如何使用mysql_fetch_assoc处理php中的记录？

Question

我正在尝试检索主题名称，有两种类型的主题可选和主要。每个学生都包含可选和主要科目。通过使用这些主题ID我可以从主题表中检索名称。 但是在这个结果中只出现了一个可选的主题名称，我的代码中是否有任何问题？请帮助我

  $subject_names = array();
for($i=0;$i<count($student_num_data);$i++)
{
    $optional_id_list = mysql_query("SELECT optional_subject_id FROM ms_student WHERE student_id = ".$student_num_data[$i]['student_id']);
    while($row = mysql_fetch_assoc($optional_id_list))
    {
        foreach ($row as $key) 
        {
            $optional_subject = mysql_query("SELECT subject_name FROM ms_subject WHERE subject_id = ".$key['optional_subject_id']); 
            $optional_subject_name = array();
            while($row1 = mysql_fetch_assoc($optional_subject))
            {
                $optional_subject_name[] = $row1;  
            }
        }
    }
    $subject_id_list = mysql_query("SELECT subject_ids FROM subject_config WHERE stream_index =".$stu_stream);  
    while($row = mysql_fetch_assoc($subject_id_list))
    {
        foreach ($row as $key) 
        {
            $main_subject = mysql_query("SELECT subject_name FROM ms_subject WHERE subject_id = ".$key['subject_ids']); 
            $main_subject_name = array();
            while($row3 = mysql_fetch_assoc($main_subject))
            {
                $main_subject_name[] = $row3;  
            }
        }
    } 
    $subject_names = $main_subject_name[0]['subject_name'].','.$optional_subject_name[0]['subject_name'];

    $small_subject_name=trim($subject_names);//remove whitespace at end of string         
    $small_subject_name = str_replace(" ", "-", $small_subject_name); // Replaces all spaces with hyphens.     
    $small_subject_name = preg_replace('/[^A-Za-z0-9\-]/', '', $small_subject_name); // Removes special chars.     
    $small_subject_name =  preg_replace('/-+/', '-', $small_subject_name); // Removes multiple hyphens.            
    $small_subject_name = str_replace("-", " ", $small_subject_name); // Replaces all hyphens with spaces.            
    $small_subject_name=strtolower(str_replace(" ","_",$small_subject_name));  

    if($stu_class == 11 || $stu_class == 12)
    {  
        $subject_statement = $subject_statement."IF(".$small_subject_name."='ab','ab',round(".$small_subject_name.",2)) AS ".$small_subject_name.",";   
        $sum_subject_statement = $sum_subject_statement.'IFNULL(`'.$small_subject_name.'`,0) + ';           
        /* start by 1022 02-09-2014 */
        $subject_pass_statement .= $small_subject_name." >= 33 AND ";       
        $subject_pass_statement = substr($subject_pass_statement,0,strlen($subject_pass_statement)-5);
        $update_pass_statement = "UPDATE ".$table_name." SET class_rank = 'P' WHERE 1=1 ".$track_div_stmnt.$id_stmnt." AND student_id = ".$student_num_data[$i]['student_id']." AND ".$subject_pass_statement;
        $update_pass_query=mysql_query($update_pass_statement);
    }
}

Answer 1

这不是关于mysql_fetch_assoc函数，而是关于代码的逻辑。你很难，主要是因为你的代码看起来如何。你一次遇到太多问题，这里最合理的解决方案是：

分解问题

1. 考虑一下这段代码的主要任务/操作是什么 让我们说：
   1. 对于每个学生，我们都会检索他们注册的可选科目。
   2. 然后我们检索他们正在服用的主要科目。
   3. 然后我们检索他们的结果以确定他们通过了哪些科目以及他们没有通过哪些科目。

2. 从表示您的意图的代码开始，使用尚不存在的函数：

    $passStatementForYear = new PassStatement($year);
    $students = getStudentsForSchoolYear($year);
    foreach ($students as $student) {
        $optionalSubjects = loadOptionalSubjectsForStudent($student);
        $mainSubjects = loadMainSubjectsForStudent($student);
        $passReport = prepareReportForStudent(
            $student, $optionalSubjects, $mainSubjects
        );
        updatePassStatementUsingStudentsReport(
            $passStatementForYear, $passReport
        );
    }
   

3. 构建缺少的部分（功能，类等）

现在您只需要处理这些独立的步骤。每个功能都引入了一个新的小问题，你可以用孤立和简单的方式来处理。只有这样，如果您发现自己遇到mysql_fetch_assoc问题，请参阅他们显示的官方文档和示例：

$sql = "SELECT id as userid, fullname, userstatus
        FROM   sometable
        WHERE  userstatus = 1";

$result = mysql_query($sql);

if (!$result) {
    echo "Could not successfully run query ($sql) from DB: " . mysql_error();
    exit;
}

if (mysql_num_rows($result) == 0) {
    echo "No rows found, nothing to print so am exiting";
    exit;
}

// While a row of data exists, put that row in $row as an associative array
// Note: If you're expecting just one row, no need to use a loop
while ($row = mysql_fetch_assoc($result)) {
    echo $row["userid"];
    echo $row["fullname"];
    echo $row["userstatus"];
}

mysql_free_result($result);