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[agarwalsrushti]

Hi,
Im just stuck up with a small query. In my database ive a table called specialization which is dependent on qualification. In specialization i ve 3 fields: QualificationId, SpecializationId,SpecializtionArea. in specialization table there are SpecializationArea with same name but with diferent SpecializationId and QualificationId. Now when i enter search page and select to search based on specialization then it doesnt displays the results for search. In the userdetails table specialization and qualification are stored with thier id values.
Ive used the query below. Is the query correct?

//Create Query for Specialization
if($dropdown == Specialization){
$query= mysql_query("SELECT QualificationId,SpecializationId FROM specialization WHERE SpecializationArea='$search'");
while($nt=mysql_fetch_array($query)){
//echo "" .$nt['QualificationId']. "" .$nt['SpecializationId']. "<BR>";
$result = "SELECT UserName FROM userdetails WHERE Qualification='$nt[QualificationId]' AND Specialization='$nt[SpecializationId]'" or die (mysql_error()); 
$data = mysql_query($result) or die(mysql_error()); 
}
}
 

Your help will be appreciated.

Thanks
Srushti

[debasisdas]

First ensure that data exists in database as per your search conditions.

It is always better to use LIKE instead of using = for search operation.

[agarwalsrushti]

First ensure that data exists in database as per your search conditions.

It is always better to use LIKE instead of using = for search operation.

Thanks. There is data in the userdetails table as per the condition.

[code green]

Your query looks fine it is the php code that looks wrong.
Some debugging statements would help!
Anyway[PHP]if($dropdown == Specialization){[/PHP] Here php will treat Specialization as a constant.
If you had error_reporting() switched on as recommended in this forum
then this would have flashed a warning
What is Specialization meant to be, a string?
[PHP]if($dropdown == 'Specialization'){[/PHP]

[agarwalsrushti]

Here Specialization is a constant. Its the value that i ve assigned to it in drop down. The other constants work fine and displays the result. Its not going in this while loop even if matches are there. I dont think theres a problem with this loop coz it works well for other search criterias and displays the results.

$resultnum = mysql_num_rows($data); // Just print $resultnum if you want to show how many results returned
if($resultnum>0) { // Echos out matches if anything was found
while($info=mysql_fetch_array($data)){
//Print it out to page 
echo "" .$info['FullName']. "";
}// end of while
}//end of if
 

[code green]

I am suprised this works [PHP]$result = "SELECT UserName FROM userdetails WHERE Qualification='$nt[QualificationId]'
AND Specialization='$nt[SpecializationId]'" or die (mysql_error()); [/PHP] Arrays in query strings always produce an error for me and require curly braces [PHP]$result = "SELECT UserName FROM userdetails WHERE Qualification='{$nt[QualificationId]}'
AND Specialization='{$nt[SpecializationId]}'" or die (mysql_error()); [/PHP] Don't understand what you mean

Here Specialization is a constant. Its the value that i ve assigned to it in drop down