php – 如何正确使用foreach()

我有以下代码

foreach ($_POST['courses'] as $course ) {
      $sql1 = "SELECT id FROM  `courses` where title='".$course."'";
      $course_id = MysqLi_query( $connection, $sql1 ); 
    $sql2=MysqLi_query($connection,"INSERT INTO `ett`.`course-teacher` (`courses_id` ,`teacher-id`) VALUES ('".$teacher_id."`,`".$course_id."`)"); }

课程是一个多选的html标签,我基本上试图选择所有已选择的课程.

但它给了我以下错误

Catchable Fatal error: Object of class MysqLi_result Could not be converted to string in C:\wamp\www\ETT logo\Controlpanel\uploadteacher.PHP on line 36

请注意,第35行是上面代码中的第2行.

解决方法:

MysqLi_query()将返回一个结果对象,而不仅仅是您想要的字段.您需要获取结果数据 – 只是隐式地将其转换为字符串将不起作用.

foreach ($_POST['courses'] as $course ) {
      $sql1 = "SELECT id FROM  `courses` where title='".$course."'";
      $result = MysqLi_fetch_assoc(MysqLi_query( $connection, $sql1 )); 
      $result2 = MysqLi_fetch_assoc(MysqLi_query($connection,"INSERT INTO `ett`.`course-teacher` (`courses_id` ,`teacher-id`) VALUES ('".$teacher_id."`,`".$result['id']."`)"); }