Query Failed: Unknown Column 'tenantid' In 'where Clause'

[sagisgirl]

I am getting the following error and can not figure out why:

 

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource

 

so, i made a bit changes to my code here

$rs_recTuser = mysql_query($query_recTuser)

 

to this

 

$rs_recTuser = mysql_query($query_recTuser) or die ('Query failed: ' . mysql_error(). "<br />\n$query_recTuser");

 

then a new error came :

 

Query failed: Unknown column 'tenantID' in 'where clause'

SELECT * FROM u_user WHERE tenantID = 9

 

here is my code:

 

$query_recTuser = "SELECT * FROM u_user WHERE tenantID = ".$tenantID;
$rs_recTuser = mysql_query($query_recTuser) or die ('Query failed: ' . mysql_error(). "<br />\n$query_recTuser");
$totalRows_recTuser = mysql_num_rows($rs_recTuser);

[Pikachu2000]

Evidently, there is no field by the name tenantID in your table.

[sagisgirl]

thank you..it worked fine now..thanks...

 

but i got another error.:

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

 

here is the code

 

$query_recTenantID = "SELECT * FROM md_tenant WHERE tenantID = ".$tenantID;
$rs_recTenantID = mysql_query($query_recTenantID);
$recTenantID = mysql_fetch_array($rs_recTenantID);

 

why is that error occur..?

Edited December 24, 2012 by sagisgirl

[MDCode]

Your query is failing add

echo mysql_error();

after mysql_query($query_recTenantID);

Edited December 24, 2012 by SocialCloud

[sagisgirl]

is it like this

 

$query_recTenantID = "SELECT * FROM md_tenant WHERE tenantID = ".$tenantID;
$rs_recTenantID = mysql_query($query_recTenantID);
echo mysql_error();
$recTenantID = mysql_fetch_array($rs_recTenantID);

 

still got the error..

 

im new in php..so please do understand me...

[Jessica]

What error did you get? Your original error was that tenantID is not a column. If it's not, either create it or figure out what column you wanted.

[sagisgirl]

i already create column for tenantID... the error is still the same even after i add

 

echo mysql_error();

 

here are the error:

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 '' at line 1

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

[Christian F.]

If you echo out the SQL query as well, you should notice what's missing from it. From there you should be able to determine what the problem is, and by tracing the problem upwards in the code you should find the cause.

[sagisgirl]

ok...i did that echo like this..

 

$query_recTenantID = "SELECT * FROM md_tenant WHERE tenantID = ".$tenantID;
$rs_recTenantID = mysql_query($query_recTenantID);
while($recTenantID = mysql_fetch_array($rs_recTenantID))
{ echo $recTenantID['tenantID']; }

 

but still i got the same error...

[Jessica]

Read the link in my signature. You're not capturing MySQL errors.

[sagisgirl]

sorry jessica..but i didn't understand you... what do you mean by i'm not capturing

mySQL errors..?

[sagisgirl]

ok i change my code to this..

 

$query_recTenantID = "SELECT * FROM md_tenant WHERE tenantID = ".$tenantID;
$rs_recTenantID = mysql_query($query_recTenantID) or die ('Query failed: ' . mysql_error(). "<br />\n $query_recTenantID");
while($recTenantID = mysql_fetch_array($rs_recTenantID))
{ echo $recTenantID['tenantID']; }

 

then it give an error :

 

Parse error: syntax error, unexpected ')'

 

why..?

[Pikachu2000]

The code you posted has no parse errors. It must be somewhere else.

[sagisgirl]

suddenly it give me a new error :

 

Query failed: 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 '' at line 1

SELECT * FROM md_tenant WHERE tenantID =

 

why...??

[sowna]

tenantID is empty here... check the tenantID variable...

[sagisgirl]

yeayh..i just realized that, the variable is empty...

 

ok,thanks for you helping...thanks guys...