellpadding="1px" cellspacing="1px">
<tr>
<td class="style19">Resource ID</td><td class="style19">ISBN</td><td class="style19">Title</td><td class="style19">Author</td><td class="style19">Publisher</td><td class="style19">Year</td><td class="style19">Edition</td><td class="style19">Language</td><td class="style19">Type</td><td class="style19">Price</td><td class="style19">Location</td><td class="style19">Abstract</td><td class="style19">Availability</td><td class="style19">Last Issued</td><td class="style19">Last Received</td><td class="style19">Loan To</td>
</tr>

<?php
while($info3 = mysql_fetch_array($result3)){
$RID = $info3["BookID"];
$ISBN = $info3["ISBN"];
$Title = $info3["Title"];
$Author = $info3["Author"];
$Publisher = $info3["Publisher"];
$Year = $info3["Year"];
$Edition = $info3["Edition"];
$Language = $info3["Language"];
$Type = $info3["Type"];
$Price = $info3["Price"];
$Location = $info3["Location"];
$Abstract = $info3["Abstract"];
$Availability = $info3["Availability"];
$Issue = $info3["IssueDate"];
$Receive = $info3["ReceiveDate"];
$User = $info3["User"];


 echo '
 <tr>
 <td class="style20">' . $RID . '</td>
 <td class="style20">' . $ISBN . '</td>
 <td class="style20">' . $Title . '</td>
 <td class="style20">' . $Author . '</td>
 <td class="style20">' . $Publisher . '</td>
 <td class="style20">' . $Year . '</td>
 <td class="style20">' . $Edition . '</td>
 <td class="style20">' . $Language . '</td>
 <td class="style20">' . $Type . '</td>
 <td class="style20">' . $Price . '</td>
 <td class="style20">' . $Location . '</td>
 <td class="style20">' . $Abstract . '</td>
 <td class="style20">' . $Availability . '</td>
 <td class="style20">' . $Issue . '</td>
 <td class="style20">' . $Receive . '</td>
 <td class="style20">' . $User . '</td>
 </tr>
 ';
}

?>

</table>

</body>

</html>

The problem is, The query is skipping (or not giving) data from the first row. Everything else works fine..

I am using

Apache/2.2.13 (Win32) PHP/5.3.0

MySQL client version: mysqlnd 5.0.5-dev - 081106 - $Revision: 1.3.2.27 $

Spiny Norman · Accepted Answer · 2010-12-30 09:24:24Z

1

It's because of the line:

($info3 = mysql_fetch_array($result3));

You are fetching the first row and not doing anything with it :)

Also, the { and } around your first block of PHP don't do anything.

EDIT:

If it was working before, except for the first row, you should be ok if you change the first PHP block to:

<?php
$username = 'root';
$bookid = "GCK";
$con = mysql_connect("localhost","root","password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("library", $con);

$query3 = "SELECT * FROM GCKcatalogue WHERE BookID LIKE'%$bookid%'";
$numresults=mysql_query($query3);
$numrows=mysql_num_rows($numresults);

$result3 = mysql_query($query3);
?>

There are some other strange things going on (for example, you never use the $username variable), but at this should give the correct output at least.

edited Dec 30, 2010 at 9:24

answered Dec 30, 2010 at 9:13

Spiny Norman

8,3271 gold badge34 silver badges57 bronze badges

if i remove that line, how can i get data?

Alfred
– Alfred

12/30/2010 09:19:31
Commented Dec 30, 2010 at 9:19
You are already getting the data in the line while($info3 = mysql_fetch_array($result3)){.

Spiny Norman
– Spiny Norman

12/30/2010 09:20:24
Commented Dec 30, 2010 at 9:20
can u modify the script for me?? i am stuck.. :P

Alfred
– Alfred

12/30/2010 09:21:48
Commented Dec 30, 2010 at 9:21

Add a comment |

Jan Hančič · Accepted Answer · 2010-12-30 09:13:38Z

1

It's becouse you are fetching the first row with this line:

($info3 = mysql_fetch_array($result3));

And this moves the array pointer to the second item.

Also you don't have to execute the query twice to get number of rows and the data. Just do something like this:

$result = MySql_Query ( 'QUERY HERE' );
$numOfRows = MySql_Num_Rows ( $result );
while ( $row = MySql_Fetch_Array ( $result ) ) {
  ...
}

I would also advise you to not use the MySql extension, use the new MySqli or PDO.

answered Dec 30, 2010 at 9:13

Jan Hančič

54k17 gold badges98 silver badges101 bronze badges

if i remove that line, how can i get data?

Alfred
– Alfred

12/30/2010 09:19:10
Commented Dec 30, 2010 at 9:19
You are retrieving data in your while loop: while($info3 = mysql_fetch_array($result3)){

Jan Hančič
– Jan Hančič

12/30/2010 09:22:15
Commented Dec 30, 2010 at 9:22

Add a comment |

Deepak · Accepted Answer · 2010-12-30 09:23:42Z

0

Try also this:

$result = MySql_Query ( 'QUERY HERE' ); $numOfRows = MySql_Num_Rows ( $result ); while ( $row = MySql_Fetch_Array ( $result ) ) { $arrData[] = $row; }

print_r($arrData);

answered Dec 30, 2010 at 9:23

Deepak

1,1054 gold badges13 silver badges22 bronze badges

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