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PHP生成jQuery数据_Php_Jquery_Mysql - Fatal编程技术网

PHP生成jQuery数据

PHP生成jQuery数据,php,jquery,mysql,Php,Jquery,Mysql,我想使用jQuery支持的图表。如果jQuery需要这样的数据: data: [{ period: '2011 Q1', sales: 1400, profit: 400 }, { period: '2011 Q2', sales: 1100, profit: 600 }, { period: '2011 Q3', sales: 1600, profit: 500 }, { period:

我想使用jQuery支持的图表。如果jQuery需要这样的数据:

 data: [{
     period: '2011 Q1',
     sales: 1400,
     profit: 400
 }, {
     period: '2011 Q2',
     sales: 1100,
     profit: 600
 }, {
     period: '2011 Q3',
     sales: 1600,
     profit: 500
 }, {
     period: '2011 Q4',
     sales: 1200,
     profit: 400
 }, {
     period: '2012 Q1',
     sales: 1550,
     profit: 800
 }],

我如何使用MySQL和PHP生成此文件?

这在很大程度上取决于数据的存储方式。简单的方法是将数据放入数组,并使用
json\u encode()
将其显示为字符串


var myData=;

从表中获取数据并使其成为JSON的示例代码

 $var = array();
 $result = mysqli_query($con, "SELECT * FROM TableName");
 while($obj = mysqli_fetch_object($result)) {
    $var[] = $obj;
 }
 echo 'users:'.json_encode($var).'}';
这是参考文献。。
您可以看到JSON的结构。

这段代码可能会对您有所帮助

$my_array = array();
$custom_variable = "anything";
$con=mysqli_connect("localhost",'username','password','databaseName');
// Check connection
if (mysqli_connect_errno())
{
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$fetch = mysqli_query($con,"SELECT * FROM your_table"); 

while ($row = mysqli_fetch_array($fetch)) {
    $my_array['col1'] = $row['col1'];
    $my_array['col2'] = $row['col2']; 
    $my_array['custom_col'] = $custom_variable;
}

echo "data:".json_encode($my_array);
mysqli_close($con);

您可以在php端使用循环生成字符串,我看,在这之后,我是否创建一个变量并将其设置为json_encode($my_数组),然后在数据中:我将其设置为变量?当然,您可以替换
echo”数据:“.json_encode($my_数组)$my\u variable=“data:.json\u encode($my\u数组)进行编码
如果您的意思是向json数组添加自定义元素:您可以使用以下命令在while循环中将自定义元素设置为
$my_数组
$my_数组['custom']=“custom data”
$my_array = array();
$custom_variable = "anything";
$con=mysqli_connect("localhost",'username','password','databaseName');
// Check connection
if (mysqli_connect_errno())
{
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$fetch = mysqli_query($con,"SELECT * FROM your_table"); 

while ($row = mysqli_fetch_array($fetch)) {
    $my_array['col1'] = $row['col1'];
    $my_array['col2'] = $row['col2']; 
    $my_array['custom_col'] = $custom_variable;
}

echo "data:".json_encode($my_array);
mysqli_close($con);