Re: [PHP] Ambiguous?

2013-08-09 Thread Stuart Dallas 
On 9 Aug 2013, at 13:00, Karl-Arne Gjersøyen karlar...@gmail.com wrote:

 1) query
 foreach($varenr_inn_pa_lager as $vnr){
include('../../tilkobling.php');
$sql = SELECT * FROM exan,dynamit WHERE varenr = '$vnr' LIMIT
 1;
$resultat = mysql_query($sql, $tilkobling) or
 die(mysql_error());
 
 2) Result:
 Column 'varenr' in where clause is ambiguous
 
 3) Tables in MySQL database
 
 mysql SELECT * FROM exan;
 ++-+---+--+--+-+
 | leverandor | valgt_lager | un_nr | varenavn | varenr   | kg_pa_lager |
 ++-+---+--+--+-+
 
 mysql SELECT * FROM dynamit;
 ++-+---++---+-++-+
 | leverandor | valgt_lager | type  | dim_mm | un_nr |
 varenavn| varenr | kg_pa_lager |
 ++-+---++---+-++-+
 
 4) My question:
 What means with ambiguous here?

The field exists in both tables, so you need to specify which one you mean.

Please learn how to use Google, it knows the answer to almost everything!

-Stuart

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Stuart Dallas
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http://3ft9.com/
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Re: [PHP] Ambiguous?

2013-08-09 Thread Floyd Resler 


On Aug 9, 2013, at 8:00 AM, Karl-Arne Gjersøyen karlar...@gmail.com wrote:

 1) query
 foreach($varenr_inn_pa_lager as $vnr){
include('../../tilkobling.php');
$sql = SELECT * FROM exan,dynamit WHERE varenr = '$vnr' LIMIT
 1;
$resultat = mysql_query($sql, $tilkobling) or
 die(mysql_error());
 
 2) Result:
 Column 'varenr' in where clause is ambiguous
 
 3) Tables in MySQL database
 
 mysql SELECT * FROM exan;
 ++-+---+--+--+-+
 | leverandor | valgt_lager | un_nr | varenavn | varenr   | kg_pa_lager |
 ++-+---+--+--+-+
 
 mysql SELECT * FROM dynamit;
 ++-+---++---+-++-+
 | leverandor | valgt_lager | type  | dim_mm | un_nr |
 varenavn| varenr | kg_pa_lager |
 ++-+---++---+-++-+
 
 4) My question:
 What means with ambiguous here?
 
 Thanks.
 Karl

Your field varenr is in both tables you are joining.  Therefore, MySQL 
doesn't know which table it should use the field from.  So, you need to qualify 
it by putting the table name in front of the field.  Such as:
SELECT * FROM exan,dynamit WHERE exan.varenr='$vnr' LIMIT 1

Take care,
Floyd


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Re: [PHP] Ambiguous?

2013-08-09 Thread richard gray 


On 09/08/2013 14:00, Karl-Arne Gjersøyen wrote:

1) query
foreach($varenr_inn_pa_lager as $vnr){
 include('../../tilkobling.php');
 $sql = SELECT * FROM exan,dynamit WHERE varenr = '$vnr' LIMIT
1;
 $resultat = mysql_query($sql, $tilkobling) or
die(mysql_error());

2) Result:
Column 'varenr' in where clause is ambiguous

3) Tables in MySQL database

mysql SELECT * FROM exan;
++-+---+--+--+-+
| leverandor | valgt_lager | un_nr | varenavn | varenr   | kg_pa_lager |
++-+---+--+--+-+

mysql SELECT * FROM dynamit;
++-+---++---+-++-+
| leverandor | valgt_lager | type  | dim_mm | un_nr |
varenavn| varenr | kg_pa_lager |
++-+---++---+-++-+

4) My question:
What means with ambiguous here?

Thanks.
Karl
a field named 'varenr' exists in both tables - your sql query needs to 
specify from which table you are selecting the data.



foreach($varenr_inn_pa_lager as $vnr){
   include('../../tilkobling.php');
   $sql = SELECT * FROM exan,dynamit WHERE varenr = '$vnr' LIMIT



do you really need to include the file 'tilkobling.php' repeatedly for 
each iteration over $varenr_inn_pa_lager??


Rich


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