我想推动JSON编码形式我的数据库选择到chart.js数据集,但即时通讯不知道如何去解决它,而没有过度工程。推动JSON编码的PHP到chart.js
下面是一个简单的选择,从它的追溯表分数:
if($teamData == 0){
$allTeams = 'All';
} else{
$sql = "SELECT * FROM compScore WHERE memberId = 1";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$jsonScores = json_encode($row, JSON_PRETTY_PRINT);
}
}
}
header('Content-type: application/json');
echo $jsonScores;
的这个输出是:
{ "id": "1", "score1": "2", "score2": "3", "score3": "5", "score4": "4", "score5": "3", "score6": "2", "score7": "1", "score8": "3", "memberId": "1" }
我有一个问题,即只取1分的记录,我也想排除该字段memberId
更何况我如何将该结果推送到:
datasets : [
{
fillColor: "rgba(102,45,145,.1)",
strokeColor: "rgba(102,45,145,1)",
pointColor : "rgba(220,220,220,1)",
pointStrokeColor : "#fff",
data : [] // HERE IS WHERE THE DATA NEEDS TO GO
}
首先从阵列中删除不必要的元素。请参阅此链接:http://stackoverflow.com/questions/4466159/delete-element-from-multidimensional-array-based-on-value –
之后,将其编码为json格式并传入js库部分。 –
@prakashtank即时通讯不知道如何将它传递到js库,这就是我有点卡住 – PhpDude