我想用PHP从html请求一个日期时间变量,
,并使用该变量从mysql数据库中选择数据。然后创建一个json来生成highchart。
成功生成json,但没有生成highchart.
我用Firebug它显示TypeError: obj is null
这是我的代码
PHP
<?php
header('Content-type: application/json');
$con = mysql_connect("localhost","root","");
mysql_select_db("project", $con);
$month=$_REQUEST['select_month'];
$sqlChart_Q="SELECT T,value FROM `Temperature`WHERE month(T)='$month'";
$sqlChart_R=mysql_query($sqlChart_Q);
while($arrChart_ROW=mysql_fetch_array($sqlChart_R)) {
$time=strtotime($arrChart_ROW[0])*1000;
$temp=(float)$arrChart_ROW[1];
$arrChart[]=array("time" => $time,"value" => $temp);
}
echo json_encode($arrChart);
?>
和js
var chart;
$(document).ready(function() {
var options = {
chart: {
renderTo: 'month',
type: 'spline',
},
title: {
},
xAxis: {
type: 'datetime'
},
yAxis: {
},
series: [{
name: 'Temperature',
data: []
}]
};
$.getJSON('PHP/db_month.php', function(json) {
temp = [];
$.each(json, function(key,value) {
temp.push([value.time,value.value]);
});
options.series[0].data = temp;
chart = new Highcharts.Chart(options);
});
});
json (select_month=5;月美元= $ _REQUEST [' select_month '])
[{"time":1400076317000,"value":50},{"time":1400076322000,"value":25},{"time":1400076327000,"value":34},{"time":1400110911000,"value":50},{"time":1400110916000,"value":43},{"time":1400110919000,"value":75},{"time":1400110920000,"value":35},{"time":1400110922000,"value":46},{"time":1400110924000,"value":66},{"time":1400110925000,"value":73},{"time":1400113566000,"value":20},{"time":1400161118000,"value":55},{"time":1400186496000,"value":50},{"time":1400193165000,"value":43},{"time":1400196532000,"value":50}]
顺便说一下,当sql函数改为
时 SELECT T,value FROM `Temperature`WHERE month(T)='5'
它将成功生成数据在五月创建的highchart。
(PS: T的类型是时间戳,我不知道是不是很重要。
我的代码有什么问题?有人能帮我解决这个问题吗?非常感谢!!
您需要在json_encode函数中设置JSON_NUMERIC_CHECK,因为您返回的是字符串,而不是数字。
当查询返回一个空集时,变量$arrChart没有定义。你必须在循环之前定义这个变量。
$arrChart = array();
也你的sql查询是不安全的。你必须这样写:
$sqlChart_Q = "SELECT T,value FROM `Temperature`WHERE month(T)=" . (int) $month;