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我使用以下 php 查询从 mySQL 获取此 JSON 结果：

/* grab the posts from the db */
 $query = "SELECT * FROM ko_timetable";
 $result = mysql_query($query,$link) or die('Errant query:  '.$query);

 /* create one master array of the records */
 $posts = array();

if(mysql_num_rows($result)) {

while($post = mysql_fetch_assoc($result)) {
      $posts[] = array('post'=>$post);
    } 
}

JSON结果：

post =         {
"CLASS_LEVEL" = "Intro/General";
"CLASS_TYPE" = "Muay Thai";
"DAY_OF_WEEK" = Sunday;
ID = 19;
"ORDER_BY" = 5;
TIME = "1:00pm - 2:30pm";
};
}
{
post =         {
"CLASS_LEVEL" = "General/Intermediate/Advanced";
"CLASS_TYPE" = "Muay Thai Spar - Competitive";
"DAY_OF_WEEK" = Sunday;
ID = 27;
"ORDER_BY" = 5;
TIME = "6:00pm - 9:00pm";
};
},
{
post =         {
"CLASS_LEVEL" = "Fighters/Advanced/Intermediate";
"CLASS_TYPE" = "Fighters Training";
"DAY_OF_WEEK" = Monday;
ID = 1;
"ORDER_BY" = 1;
TIME = "9:30am - 11:00pm";
};

但是如何更改此查询以按“DAY_OF_WEEK”获取此结果组。请参阅下面的示例，感谢您的帮助：

{
Sunday =         {
"CLASS_LEVEL" = "Intro/General";
"CLASS_TYPE" = "Muay Thai";
"DAY_OF_WEEK" = Sunday;
ID = 19;
"ORDER_BY" = 5;
TIME = "1:00pm - 2:30pm";
};
{
"CLASS_LEVEL" = "General/Intermediate/Advanced";
"CLASS_TYPE" = "Muay Thai Spar - Competitive";
"DAY_OF_WEEK" = Sunday;
ID = 27;
"ORDER_BY" = 5;
TIME = "6:00pm - 9:00pm";
};
},
{
Monday =         {
"CLASS_LEVEL" = "Fighters/Advanced/Intermediate";
"CLASS_TYPE" = "Fighters Training";
"DAY_OF_WEEK" = Monday;
ID = 1;
"ORDER_BY" = 1;
TIME = "9:30am - 11:00pm";
};

谢谢

4

2 回答 2

2

您可以像这样使用 DAY_OF_WEEK 作为数组索引，

while($post = mysql_fetch_assoc($result)) {
      $posts[$posts["DAY_OF_WEEK"]] = array('post'=>$post);
    }

但请记住，在示例中，您显示数组的索引是日期的名称，然后上周日的结果将被本周周日替换，依此类推。您还可以使用计数变量来避免用重复名称替换键，就像这样。

$count = 1;
while($post = mysql_fetch_assoc($result)) {
      $posts[$posts["DAY_OF_WEEK"].$count] = array('post'=>$post);
      $count++;
    }
于 2012-05-04T09:56:39.430 回答
1

改变这个：

while($post = mysql_fetch_assoc($result)) {
    $posts[ $post['DAY_OF_WEEK'] ] = $post;
}
于 2012-05-04T08:42:39.717 回答