我将不胜感激这方面的一些帮助。我已经实现了我想要的,但我不知道这是否是最好的方法。
第一次我查询了数据库。
mysql_select_db($database);
$sql_many_sections ="
SELECT sections.subject, sections.section_number, sections.classroom,
sections.level_id, sections.course, sections.campus, sections.times,
sections.shift
FROM sections
LEFT JOIN section_teacher
ON sections.section_number = section_teacher.section_number
LEFT JOIN profile
ON profile.teacher_number = section_teacher.teacher_number
WHERE profile_id ='{$colname_teacher}'
ORDER BY sections.section_number ASC";
// bring out the results as an array so I can loop through them later
$result = mysql_query($sql_many_sections);
confirm_query($result);
while ($teaching_sections = mysql_fetch_array($result)) {
$teaches[] = $teaching_sections;
}
然后我 print_r 的结果是哪个。
Array
(
[0] => Array
(
[0] => English
[subject] => English
[1] => 32346
[section_number] => 32346
[2] => G635
[classroom] => G635
[3] => 2
[level_id] => 2
[4] => 236
[course] => 236
[5] => Male Science
[campus] => Male Science
[6] => 8 - 9:40 am, 10 - 11:40 am
[times] => 8 - 9:40 am, 10 - 11:40 am
[7] => AM
[shift] => AM
)
[1] => Array
(
[0] => English
[subject] => English
[1] => 49493
[section_number] => 49493
[2] => 1328
[classroom] => 1328
[3] => 2
[level_id] => 2
[4] => 236
[course] => 236
[5] => Male Humanities
[campus] => Male Humanities
[6] => 8 - 9:40 am, 10 - 11:40 am
[times] => 8 - 9:40 am, 10 - 11:40 am
[7] => AM
[shift] => AM
)
)
当我现在想像这样返回这些数据时
• Section: 32346
• Subject: English
• Classroom: G635
• Level: 2
• Course: 236
• Campus: Male Science
• Shift: AM
• Shift times: 8 - 9:40 am, 10 - 11:40 am
我可以让它工作的唯一方法是执行以下操作,这是一个意外。实现这些相同结果的正确方法是什么。谢谢
<?php
foreach($teaches as $value)
{ ?>
<ul>
<li>Section: <? echo $value[1];?></li>
<li>Subject: <? echo $value[0];?></li>
<li>Classroom: <? echo $value[2];?></li>
<li>Level: <? echo $value[3];?></li>
<li>Course: <? echo $value[4];?></li>
<li>Campus: <? echo $value[5];?></li>
<li>Shift: <? echo $value[7];?></li>
<li>Shift times: <? echo $value[6];?></li>
</ul>
<?
}
?>