Unknown column 'priser.pr_id' in 'on clause'

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October 5th, 2012, 03:15 AM

rootKID

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Unknown column 'priser.pr_id' in 'on clause'

Here is my SQL code:

SELECT
p1.p_id,
p1.p_titel,
p1.p_kunstner,
p1.p_oprettet,
image_url.img_url,
genre.g_navn,
priser.pr_pris
FROM
produkter AS p1
INNER JOIN produkter ON p1.p_FK_pris = priser.pr_id
INNER JOIN image_url ON image_url.img_FK_products = p1.p_id
INNER JOIN genre ON p1.p_FK_genre = genre.g_id

and im getting this error:

Unknown column 'priser.pr_id' in 'on clause'

and im having no clue at all what the (crap) is wrong here... ideas?

-thanks!

...

October 5th, 2012, 03:42 AM

Problem Solver

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Location: Stockholm, Sweden

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There is no "priser" table present in your query hence the error message -"Unknown column 'priser.pr_id' ".

My guess is that you don't want to perform a self join so the table in the second join should be "priser" like this instead:

Code:

SELECT
p1.p_id,
p1.p_titel,
p1.p_kunstner,
p1.p_oprettet,
image_url.img_url,
genre.g_navn,
priser.pr_pris
FROM
produkter AS p1
INNER
  JOIN
    priser
    ON p1.p_FK_pris = priser.pr_id
INNER
  JOIN
    image_url
    ON image_url.img_FK_products = p1.p_id
INNER
  JOIN
    genre
    ON p1.p_FK_genre = genre.g_id

__________________
/Stefan

October 5th, 2012, 03:58 AM

rootKID

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Join Date: Oct 2012

Posts: 3

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hmm....
yeah, working now... lol..

should have seemed that one comming...
anyways... working now, almost...

now they WONT getting out...
freaking annoying...

CODE:

PHP Code:


		
			
<?php

$bg = '';// variablen $bg oprettes og er tom.

/*p1 = Aliases instead of (produkter)...*/

$query_produkter = "

SELECT

    p1.p_id,

    p1.p_titel,

    p1.p_kunstner,

    p1.p_oprettet,

    image_url.img_url,

    genre.g_navn,

    priser.pr_pris

FROM

    produkter AS p1

        INNER JOIN priser ON p1.p_FK_pris = priser.pr_id

        INNER JOIN image_url ON image_url.img_FK_products = p1.p_id

        INNER JOIN genre ON p1.p_FK_genre = genre.g_id";

$result_produkter = mysql_query($query_produkter)or die(mysql_error());

?>

<table class="butikliste_tableshow">

<tr>

<td>&nbsp;</td>

<td>KUNSTNER</td>

<td>TITEL</td>

<td>GENRE</td>

<td>OPRETTET</td>

<td>PRIS</td>

<td>KØB</td>

</tr>

<?php

while($produkter_row = mysql_fetch_array($result_produkter))

{

    $bg = ($bg == '' ? '#999999' : '');//skal antages som en if/else saetning...

    echo "<tr bgcolor='$bg'>";

    

    echo "<td>"."<img src='50x50/".$produkter_row['img_url']."' />"."</td>";

    echo "<td>"."<a href='butik_detalje.php?pid=".$produkter_row['p_id']."'>".$produkter_row['p_kunstner']."</a>"."</td>";

    echo "<td>".$produkter_row['p_titel']."</td>";

    echo "<td>".$produkter_row['g_navn']."</td>";

    echo "<td>".$produkter_row['addeddate']."</td>";

    echo "<td>".$produkter_row['pr_pris']."</td>";

    echo "<td>"."<a href='#'><img src='styles/images/kurv.png' /></a>"."</td>";

    

    echo "</tr>";

}//While Statement ends...

?>

</table>

Ideas?...

October 5th, 2012, 04:07 AM

cafelatte

Contributing User

Dev Shed Intermediate (1500 - 1999 posts)

Join Date: Mar 2008

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"now they WONT getting out..."

I'm not sure what this means, so not sure how to respond.

For problems with MySQL queries, it's often helpful if we're able to replicate your scenario, so consider providing CREATE and INSERTs for each of the relevant tables, together with the result set you'd expect from your query.

October 5th, 2012, 04:48 AM

rootKID

Registered User

Join Date: Oct 2012

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ok lol... i should take a knife and put my heart out of my boddy...

i had no SQL rows inside produkter table... lol...
working 100% now -.-'...

sorry for waste of time :P...

October 5th, 2012, 05:10 AM

cafelatte

Contributing User

Join Date: Mar 2008

Posts: 1,923

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Reputation Power: 377

On this occasion, that might be a step too far.

Unknown column 'priser.pr_id' in 'on clause'

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