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我想从一个MySQL查询建立一个JSON对象。我正在努力争取的JSON结构应该是这样的:PHP的阵列到JSON
{
"a": [
{
"user": "alb",
"time": "2011-09-12 05:56:36"
},
{
"user": "arg",
"time": "2011-09-12 05:56:36"
}
]
"b": [
{
"user": "blah",
"time": "2011-09-12 05:56:36"
},
{
"user": "bleh",
"time": "2011-09-12 05:56:36"
}
]
}
然而,我的代码总是返回包裹在一个阵列,像这样一个JSON对象:
[
{
"a": [
{
"user": "alb",
"time": "2011-09-12 05:56:36"
},
{
"user": "arg",
"time": "2011-09-12 05:56:36"
}
]
"b": [
{
"user": "blah",
"time": "2011-09-12 05:56:36"
},
{
"user": "bleh",
"time": "2011-09-12 05:56:36"
}
]
}
]
这里是PHP代码我使用:
<?php
$json_data = array();
foreach($blogs as $blog)
{
$sql = "SELECT * from users";
$query = mysql_query($sql);
$ablog = array();
while ($row = mysql_fetch_assoc($query))
{
$json_element = array(
"user"=> $row[username] ,
"time"=> $row[time]
);
array_push($ablog,$json_element);
}
$eblog = array($blog => $ablog);
array_push($json_data,$eblog);
}
$json_output = json_encode($json_data);
print $json_output;
?>
我想知道:为什么我得到一个数组中包装的JSON对象?我在上面的代码中错误地做了什么?
谢谢。
这两个json都是无效的,你需要在**“b”之前加逗号:** – Gowri
你应该发布程序的* actual *输出 – meagar