[SOLVED] Unknown column '06JU06E1' in 'where clause'

[Clinton]

I've googled the problem but everybody else's solutions is working for mine. Not sure what's wrong.

 

Error: Unknown column '06JU06E1' in 'where clause' (06JU06E1 is $datecode)

 

$sumadd = mysql_query("SELECT datecode,SUM(signin),SUM(signout) FROM transactionlog WHERE datecode = $datecode GROUP BY `datecode`") or die(mysql_error());


$num = mysql_num_rows($sumadd);

if(!$num)

{

echo "No entries have been found for this datecode";

}
else

{

while($row = mysql_fetch_array($sumadd)){

$totaladd = $row['SUM(signin)'];
$totalsubtract = $row['SUM(signout)'];

$grandtotal = ($totaladd - $totalsubtract);



echo "<center>Amount in Magazine = <font color=purple>$grandtotal</font> </center>";

}

}

[DarkWater]

You need ' ' around $datecode inside the query because it's a string.

 

(I knew the problem before I even clicked the title.  Score!)

[Clinton]

LoL. You're good. I don't care what the rest of the people write about you here... you are good. 

[bluejay002]

Homerun! score goes to DarkWater!

[DarkWater]

@hunna03:  Thanks!  I think. (Click Topic Solved, btw)

 

@bluejay002:

DarkWater: 1

The Rest of Planet Earth: 0

[MasterACE14]

show of DarkWater 

[Clinton]

It was a good thing...

 

... and sorry... I forgot. It's early.

[DarkWater]

By the way, hunna, instead of the messy $row['SUM(signin)'], you should alias the functions in the query so it's easier.

 

$sumadd = mysql_query("SELECT datecode,SUM(signin) as in,SUM(signout) as out FROM transactionlog WHERE datecode = '$datecode' GROUP BY `datecode`") or die(mysql_error());

 

Then $row['in'] and $row['out'] will hold the correct values.