我从数据库查询中返回以下数据:
+---------------+-----------+------------------+--------------+-------+ | district_name | school_id | school_name | section | score | +---------------+-----------+------------------+--------------+-------+ | My ISD | 11 | My First School | English | 20 | | My ISD | 11 | My First School | Math | 23 | | My ISD | 11 | My First School | Reading | 24 | | My ISD | 11 | My First School | Science | 23 | | My ISD | 12 | My Second School | English | 11 | | My ISD | 12 | My Second School | Math | 19 | | My ISD | 12 | My Second School | Reading | 22 | | My ISD | 12 | My Second School | Science | 26 | +---------------+-----------+------------------+--------------+-------+
我需要将这些数据放入一个数组中,以便通过学校轻松输出得分表:
School English Math Reading Science ------------------------------------------------------- My First School 20 23 24 23 My Second School 11 19 22 26
我无法将此数据格式化为完成此操作的数组。理想的结构是:
array(
$schoolName => array(
'results' => array(
'section' => $section_name
'score' => $score
),
),
);
我尝试了几种方法,但无法让它们正常工作。
答案 0 :(得分:0)
$table_rows = array();
while ($row = mysql_fetch_assoc($mysql_result)) {
$school_name = $row['school_name'];
$table_rows[$school_name]['results'][$row['section']] = $row['score'];
// adding the other stuff to $table_rows...
}
我希望这会有所帮助。
答案 1 :(得分:0)
while ($row = mysql_fetch_array($query)) { //you can define the query, right?
$array[$row['school_name']]['results'][$row['section']] = $row['score'];
}
虽然我不支持你设置数据库的方式,但我不打算给你讲讲。我认为这是一个学校项目或其他什么。