PHP programming forum. Ask questions or help people concerning PHP code. Don't understand a function? Need help implementing a class? Don't understand a class? Here is where to ask. Remember to do your homework!
i have run the script and i am getting the following error in the output
<b>Warning</b>: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /usr/users1/jonfort/public_html/designfreak/pooch/product-detail.php on line 132
<?php $results2 = mysql_query($query2) or die ("Error in query: $query2. ".mysql_error()); ?>
and got the following results
Error in query: SELECT DISTINCT stock.stockID, size.Size FROM poochieProd AS prod LEFT JOIN poochieStock AS stock ON Prod.ID = stock.ID LEFT JOIN poochieSizes AS size ON stock.SizeID = size.SizeID WHERE Prod.ID = '-1' AND stock.stock > 0 ORDER BY size.SizeID ASC. Unknown table 'Prod' in where clause
<select name="os0" class="text" id="selectSize">
<option value="Select Size">Select Size</option>
<?php
$query2 = sprintf("
SELECT DISTINCT
stock.stockID, size.Size
FROM
poochieProd AS prod
LEFT JOIN poochieStock AS stock ON Prod.ID = stock.ID
LEFT JOIN poochieSizes AS size ON stock.SizeID = size.SizeID
WHERE
prod.prodID = '%s' AND stock.stock > 0
ORDER BY
size.SizeID ASC", GetSQLValueString($var3_Recordset1, "int"));
$results2 = mysql_query($query2);
while($row2 = mysql_fetch_array($results2)){
?>
<option value="<?php echo $row2['Size']; ?>"><?php echo $row2['Size']; ?></option>
<?php
}
?>
</select>
i have been looking but cant seem to see
thanks in advance
Last edited by Benjamin on Wed Oct 03, 2012 2:14 pm, edited 1 time in total.
Reason:Added [syntax=php||htm||css||javascript||sql||etc] - Please use [syntax] tags when posting code in the forums! Thanks.
<?php
$query2 = sprintf("
SELECT DISTINCT stock.stockID, size.Size
FROM poochieProd AS prod
LEFT JOIN poochieStock AS stock ON prod.ProdID = stock.stockID
LEFT JOIN poochieSizes AS size ON stock.SizeID = size.SizeID
WHERE prod.ProdID = '%s' AND stock.stock > 0
ORDER BY size.SizeID ASC", GetSQLValueString($var3_Recordset1, "int"));
$results2 = mysql_query($query2);
while($row2 = mysql_fetch_array($results2)){
?>
Error
There seems to be an error in your SQL query. The MySQL server error output below, if there is any, may also help you in diagnosing the problem
ERROR: Unknown Punctuation String @ 7
STR: <?
SQL: <?php
$query2 = sprintf("
SELECT DISTINCT stock.stockID, size.Size
FROM poochieProd AS prod
LEFT JOIN poochieStock AS stock ON prod.ProdID = stock.stockID
LEFT JOIN poochieSizes AS size ON stock.SizeID = size.SizeID
WHERE prod.ProdID = '%s' AND stock.stock > 0
ORDER BY size.SizeID ASC", GetSQLValueString($var3_Recordset1, "int")); <?php
$query2 = sprintf("
SELECT DISTINCT stock.stockID, size.Size
FROM poochieProd AS prod
LEFT JOIN poochieStock AS stock ON prod.ProdID = stock.stockID
LEFT JOIN poochieSizes AS size ON stock.SizeID = size.SizeID
WHERE prod.ProdID = '%s' AND stock.stock > 0
ORDER BY size.SizeID ASC", GetSQLValueString($var3_Recordset1, "int")); <?php
$query2 = sprintf("
SELECT DISTINCT stock.stockID, size.Size
FROM poochieProd AS prod
LEFT JOIN poochieStock AS stock ON prod.ProdID = stock.stockID
LEFT JOIN poochieSizes AS size ON stock.SizeID = size.SizeID
WHERE prod.ProdID = '%s' AND stock.stock > 0
ORDER BY size.SizeID ASC", GetSQLValueString($var3_Recordset1, "int")); <?php
$query2 = sprintf("
SELECT DISTINCT stock.stockID, size.Size
FROM poochieProd AS prod
LEFT JOIN poochieStock AS stock ON prod.ProdID = stock.stockID
LEFT JOIN poochieSizes AS size ON stock.SizeID = size.SizeID
WHERE prod.ProdID = '%s' AND stock.stock > 0
ORDER BY size.SizeID ASC", GetSQLValueString($var3_Recordset1, "int"));
SQL query:
<?php $query2 = sprintf(" SELECT DISTINCT stock.stockID, size.Size FROM poochieProd AS prod LEFT JOIN poochieStock AS stock ON prod.ProdID = stock.stockID LEFT JOIN poochieSizes AS size ON stock.SizeID = size.SizeID WHERE prod.ProdID = '%s' AND stock.stock > 0 ORDER BY size.SizeID ASC", GetSQLValueString($var3_Recordset1, "int"));
MySQL said:
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '<?php
$query2 = sprintf("
SELECT DISTINCT stock.stockID, size.Size
' at line 1
SELECT DISTINCT stock.stockID, size.Size
FROM poochieProd AS prod
LEFT JOIN poochieStock AS stock ON prod.ProdID = stock.stockID
LEFT JOIN poochieSizes AS size ON stock.SizeID = size.SizeID
WHERE prod.ProdID = '%s' AND stock.stock > 0
ORDER BY size.SizeID ASC
and replaced the %s with a 5 because i know there is a product id in the DB with the value of 5
i got these results
MySQL returned an empty result set (i.e. zero rows). (Query took 0.0004 sec)
SELECT DISTINCT stock.stockID, size.Size
FROM poochieProd AS prod
LEFT JOIN poochieStock AS stock ON prod.ProdID = stock.stockID
LEFT JOIN poochieSizes AS size ON stock.SizeID = size.SizeID
WHERE prod.ProdID = '5'
AND stock.stock >0
ORDER BY size.SizeID ASC
LIMIT 0 , 30
so its saying its empty even though there is a record in there?
<?php
$query2 = sprintf("
SELECT DISTINCT stock.stockID, size.Size
FROM poochieProd AS prod
LEFT JOIN poochieStock AS stock ON prod.ProdID = stock.ProdID
LEFT JOIN poochieSizes AS size ON stock.SizeID = size.SizeID
WHERE prod.ProdID = '%s' AND stock.stock > 0
ORDER BY size.SizeID ASC", GetSQLValueString($var3_Recordset1, "int"));
$results2 = mysql_query($query2);
while($row2 = mysql_fetch_array($results2)){
?>
but am still not getting the sizes?. with regards to my naming convention do you mean just change everything to lowercase?
Last edited by Benjamin on Thu Oct 04, 2012 5:19 pm, edited 1 time in total.
Reason:Added [syntax=php||htm||css||javascript||sql||etc] - Please use [syntax] tags when posting code in the forums! Thanks.