Searching Database

[coderphp]

Howdy friends,

I am trying to display multiple rows from MySQL in Html table and I am using while() loop for that purpose but its not working can any one help me...

here's the script..

 

 

echo " <h1><strong><center><font color='#663300'>Search Result/s for Name</font></center></strong></h1>";

 

echo "<table border='1' align='center' font color='#663300'>

<tr>

<th>Name</th>

<th>City</th>

<th>country</th>

<th>Zimmedar/Sub Ordinate</th>

<th>Course</th>

 

</tr>";

 

while($far = mysql_fetch_array($query))

{

 

 

 

echo "<tr class=\"gray\" font color='#663300'>";

echo " <td align='center' width='15%' font color='#663300'>".$far['name']."</td> ";

echo " <td align='center' width='20%' font color='#663300'>".$far['city']."</td> ";

echo " <td align='center' width='25%' font color='#663300'>".$far['country']."</td> ";

echo " <td align='center' width='20%' font color='#663300'>".$far['z/s']."</td> ";

echo " <td align='center' width='20%' font color='#663300'>".$far['course']."</td> ";

 

echo "</tr>";

 

echo"</table>";}

 

but the data is not displayed in table:-((

[kimi2k]

maybe you have empty query, or this columns no available...

make for test:

 

while($far = mysql_fetch_array($query))

 

{

var_dump($far);

break;

...

[coderphp]

The problem is the data is shown but not in table only one column is shown in table and the rest are shown in junk...

I don't understand why is this so??

[kimi2k]

Show your query and database structure

Edited December 18, 2012 by kimi2k

[Barand]

Turn on error reporting and check for mysql errors.

 

What does mysql_error() return?

[coderphp]

here is the query Kimi2k

 

if($country=mysql_real_escape_string($_POST['country']) or DIE('please choose a country'))

 

{

$query=mysql_query("SELECT * FROM `info` WHERE `country` LIKE '$country'");

}

 

it returns no error Barand

[Barand]

How do you know that if you are not checking for errors?

 

$query=mysql_query("SELECT * FROM `info` WHERE `country` LIKE '$country'");
if (!$query) echo mysql_error();

[coderphp]

No I have checked it Barand it gives no error,instead the data is shown but not in table but in junk form...

[Barand]

What did the var_dump($far) show? (as Kimi2k suggested)