PHP:
session_start();
$school_id = $_SESSION['school_id'];
$faculty_id = $_SESSION['user_id_fac'];
$subject_id = $_POST['subject_id'];
$year_grade_level = $_POST['year_level'];
$subject_handeler_id = $_POST['subject_handler_id'];
$student_grades_boy = $_POST['student_grades_boy'];
$student_grades_girl = $_POST['student_grades_girl'];
$update_grades_boys = "UPDATE registrar_grade_archive SET";
//SET status = '0' WHERE subject_id = '$subject_id'"
$vaues_girl = array();
$values_boy = array();
foreach ($student_grades_boy as $key=>$data) {
$student_id_B= $data['studnt_B_id'];
$grade_B = $data['studnt_grade_B'];
$values_boy[$key] = 'grade = \''.$grade_B.'\' WHERE student_id = \''.$student_id_B.'\' AND subject_id = \''.$subject_id.'\' AND school_id = \''.$school_id.'\' AND advisor_faculty_id = \''.$faculty_id.'\' AND subject_handler_id = \''.$subject_handeler_id.'\' ' ;
}
$values_boy = implode(', ', $values_boy);
$ready_edit_grades_boy = $update_grades_boys . $values_boy;
$save_grades_boy = mysql_query($ready_edit_grades_boy) or die(mysql_error());
请帮助伙计们。提前谢谢
答案 0 :(得分:2)
这里有一些问题:
$student_grades_boy
包含多个项目,则您的sql将有多个WHERE
语句(您只能有1个); SET
与列名称之间的空格; mysql_
函数已弃用。答案 1 :(得分:1)
您似乎在SET
和grade
之间没有空格。
在这里添加空格应该可以解决问题:
$update_grades_boys = "UPDATE registrar_grade_archive SET ";
如果不这样做,如果您可以发布echo $ready_edit_grades_boy;
的结果并更新您的问题,那将会非常有用。
答案 2 :(得分:1)
尝试
$update_grades_boys = "UPDATE registrar_grade_archive SET ";
SET
之后需要一个空格..
答案 3 :(得分:0)
你没有逃避变种,所以它可能是你的价值观中的一些“或”。