来自返回的 json 数据的 jquery 图表答案

【问题标题】：jquery charts from returned json data来自返回的 json 数据的 jquery 图表
【发布时间】：2014-11-05 14:15:25
【问题描述】：

我已经使用 mysql 和 php 从我的数据库中获取了 json 数据，现在想使用 $.ajax 函数将其传回并动态地创建一个样条图，任何帮助或示例将不胜感激。同样对于多次验血日期重复，我怎样才能只显示一个日期与多个验血结果。

这是我的php代码

header('Set Access-Control-Allow-Origin: *');
header('Content-type: application/json');
require_once"conn.php";

if($_SERVER['REQUEST_METHOD']=='GET'){

//recieve credentials from the user 
$nhsno = mysqli_real_escape_string($conn, $_GET['pnhsno']);

//check the variables recieved are not empty
    if($nhsno != ''){

        $sql = "SELECT date,name,tname,value FROM examination e, testresults t, examtype ex, testname tn
        WHERE e.patientnhs_no = '$nhsno' and e.etype_id = '1' and
        e.examination_id = t.examination_id and e.etype_id = ex.etype_id and t.tname_id = tn.tname_id ";//create an sql statement 
        $result = $conn->query($sql);//run sqlm statement 
        $row = $result->fetch_assoc(); //fetch row of data


    foreach ($result as $row){
        $return[]=array('date'=>$row['date'],
                        'name'=>$row['name'],
                        'test name'=>$row['tname'],
                        'value'=>$row['value']);
    }
}
}
$conn = null;
echo json_encode($return);


?>

这是我的 json 输出

[{"date":"2004-07-05","name":"blood test","test name":"t3","value":"6.8"},  {"date":"2004-07-05","name":"blood test","test name":"t4","value":"29"},{"date":"2004-07-05","name":"blood test","test name":"tsh","value":"0.01"},{"date":"2004-07-05","name":"blood test","test name":"thyroglobulin level","value":"0.5"},{"date":"2005-06-15","name":"blood test","test name":"t3","value":"5.2"},{"date":"2005-06-15","name":"blood test","test name":"t4","value":"30"},{"date":"2005-06-15","name":"blood test","test name":"tsh","value":"0.02"},{"date":"2005-06-15","name":"blood test","test name":"thyroglobulin level","value":"0.5"}]

【问题讨论】：

您使用哪个库来创建图表...
canvas jquery 图表
charts.js 或任何其他我能得到样条图的东西
您需要根据您要使用的 javascript 库来构建您的 json 响应...请参阅此示例：jsfiddle.net/gh/get/jquery/1.9.1/highslide-software/…
非常感谢您的指导

标签： php jquery mysql json

【解决方案1】：

因为你混合了很多东西，并且你在循环之前获取 1 行。

应该是这样的：

$result = $conn->query($sql); //run sqlm statement 
//Do not need this!
//$row = $result->fetch_assoc(); //fetch row of data
//And $row = $result... not opposite.
while ($row = $result->fetch_assoc()) {
    $return[] = array('date' => $row['date'],
        'name' => $row['name'],
        'test name' => $row['tname'],
        'value' => $row['value']);
}

【讨论】：