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mysql_fetch_array()

 
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mysql_fetch_array(), trying to join 2 tables in 2 different DBs together
Ladydice
post 9 Jul, 2008 - 12:07 AM
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i've been having a hard time trying to join 2 tables in 2 different DBs together. This is the first time i'm doing this so please tell me if there's something wrng.

CODE

$result2=mysql_query("SELECT * FROM Cash.Payment
JOIN Contest.Details
ON Cash.Payment.INVOICE = Contest.Details.INVOICE  
WHERE Cash.Payment.USER_NAME = '$_COOKIE[ID_my_site]'
LIMIT $offset, $rowsPerPage");

while($row= mysql_fetch_array($result2)) {

echo "<td>" . $row['INVOICE']. "</td>";
echo "<td>" . $row['PAYMENT_STATUS']. "</td>";
echo "<td>" . $row['PAYMENT_DATE']. "</td>";
echo "<td>" . $row['AMOUNT']. "</td>";
echo "<td>" . $row['TRANS_ID']. "</td>";
echo "<td>" . $row['PAYER_STATUS']. "</td>";
echo "<td>" . $row['PAYER_STATUS']. "</td>";


i keep getting an error on my while statement.
error: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

help would be much appreciated smile.gif
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EvinOwen
post 11 Jul, 2008 - 01:40 PM
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the error you are getting is because there is a problem with your query. it means that you are passing a value to your mysql_fetch_array() function that is not the result of a mysql_query(), meaning that the query failed and has returned null. i would suggest checking to see if your result is null before attempting to access it, though its not necessary.

add 'or die(mysql_error())' after your mysql_query() function, and you can get the error from your query...

like this:

CODE

$result2 = mysql_query("SELECT * FROM Cash.Payment JOIN Contest.Details ON Cash.Payment.INVOICE = Contest.Details.INVOICE WHERE Cash.Payment.USER_NAME = '$_COOKIE[ID_my_site]' LIMIT $offset, $rowsPerPage") or die(mysql_error());
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Ladydice
post 21 Jul, 2008 - 02:24 AM
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QUOTE(EvinOwen @ 11 Jul, 2008 - 01:40 PM) *

the error you are getting is because there is a problem with your query. it means that you are passing a value to your mysql_fetch_array() function that is not the result of a mysql_query(), meaning that the query failed and has returned null. i would suggest checking to see if your result is null before attempting to access it, though its not necessary.

add 'or die(mysql_error())' after your mysql_query() function, and you can get the error from your query...

like this:

CODE

$result2 = mysql_query("SELECT * FROM Cash.Payment JOIN Contest.Details ON Cash.Payment.INVOICE = Contest.Details.INVOICE WHERE Cash.Payment.USER_NAME = '$_COOKIE[ID_my_site]' LIMIT $offset, $rowsPerPage") or die(mysql_error());



thanks alot for the tip.... biggrin.gif icon_up.gif
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