我正在处理一个php代码,我插入两个表并希望从插入的第一个表中获取id,现在我得到此错误:Call to undefined function sqlsrv_field()
。我正试图从表routines
中获取routine_id
。抓取最后一个插入ID sqlsrv
代码:
date_default_timezone_set('Europe/Oslo');
$date = strftime ('%Y-%m-%d');
$time = strftime('%H:%M:%S');
$value = $_GET['Temp'];
$conn = sqlsrv_connect('BILAL' , $conn_array);
$sql = "Insert into routines (date, time, value, emp_id) values ('$date', '$time', '$value', (SELECT id FROM emps WHERE user_name='Arduino'))";
if (sqlsrv_begin_transaction($conn) === false) {
die(print_r(sqlsrv_errors(), true));
}
$query = sqlsrv_query($conn, $sql);
if($query === false) {
die(print_r(sqlsrv_errors(), true));
}
sqlsrv_next_result($query);
sqlsrv_fetch($query);
$id = sqlsrv_field($query,0);
$sql2 = "Insert into measure_routines (routine_id, measure_id, pool_id) values ('$id', (Select id from measurements where title='A_Auto_Temperatur'), 1)";
$stmt = sqlsrv_query($conn, $sql2);
if($stmt === false) {
die(print_r(sqlsrv_errors(), true));
}