Question

我正在尝试使用Google Visualization API创建一个图表，PHP＆amp; MySQL在后台。

我正在做的是：

使用PHP / SQL从db获取数据

$sth = mysql_query("SELECT * FROM Chart");

使用PHP创建JSON

$rows = array();

while($r = mysql_fetch_assoc($sth)) {
  $rows[] = $r;
}

$jdata = json_encode($rows);

然后使用JSON

提供Google Visualization API

var data = new google.visualization.DataTable(<?php echo $jdata ?>);

只是为了确保JSON实际上采用了我所做的正确格式：

$jdata = json_encode($rows);
print $jdata;

返回：

[{"id":"1","quarters":"1","salary":"1250"},{"id":"2","quarters":"2","salary":"2500"},{"id":"3","quarters":"3","salary":"4526"},{"id":"4","quarters":"4","salary":"4569"}]

所以，

数据库连接正常。
从PHP数组创建JSON就行了。
JSON格式没问题。

Firebug返回错误说：

表没有列。 [打破此错误] b，S1），[b]}函数Zq（a，b）{var c = a [xc] ...“]。”））：d（l（“表没有列。 “））}

问题是如何从JSON数据创建列？

更新

用于创建下图的代码：

// SQL Query
$sth = mysql_query("SELECT * FROM Chart");
//$rows = array();

while($r = mysql_fetch_assoc($sth)) {
   if(!isset($google_JSON)){    
     $google_JSON = "{cols: [";    
     $column = array_keys($r);
     foreach($column as $key=>$value){
         $google_JSON_cols[]="{id: '".$key."', label: '".$value."'}";
     }    
     $google_JSON .= implode(",",$google_JSON_cols)."],rows: [";       
   }
   $google_JSON_rows[] = "{c:[{v: '".$r['id']."'}, {v: '".$r['quarters']."'}, {v: '".$r['salary']."'}]}";
}    

// you may need to change the above into a function that loops through rows, with $r['id'] etc, referring to the fields you want to inject..

$data = $google_JSON.implode(",",$google_JSON_rows)."]}";

输出HTML代码：

        <!-- load Google AJAX API -->
    <script type="text/javascript" src="http://www.google.com/jsapi"></script>
    <script type="text/javascript">  
        //load the Google Visualization API and the chart  
        google.load('visualization', '1', {'packages':['columnchart']});  

        //set callback  
        google.setOnLoadCallback (createChart);  

        //callback function  
        function createChart() {  

            //create data table object  
            var data = new google.visualization.DataTable({cols: [{id: '0', label: 'id'},{id: '1', label: 'quarters'},{id: '2', label: 'salary'}],rows: [{c:[{v: '1'}, {v: '1'}, {v: '1250'}]},{c:[{v: '2'}, {v: '2'}, {v: '2500'}]},{c:[{v: '3'}, {v: '3'}, {v: '4526'}]},{c:[{v: '4'}, {v: '4'}, {v: '4569'}]}]});  

            //instantiate our chart objects  
            var chart = new google.visualization.ColumnChart (document.getElementById('chart'));  

            //define options for visualization  
            var options = {width: 400, height: 240, is3D: true, title: 'Company Earnings'};  

            //draw our chart  
            chart.draw(data, options);  

        }  
    </script>

    <div id="chart"></div>

When using the code above the script is creating the graph, but something is wrong there

Answer 1

根据docs，您是否尝试单独建立列引用和数据？

var data = new google.visualization.DataTable();
data.addColumn('string', 'Task');
data.addColumn('number', 'Hours per Day');
data.addRows([
  ['Work', 11],
  ['Eat', 2],
  ['Commute', 2],
  ['Watch TV', 2],
  ['Sleep', {v:7, f:'7.000'}]
]);

要格式化为对象的正确JSON，您可以按如下方式进行设置：

while($r = mysql_fetch_assoc($sth)) {
   if(!isset($google_JSON)){    
     $google_JSON = "{cols: [";    
     $column = array_keys($r);
     foreach($column as $key=>$value){
         $google_JSON_cols[]="{id: '".$key."', label: '".$value."'}";
     }    
     $google_JSON .= implode(",",$google_JSON_cols)."],rows: [";       
   }
   $google_JSON_rows[] = "{c:[{v: '".$r['id']."'}, {v: ".$r['quarters']."}, {v: ".$r['salary']."}]}";
}    
// you may need to change the above into a function that loops through rows, with $r['id'] etc, referring to the fields you want to inject..
echo $google_JSON.implode(",",$google_JSON_rows)."]}";

Answer 2

Complete Working Example: PHP/MYSQL/Google Chart/JSON

PHP MySQL Google Chart JSON - Complete Example

带PHP和PHP的Google图表工具MYSQL

2 个答案: