PHP mysql query returns a blank screen

Question

I've commented out the bottom part, and the SQL query works fine. Its the displaying of the query where the error is coming from i believe.

$host = "127.0.0.1"; 
$user = "root"; 
$pass = "Toom13371!";

$connection = mysql_connect($host,$user,$pass) or die (mysql_errno().":     ".mysql_error()."<BR>");

// 2. Selecting DB.

$dbname = "filters"; 
mysql_select_db($dbname);

// 3. Build/Test SQL Query

$sql = ("select * from filter_bandpass where start_passband=" . $_POST['Lowfreq'] . "         and stop_passband='" . $_POST['Highfreq'] . "'");
//echo $sql; //Comment/Uncomment to test sql query.

// 4. Retrieve info from MySQL.

$query = mysql_query($sql);

// 5. Display Query.

echo "<table border='1'>
<tr>
<th>Low Frequency</th>
<th>High Frequency</th>
</tr>";

while($row = mysql_fetch_array($query)) {
  echo "<tr>";
  echo "<td>" . $row['Lowfreq'] . "</td>";
  echo "<td>" . $row['Highfreq'] . "</td>";
  echo "</tr>";
}

echo "</table>";
  ?>

Any help would be appreciated, I'm sure it's going to be some small stupid error i've over looked.

Thanks in advance :)

This looks terrifyingly insecure. Are you sure your user parameters are properly escaped? $_POST data should NEVER go directly into a query. mysql_query is an obsolete interface and should not be used in new applications and will be removed in future versions of PHP. A modern replacement like PDO is not hard to learn. If you're new to PHP, a guide like PHP The Right Way can help explain best practices. — tadman
– tadman, Commented Aug 21, 2014 at 10:11
@thomasward1212 - It is good practice on stackoverflow to accept an answer from given answers. Please select an answer — Lemuel Botha
– Lemuel Botha, Commented Aug 22, 2014 at 7:27

Lemuel Botha · Accepted Answer · 2014-08-21 10:25:18Z

1

I'm guessing, based on your query, that you need to change this

mysql_select_db($dbname);

to

mysql_select_db($dbname, $connection);

and

while($row = mysql_fetch_array($query)) {
  echo "<tr>";
  echo "<td>" . $row['Lowfreq'] . "</td>";
  echo "<td>" . $row['Highfreq'] . "</td>";
  echo "</tr>";
}

to

while($row = mysql_fetch_array($query)) {
  echo "<tr>";
  echo "<td>" . $row['start_passband'] . "</td>";
  echo "<td>" . $row['stop_passband'] . "</td>";
  echo "</tr>";
}

edited Aug 21, 2014 at 10:25

answered Aug 21, 2014 at 10:12

Lemuel Botha

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1 Comment

thomasward1212 Over a year ago

What i'm getting now is the top part of the table, this bit; echo "<table border='1'> <tr> <th>Low Frequency</th> <th>High Frequency</th> </tr>"; But not the query generated part.

MD SHAHIDUL ISLAM · Accepted Answer · 2014-08-21 10:25:18Z

0

Change line

mysql_select_db($dbname);

to

mysql_select_db($dbname, $connection);

Also before query check

$_POST['Lowfreq'] and $_POST['Highfreq']

If there is no value in these variables query will must return empty.

edited Aug 21, 2014 at 10:25

answered Aug 21, 2014 at 10:11

MD SHAHIDUL ISLAM

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Comments

Bud Damyanov · Accepted Answer · 2014-08-21 10:49:43Z

0

In your query there should not brackets for string.

 $sql = ("select * from filter_bandpass where start_passband=" . $_POST['Lowfreq'] . "         and stop_passband='" . $_POST['Highfreq'] . "'");

should be:

 $sql = "select * from filter_bandpass where start_passband=" . $_POST['Lowfreq'] . "         and stop_passband='" . $_POST['Highfreq'] . "'";

edited Aug 21, 2014 at 10:49

Bud Damyanov

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answered Aug 21, 2014 at 10:32

Kunj Patel

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