php 如何从mysql数据库构建JSON数组
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How to build a JSON array from mysql database
提问by Stan Forrest
Okay I have been racking my brain trying to build a JSON array from mysql. The array MUST be in the following format. I am using fullcalendar and want to make the events on the calendar dynamic. Below is the code that builds the array, but currently it does not get the information from mysql
好吧，我一直在绞尽脑汁试图从 mysql 构建一个 JSON 数组。数组必须采用以下格式。我正在使用 fullcalendar 并希望使日历上的事件动态化。下面是构建数组的代码，但目前它没有从mysql获取信息
$year = date('Y');
$month = date('m');
echo json_encode(array(
//Each array below must be pulled from database
//1st record
array(
'id' => 111,
'title' => "Event1",
'start' => "$year-$month-10",
'url' => "http://yahoo.com/"
),
//2nd record
array(
'id' => 222,
'title' => "Event2",
'start' => "$year-$month-20",
'end' => "$year-$month-22",
'url' => "http://yahoo.com/"
)
));
回答by jk.
Is something like this what you want to do?
你想做这样的事情吗？
$return_arr = array();
$fetch = mysql_query("SELECT * FROM table");
while ($row = mysql_fetch_array($fetch, MYSQL_ASSOC)) {
$row_array['id'] = $row['id'];
$row_array['col1'] = $row['col1'];
$row_array['col2'] = $row['col2'];
array_push($return_arr,$row_array);
}
echo json_encode($return_arr);
It returns a json string in this format:
它以这种格式返回一个 json 字符串：
[{"id":"1","col1":"col1_value","col2":"col2_value"},{"id":"2","col1":"col1_value","col2":"col2_value"}]
OR something like this:
或者像这样：
$year = date('Y');
$month = date('m');
$json_array = array(
//Each array below must be pulled from database
//1st record
array(
'id' => 111,
'title' => "Event1",
'start' => "$year-$month-10",
'url' => "http://yahoo.com/"
),
//2nd record
array(
'id' => 222,
'title' => "Event2",
'start' => "$year-$month-20",
'end' => "$year-$month-22",
'url' => "http://yahoo.com/"
)
);
echo json_encode($json_array);
回答by Wrikken
The PDO solution, just for a better implementation then mysql_*:
PDO 解决方案，只是为了更好的实现mysql_*：
$array = $pdo->query("SELECT id, title, '$year-month-10' as start,url
FROM table")->fetchAll(PDO::FETCH_ASSOC);
echo json_encode($array);
Nice feature is also that it will leave integers as integers as opposed to strings.
不错的功能还在于它将整数保留为整数而不是字符串。
回答by AndroLogiciels
Just an update for Mysqli users :
只是 Mysqli 用户的更新：
$base= mysqli_connect($dbhost, $dbuser, $dbpass, $dbbase);
if (mysqli_connect_errno())
die('Could not connect: ' . mysql_error());
$return_arr = array();
if ($result = mysqli_query( $base, $sql )){
while ($row = mysqli_fetch_assoc($result)) {
$row_array['id'] = $row['id'];
$row_array['col1'] = $row['col1'];
$row_array['col2'] = $row['col2'];
array_push($return_arr,$row_array);
}
}
mysqli_close($base);
echo json_encode($return_arr);
回答by Bijender Singh Shekhawat
Use this
用这个
$array = array();
$subArray=array();
$sql_results = mysql_query('SELECT * FROM `location`');
while($row = mysql_fetch_array($sql_results))
{
$subArray[location_id]=$row['location']; //location_id is key and $row['location'] is value which come fron database.
$subArray[x]=$row['x'];
$subArray[y]=$row['y'];
$array[] = $subArray ;
}
echo'{"ProductsData":'.json_encode($array).'}';
