Question

我有更新学生记录的代码。

$select_query="select student_id from student_information where student_id = '$student_id'";
$result_set = mysql_query($select_query,$link_id);

if($row = mysql_fetch_array($result_set)){
    $flag="exists";
    header("location:Admin_Home.php?flag=$flag&student_id=$student_id");
    die();
}
else{
    /*
        This block is used to insert the learners record in database 
        if the student_id is not yet registered in the database. 
    */
    mysql_query("SET AUTOCOMMIT = 0 ");
    if(mysql_error() != null){
        die(mysql_error());
    }
    $query = "insert into student_information(student_id,student_password,first_name,last_name,registration_date,gender,date_of_birth,";
    $query .= "contact_no,grade,section,LRN,email1,email2,address,description,learner_id)";
    $query .= " values('$student_id','$student_password','$first_name','$last_name',now(),'$gender','$date_of_birth',";
    $query .= "'$contact_no','$grade','$section','$LRN','$email1','$email2','$address','$description','$learner_id')";
    $result = mysql_query($query,$link_id);
    if(mysql_error() != null){
        die(mysql_error());
    }
}

现在，我的问题是放置一个脚本，提醒“记录已更新”，并插入此代码的哪一部分。请指教。

Answer 1

改变这个：

if(mysql_error() != null){
    die(mysql_error());
}

对此：

if (mysql_error() != null) {
    die(mysql_error());
}
echo "Records updated.";

添加一个表示RECORDS UPDATED的脚本

1 个答案: