$sql = "SELECT * FROM `demo_viewer` WHERE 1 ";
$viewer = mysqli_query($mysqli,$sql);
$view = array();
while ($row = mysqli_fetch_array($viewer)){
$view[] = $row;
}
$viewer = json_encode($view,JSON_NUMERIC_CHECK);为什么这是代码没有显示数据..。db连接正常,但是如果代码下面的数据显示..。
$sql = "SELECT SUM(numberofview) as count FROM demo_viewer
GROUP BY YEAR(created_at) ORDER BY created_at";
$viewer = mysqli_query($mysqli,$sql);
$viewer = mysqli_fetch_all($viewer,MYSQLI_ASSOC);
$viewer = json_encode(array_column($viewer, 'count'),JSON_NUMERIC_CHECK);发布于 2017-06-06 07:31:17
尝尝这个,
mysqli_query($mysqli,'SET NAMES utf8mb4');
$sql = "SELECT * FROM `demo_viewer` WHERE 1 ";
$viewer = mysqli_query($mysqli,$sql);
$view = array();
while ($row = mysqli_fetch_array($viewer,MYSQLI_ASSOC)){
$view[] = $row;
}
$viewer = json_encode($view, JSON_PARTIAL_OUTPUT_ON_ERROR | JSON_FORCE_OBJECT | JSON_NUMERIC_CHECK);编辑
mysqli_query($mysqli,'SET NAMES utf8mb4');
$sql = "SELECT numberofclick FROM `demo_viewer` GROUP BY YEAR(created_at) ORDER BY created_at ";
$viewer = mysqli_query($mysqli,$sql);
$view = array();
while ($row = mysqli_fetch_array($viewer,MYSQLI_ASSOC)){
$view[] = $row['numberofclick'];
}
$viewer = json_encode($view, JSON_PARTIAL_OUTPUT_ON_ERROR | JSON_FORCE_OBJECT | JSON_NUMERIC_CHECK);PS:您不应该使用select * from表。因为High图表需要一个列值。不是所有的列。
https://stackoverflow.com/questions/44382845
复制相似问题