[SOLVED] How to retrieve mysql distinct query value in php?

[Giri J]

Hi,

 

My code is as follows:

 

$qry1="select distinct (coder) from daily_reports where week_no=$wkno";

$exeQ1=mysql_query($qry1) or die ("Q1 :: Error ::: ".mysql_error());

$num1=mysql_num_rows($exeQ1);

 

 

I need some help / syntax on how to retrieve the results of mysql query (select distinct (coder) from daily_reports where week_no=$wkno)

 

The query is working fine on mysql it retrieves 4 rows. I want to know how to get the results in php.

 

 

I tried mysql_fetch_array().

I get the following error:

Notice: Undefined offset: 1 in D:\wamp\www\Reports\qau\processGU.php on line 27

 

 

I tried searching but didn't exactly get the relevant solution.

Please post if there are any existing threads.

 

 

Thank you very much !

 

 

Cheers

G.

[Ken2k7]

Can you post your mysql_fetch_array() part of the code?

[GingerRobot]

Arrays in all (sensible) programming languages index from 0. You've only selected one column from the table, so it's at index 0 not 1. You'd probably find it easier to give that field an alias an use that however:

 

$qry1="select distinct (coder) AS coder from daily_reports where week_no=$wkno";
$exeQ1=mysql_query($qry1) or die ("Q1 :: Error ::: ".mysql_error());
while($row=mysql_fetch_assoc($exeQ1) ){
    echo $row['coder'].'<br />;
}
?>

[JonnoTheDev]

Sorry for changing your variables slightly, force of habbit. Here you go:

<?php
$result  = mysql_query("SELECT DISTINCT coder FROM daily_reports WHERE week_no='".mysql_real_escape_string($wkno)."'") or die(mysql_error());
$numRows = mysql_num_rows($result);
while($row = mysql_fetch_assoc($result)) {
print $row['coder']."<br />";
}
?>

[GingerRobot]

Whoops - neil's right there. No need for an alias. Excuse the brain fart.

[Giri J]

Hi guys,

 

That's like a rocket response.

 

Here is my test code.

 

<?php

 

//Connections

mysql_connect("localhost", "root", "giri") or die(mysql_error());

mysql_select_db("hexl_data") or die(mysql_error());

 

//Get Week No.

//$wkno=$_POST['wkno'];

 

 

//Mysql Querys

$qry1="select distinct (coder) from daily_reports where week_no=2";

$exeQ1=mysql_query($qry1) or die ("Q1 :: Error ::: ".mysql_error());

$num1=mysql_num_rows($exeQ1);

 

//Result set

if($num1<1)

{

echo "Unable to retreive data. Please try again.";

}

else

{

while($rows = mysql_fetch_array($exeQ1, MYSQL_ASSOC))

{

echo "Coders:".$rows['Coder'];

}

}

 

?>

 

 

and here are my errors:

 

 

 

Notice: Undefined index: Coder in D:\wamp\www\Reports\qau\test.php on line 25

Coders:

Notice: Undefined index: Coder in D:\wamp\www\Reports\qau\test.php on line 25

Coders:

Notice: Undefined index: Coder in D:\wamp\www\Reports\qau\test.php on line 25

Coders:

Notice: Undefined index: Coder in D:\wamp\www\Reports\qau\test.php on line 25

Coders:

 

 

Let me know if anything is confusing.

 

Cheers

G.

[Giri J]

The following code perfectly works.

u ROCK man \m/.

 

tons and tons of thanks to ya and thank you every one for your super fast responses. much much appreciated

 

 

 

Arrays in all (sensible) programming languages index from 0. You've only selected one column from the table, so it's at index 0 not 1. You'd probably find it easier to give that field an alias an use that however:

 

$qry1="select distinct (coder) AS coder from daily_reports where week_no=$wkno";
$exeQ1=mysql_query($qry1) or die ("Q1 :: Error ::: ".mysql_error());
while($row=mysql_fetch_assoc($exeQ1) ){
    echo $row['coder'].'<br />;
}
?>

[JonnoTheDev]

while($row=mysql_fetch_assoc($exeQ1) ){