best way to tell is to place .....'or die(mysql_error())' after any call to
mysql_query() to see what mysql is complaining about. personally I
sometimes declare the query first so i can echo that as well after the
msyql_error() function, so i can see the query that has just been
excecuted. helpful with dynamic queries.
$qry = "SELECT * FROM $tableName where $fieldName='$fieldValue'";
$result = mysql_query($qry, $conn) or die(mysql_error() . $qry);
hth
jeff
"Mike Klein"
<mikeklein@ieee.o To: php-db@lists.php.net
rg> cc:
Subject: Help needed with variable scoping? or mysql problem
08/14/2003 08:58
AM
I've been using JSP for some time now, and thought I'd write a couple of
rdbms explorers (flat tables, simple master/detail paradigm) using other
technologies, like PHP, Cocoon, etc. Well...it's been fun. I have more
working than not. But now I am hung up. I am getting wierd results in a php
script when I simply change the order of some mysql calls.
I'm using php-4.3.2, rh9, and mysql 3.23.
My error is the following (the first line above the Warning below is a dump
of parameters to the showMaster function which is what's bombing):
conn=Resource id #3 databaseName=info tableName=movies fieldName=title
fieldValue=36th Chamber
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result
resource in /foo/bar/mywebsite/private/database.php on line 107
Line 107 is the <first> mysql_num_rows call below.
function showMaster($conn, $databaseName, $tableName, $fieldName,
$fieldValue)
{
echo "conn=$conn\n";
echo "databaseName=$databaseName\n";
echo "tableName=$tableName\n";
echo "fieldName=$fieldName\n";
echo "fieldValue=$fieldValue\n";
========This code fails when placed here==========
$result = mysql_query("SELECT * FROM $tableName where
$fieldName='$fieldValue'", $conn);
$numRows = mysql_num_rows($result);
if(mysql_num_rows($result) == 1)
{
showDetail($conn, $databaseName, $tableName, $fieldName,
$fieldValue, $result);
exit;
}
echo "numRows=$numRows\n";
========================================
$fields = mysql_list_fields($databaseName, $tableName, $conn);
$numFields = mysql_num_fields($fields);
echo "<TABLE border=1 width=100%>\n";
echo "<tr>";
for($i = 0; $i < $numFields; $i++)
{
printf("<td>%s</td>", mysql_field_name($fields, $i));
}
echo "</tr>";
========The SAME code succeeds when placed here!!!!!==========
$result = mysql_query("SELECT * FROM $tableName where
$fieldName='$fieldValue'", $conn);
$numRows = mysql_num_rows($result);
if(mysql_num_rows($result) == 1)
{
showDetail($conn, $databaseName, $tableName, $fieldName,
$fieldValue, $result);
exit;
}
echo "numRows=$numRows\n";
========================================
while($myrow = mysql_fetch_array($result))
...rest of method...
}
Any ideas on why this is? When I move the lines above (from
$result=...to...echo 'numRows') down a few lines to just before the
mysql_fetch_array, then everything works. Whazzup?!?
mike klein
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