CODEWITHSUNDEEP
phpsql-serversessionstored-proceduressqlsrv
cs-87
cs-87

Reputation: 216

SQL Query Execute error when using SQL SRV (Php)

I am trying to execute following query, but it provides a syntax error which i couldn't find. Appreciate your help to resolve the issue.

Error

Array ( [0] => Array ( [0] => 07002 [SQLSTATE] => 07002 [1] => 0 [code] => 0 [2] => [Microsoft][ODBC Driver 17 for SQL Server]COUNT field incorrect or syntax error [message] => [Microsoft][ODBC Driver 17 for SQL Server]COUNT field incorrect or syntax error ) )

<?php

session_start();
require_once '../../../config/dbconfig.php';

$sql = "EXEC    smeProBizInspecInsert
@strLeadRefNo   =  ? ,
@strCustName =  ? ,                 
@strEnteredBy  =  ? , 
@dtmEnteredDate =  '',
@strInspectStatus = 'SMS06',
@strBranch = ? ";

$stmt=sqlsrv_prepare($conn,$sql,array(&$_POST['strLeadRefNo'],&$_POST['customerName'],&$_SESSION['NAME'], &$_SESSION['BRANCH'])); 
sqlsrv_execute($stmt);

$stmt = sqlsrv_prepare($conn, $sql);
if (!sqlsrv_execute($stmt)) {die( print_r( sqlsrv_errors(), true) );}
else {echo json_encode("success");} 

?>

Upvotes: 1

Views: 158

Answers (1)

cs-87
cs-87

Reputation: 216

The issue was resolved, when i passed @dtmEnteredDate and @strInspectStatus through variables. Also i have mistakenly kept the following, which caused the error.

sqlsrv_execute($stmt); $stmt = sqlsrv_prepare($conn, $sql);

<?php

session_start();
require_once '../../../config/dbconfig.php';


$sql = "EXEC    smeProBizInspecInsert
@strLeadRefNo   =  ? ,
@strCustName =  ? ,                 
@strEnteredBy  =  ? , 
@dtmEnteredDate =  ? ,
@strInspectStatus = ?,
@strBranch = ? ";

$edate = '';
$status = 'SMS06';

$stmt=sqlsrv_prepare($conn,$sql,array(&$_POST['strLeadRefNo'],&$_POST['customerName'],&$_SESSION['NAME'], 
&$edate, &$status ,&$_SESSION['BRANCH'])); 

if (!sqlsrv_execute($stmt)) {

  die( print_r( sqlsrv_errors(), true) );


}


else 
{

  echo json_encode("success");

} 


?>

Upvotes: 1

Related Questions