我想尝试运行以下PHP代码:
$con = mysql_connect("localhost", "root", "") or die('connection not made');
$db = mysql_select_db('name', $con) or die('db not selected');
$query1 = "SELECT * FROM nodesensors WHERE NodeID=2";
$result1 = mysql_query($query1, $con);
$sensorids = mysql_fetch_array($result1);
$query2 = "SELECT SensorID, Variable FROM sensors WHERE SensorID IN($sensorids)";
$result2 = mysql_query($query2, $con) or die('query not made');
$sensors = mysql_fetch_array($result2);
echo $sensors;
我想要只获得具有SensorID的传感器,这也是'sensorids'数组中的值.当我运行代码时,我得到以下内容:
Notice: Array to string conversion in C:\...\test.php on line 10
query not made
当我删除"$"时如下:
$query2 = "SELECT SensorID, Variable FROM sensors WHERE SensorID …