我正在使用SQLSRV 3.0进行PHP并针对MSSQL2008工作。
我遇到的问题是我无法让查询中的 LIKE 语句正常工作?
$connectionOptions = array( "Database"=>$myDB, "UID"=>$myUser, "PWD"=>$myPass);
$conn = sqlsrv_connect( $myServer, $connectionOptions);
$params = array(
"$sID%",
"$sUser%"
);
$sql = "select * from tbl where col2 LIKE ? or col2 LIKE ?";
$stmt = sqlsrv_query($conn, $sql, $params);
if( $stmt === false) {
die( print_r( sqlsrv_errors(), true) );
}
问题与参数数组中项目的设计有关,有什么建议吗?
问候,贾基姆
将代码调整为以下内容:
$sql = "select * from tbl where col2 LIKE ? or col2 LIKE ?";
$params = array($sID."%", $sUser."%");
$stmt = sqlsrv_query( $conn, $sql, $params);
尝试这样做:
$strSQL = "SELECT * FROM tbl WHERE col2 like '%' + CONVERT(NVARCHAR, ?) + '%' or col2 like '%' + CONVERT(NVARCHAR, ?) + '%' ";
$searchTerm = "<<some multibye characters>>";
$searchTerm = iconv('utf-8', 'utf-16le', $searchTerm);
$params = array( array($searchTerm, NULL, SQLSRV_PHPTYPE_STRING(SQLSRV_ENC_BINARY)));