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我想使用以下代码以图形的形式从 mysql 中检索数据。但是，什么都没有显示。有谁能够帮我？

<script type="text/javascript" src="https://www.google.com/jsapi"></script>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/
<script type="text/javascript">
  google.load("visualization", "1", {packages:["corechart"]});
  google.setOnLoadCallback(drawChart);

  function drawChart() {
      var jsonData = $.ajax({
          url: "graphdata.php",
          dataType:"json",
          async: false
      }).responseText;

      // Create our data table out of JSON data loaded from server.
      var data = new google.visualization.DataTable(jsonData);

      var options = {'title':'Ticket Sales','width':500,'height':400};

      // Instantiate and draw our chart, passing in some options.
      var chart = new google.visualization.ColumnChart(document.getElementById('chart_div'));
      chart.draw(data,options); 
  }
</script>
<?php
echo("<div id='chart_div'></div>");
?>

这是我的getdata.php：

$sql = "SELECT count(`booth_number`),`sold_by` FROM `registration1` where week(`Date`) =     week(curdate())  GROUP BY `sold_by`";

$result = mysql_query($sql, $conn) or die(mysql_error());

//start the json data in the format Google Chart js/API expects to recieve it
$data = array(
    'cols' => array(
        array('label' => 'Month', 'type' => 'string'),
        array('label' => 'Ticket Sales', 'type' => 'string')
      ),
    'rows' => array()
);

while($row = mysql_fetch_row($result)) {
    $data['rows'][] = array('c' => array(array('v' => $row[0]), array('v' => $row[1])));
}

echo json_encode($data);
4

1 回答 1

1

如前所述，我认为您需要将代码移动到成功回调中：

function drawChart() {
    $.ajax({
        url: "graphdata.php",
        type: "GET",
        dataType:"json",
        success: function(jsonData, textStatus, jqXHR) {
            // Create our data table out of JSON data loaded from server.
            var data = new google.visualization.DataTable(jsonData);

            var options = {'title':'Ticket Sales','width':500,'height':400};

            // Instantiate and draw our chart, passing in some options.
            var chart = new google.visualization.ColumnChart(document.getElementById('chart_div'));
            chart.draw(data,options); 
        }
    });
}

[编辑] 可以在jsBin上找到一个工作示例（带有一些模拟数据） 。

[edit2]使用重复问题中提供的输出更新jsBin上的示例：

{
    "cols": [
        {
            "label":"Month",
            "type":"string"
        },{
            "label":"Weekly Sales",
            "type":"string"
        }
    ],
    "rows": [
        {
            "c":[
                {"v":"3"},
                {"v":"jaspreet singh "}
            ]
        },{
            "c":[
                {"v":"3"},
                {"v":"joseph swanson"}
            ]
        }
    ]
}

服务器返回的 JSON 是错误的，对象应该是这样的：

  "cols": [
    {
      "label":"Month",
      "type":"string"
    },{
      "label":"Weekly Sales",
      "type":"number"
    }
  ],
  "rows": [
    {
      "c":[
        {"v":"jaspreet singh "},
        {"v":"3"}
      ]
    },{
      "c":[
        {"v":"joseph swanson"},
        {"v":"3"}
      ]
    }
  ]
};

所需的更改现在应该很明显了。希望这可以帮助。

[edit3]getdata.php应包含以下片段恕我直言：

$data = array(
    'cols' => array(
        array('label' => 'Month', 'type' => 'string'),
        array('label' => 'Ticket Sales', 'type' => 'number')
    ),
    'rows' => array()
);

while($row = mysql_fetch_row($result)) {
    $data['rows'][] = array(
        'c' => array(
            array('v' => $row[1]),
            array('v' => $row[0])
        )
    );
}
于 2013-10-05T09:24:17.590 回答