CODEWITHSUNDEEP
phpsql-serversqlsrv
Rubin Anbin
Rubin Anbin

Reputation: 328

sqlsrv_query() expects parameter 1

I was trying to run the following query. 

SELECT 
   min(rd.region) AS region, 
   min(ri.ri_id) AS ri_id, 
   ri.service_date, 
   min(ri.tt_route_start_time) AS tt_route_start_time, 
   min(ri.tt_route_end_time) AS tt_route_end_time, 
   min(ri.tgqh_exit_time) AS tgqh_exit_time, 
   min(ri.violation_vehicle_details) AS violation_vehicle_details, 
   min(ri.tt_sop_violation_start_route) AS tt_sop_violation_start_route, 
   min(ri.crew_violation_details) AS crew_violation_details, 
   rd.route_name, 
   vd.v_no, 
   min(ri.vehicle_violation_justify) AS vehicle_violation_justify, 
   min(ri.crew_violation_justify) AS crew_violation_justify, 
   min(ri.start_time_violation_justify) AS start_time_violation_justify, 
   min(ri.v_justification) AS v_justification, 
   min(ri.c_justification) AS c_justification, 
   min(ri.st_justification) AS st_justification, 
   min(vd.v_type) AS v_type 
FROM vehicles3.ride_details AS ri 
INNER JOIN vehicles3.route_details AS rd ON ri.route_id = rd.r_id 
INNER JOIN vehicles3.vehicle_details AS vd ON vd.v_id = ri.v_id 
WHERE 
   ri.status = 1 AND 
   rd.r_id = '93'AND 
   ri.service_date = '2019-10-14' 
GROUP BY ri.service_date, rd.route_name, vd.v_no, rd.r_id 
ORDER BY ri.service_date, rd.route_name, vd.v_no, rd.r_id

But when I run it shows the following errors:

Warning: sqlsrv_query() expects parameter 1 to be resource,

Warning: sqlsrv_num_rows() expects parameter 1 to be resource

But, when I echoed run the query and run it in SQL Server it works perfectly.

Upvotes: 2

Views: 11922

Answers (1)

user2342558
user2342558

Reputation: 6693

expects parameter 1 to be resource means that you are passing to that function a null resource, maybe because the connection to database failed.

The same issue about the sqlsrv_num_rows error.

You must check for the connection status before performing queries.

From the official documentation:

$serverName = "serverName\sqlexpress";
$connectionInfo = array( "Database"=>"dbName", "UID"=>"username", "PWD"=>"password" );
$conn = sqlsrv_connect( $serverName, $connectionInfo);
if( $conn === false ) {
     die( print_r( sqlsrv_errors(), true));
}

Upvotes: 2

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