我已经从php触发了存储过程。我还传递了输入参数,如下所示。
$id = 1;
$nameDetail = 'raj';
$result = mysqli_query('CALL InsertDetails($id,$nameDetail)');但是得到了低于错误。
mysqli_query() expects at least 2 parameters, 1 given ...请给出一个解决方案。
发布于 2017-03-07 20:58:51
问题是您没有设置mysqli connection.please,试试这个
$connection = mysqli_connect('localhost','username','password','db');
$result = mysqli_query($connection,'CALL InsertDetails($id,$nameDetail)');发布于 2017-03-07 20:59:41
您需要将连接作为第一个参数传入:
// connect to DB
$connect = mysqli_connect("127.0.0.1", "my_user", "my_password", "my_db");
// run procedure
$result = mysqli_query($connect, 'CALL InsertDetails($id,$nameDetail)');https://stackoverflow.com/questions/42648828
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