Javascript 来自mysql、php/JQuery的Google线形图

我已经学习了一些程序，现在我需要从MySQL获取数据并用HTML/PHP显示它。我使用以下MySQL.php文件从MySQL获取数据：

<?php
    $hostname = "localhost";
    $database = "database";
    $username = "username";
    $password = "password";
    $connect = mysql_connect($hostname, $username, $password)

    or die('Could not connect: ' . mysql_error());
    $bool = mysql_select_db($database, $connect);
    if ($bool === False){
       print "can't find $database";
    }

    $query = "SELECT * FROM  `table` ORDER BY timestamp LIMIT 0 , 100";
    $result = mysql_query($query) or die("SQL Error 1: " . mysql_error());
    // get data and store in a json array
    while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
        $values[] = array(
            'timestamp' => $row['timestamp'],
            'temperature' => $row['temperature'],
          );
    }

    echo json_encode($values);
?>

---

您的JSON格式应如下所示

{
"cols": [
    {"id":"","label":"Topping","pattern":"","type":"string"},
    {"id":"","label":"Slices","pattern":"","type":"number"}
  ],
"rows": [
    {"c":[{"v":"Mushrooms","f":null},{"v":3,"f":null}]},
    {"c":[{"v":"Onions","f":null},{"v":1,"f":null}]},
    {"c":[{"v":"Olives","f":null},{"v":1,"f":null}]},
    {"c":[{"v":"Zucchini","f":null},{"v":1,"f":null}]},
    {"c":[{"v":"Pepperoni","f":null},{"v":2,"f":null}]}
  ]
}

因此，当您试图将JSON转换为数据表时，您不会得到任何返回，因为它的格式不正确。这里有更多内容

尝试格式化JSON格式

var jsonData = {}
jsonData.cols = [
      {"label":"Date w/ time", "type":"date"},
      {"label":"Temp", "type":"number"}
  ]
jsonData.rows = [
     {"c":[{"v":new Date(2015, 11, 1, 12, 23)}, {"v":50}]},
     {"c":[{"v":new Date(2015, 11, 1, 12, 24)}, {"v":75}]}
]

您尚未定义DataTable列。请尝试按以下方式或
检查

var jsonData = {}
jsonData.cols = [
      {"label":"Date w/ time", "type":"date"},
      {"label":"Temp", "type":"number"}
  ]
jsonData.rows = [
     {"c":[{"v":new Date(2015, 11, 1, 12, 23)}, {"v":50}]},
     {"c":[{"v":new Date(2015, 11, 1, 12, 24)}, {"v":75}]}
]

// Create data table
var data = new google.visualization.DataTable();

data.addColumn('string', 'Timestamp');
data.addColumn('string', 'Temperature');