Mysql无法保存一些记录 [英] Mysql fails to save some records
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问题描述
我有一个用PHP开发的多用户应用程序来管理学生记录。该应用程序使用foreach循环来更新每个学生记录。在大多数情况下，应用程序工作正常，但有几次用户报告他们的记录尚未保存可能是这个问题的原因？
我尝试了什么：
I have a multi-user application developed in PHP to manage student records. The application uses a foreach loop to update each student record. In most instances, the application works just fine but there are a few times when users report that their records have not been saved What could be the cause of this problem?
What I have tried:
<pre>foreach($ExamMarks as $Key=> $Value)
{
$Student = array(explode(" ",$Value)); //This will convert the Values into a multi-dimensional array and store the new values in $Student
if(trim($_POST[$Student[0][1]]) == "Null" || trim($_POST[$Student[0][1]]) == " ")//If the student has a mark
{
$Grade = "";
$Comments = "";
$Points = "";
}
else if(trim($_POST[$Student[0][1]]) >= 80)
{
$Grade = "A";
$Comments = "Outstanding performance. Keep it up!";
$Points = 9;
}
else if (trim($_POST[$Student[0][1]]) >= 70)
{
$Grade = "B";
$Comments = "Good performance. Aim higher.";
$Points = 7;
}
else if (trim($_POST[$Student[0][1]]) >= 60)
{
$Grade = "C";
$Comments = "Fairly good performance. Work hard.";
$Points = 5;
}
else if (trim($_POST[$Student[0][1]]) >= 40)
{
$Grade = "D";
$Comments = "Average performance. Work harder.";
$Points = 3;
}
else if (trim($_POST[$Student[0][1]]) >= 20)
{
$Grade = "E";
$Comments = "Below Average. Put more effort.";
$Points = 1;
}
else
{
$Grade = "U";
$Comments = "Ungraded";
$Points = 0;
}
$updateSQL = "UPDATE student_exam_marks_exam SET
ExamMark = " . preg_replace('#[^0-9]#i', '',trim($_POST[$Student[0][1]])) . ",
ExamGrade = '" . $Grade . "',
Comments = '" . $Comments . "',
Points = " . $Points . " ,
TeacherID = '" . $row_Select_Teacher["TeacherID"] . "'
WHERE ExamID = " . $ExamID . " AND StudentID = " . trim($Student[0][0]) . "
AND SubjectID = " . $_GET["SubjectID"] . " AND AcademicYear = " . $row_Select_Exams["AcademicYear"].";" ;
mysqli_select_db($con, $database);
$Result1 = mysqli_query($con,$updateSQL) or die(mysql_error($con));
}
推荐答案
ExamMarks as
ExamMarks as
Key =>
Key=>
Value）
{
Value) {
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