[英]JSON Multi-dimensional Array using mysql_fetch_assoc
我的Mysqli更新查询,并输出此
SELECT milestone.id, milestone.name, milestone.date, milestone.location, milestone.story_body, milestone.short_link, milestone.created_at,
GROUP_CONCAT(images.path) as path , images.update_type
FROM milestone
INNER JOIN images ON milestone.id = images.update_id
WHERE milestone.business_id = '1' && milestone.status = '1' && images.update_type = '3'
GROUP BY milestone.id
如何使用上述查询形成JSON对象?
我试过下面的方法,不会产生任何结果
$result_array = array();
while($row = mysql_fetch_assoc($result))
{
$result_array[] = $row;
}
我想要这样的东西-
[
{
"id":"4",
"name":"2nd anniversary",
"date":"2015-12-17",
"location":"Mumbai",
"story_body":"Gzjjs jdk djks jdks jdkd jx djdb djd JD djbd djdj d",
"short_link":"izWfs",
"created_at":"2015-12-11 03:49:52",
"path":
[
{"\/SupportData\/ImpalzB2B\/uploads\/90294930451448437444826.jpg"},
{"\/SupportData\/ImpalzB2B\/uploads\/90294930451449758248579.jpg"}
],
"update_type":"3"
},
{
"id":"7",
"name":"#1styearAnniversary",
"date":"2016-01-20",
"location":"Mumbai",
"story_body":"Bsjsj jdkdk djdkdk dkdkf kdkf dkfj fjfj fjfkjdd djkd",
"short_link":"FHXh0",
"created_at":"2016-01-20 23:10:54",
"path":"\/SupportData\/ImpalzB2B\/uploads\/11453356652175.jpg",
"update_type":"3"
}
]
注意:我知道Mysql在PHP 7中没有使用。我需要用PDO和Mysqli替换它,因此请忽略该错误。 我正在同一个工作,与此同时面对这个查询。
您现在可以执行此操作-
$result_array = array();
while($row = mysql_fetch_assoc($result))
{
$temp= explode(',', $row['path']); // explode by ,
if(count($temp) > 1) { // More than one element then assign
$row['path']= $temp;
}
$result_array[] = $row;
}
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