javascript - 将变量从 Google Chart 脚本传递到 MySql PHP

Question

我需要在Google图表脚本之前从$_GET['variableID']传递一个变量，然后传递到目标页面PHP以确定我想要哪个客户端进行查询。我怎么能？ (注意:我不是 javascript 天才)。非常感谢!!

这是我定义的chart.js

/* Get data from the database */
function getData() {
    jQuery.ajax({
        url: 'get.php',
        type: 'GET',
        dataType: 'json',
        mimeType: 'multipart/form-data',
        contentType: false,
        cache: false,
        processData: false,
        success: function( data, jqXHR ) {
            if( data == "null" ) {
                // just in case
            } else {
                drawGraph( data );
            }
        },
        error: function( textStatus ) {
            console.log(" error. damm. ");
        }
    });
}

/* Initialization of Google Charts API */
google.load( "visualization" , "1", { packages: [ "corechart" ] });
google.setOnLoadCallback( getData );

/* Chart to render time-user graphs */
function drawGraph( data ) {
    for( var i = data.length; i > 0; i-- ) {
        data[i] = data[i - 1];
    }
    data[0] = [ 'Data', 'Comandes' ];
    console.log( data );
    var chartData = google.visualization.arrayToDataTable( data );var options = {
      title: '','legend':'none','chartArea': {'width': '100%', 'height': '80%'},

backgroundColor: {
    stroke: '#a5a5a5',
    strokeWidth: 1
},


      hAxis: {
        format: 'M/d/yy',
        gridlines: {count: 15}
      },
      vAxis: {
        gridlines: {color: '#d0d0d0'},
        minValue: 0
      }
    };

    var chart = new google.visualization.LineChart( document.getElementById( 'chart_div' ) );

    chart.draw( chartData , options );
}

这是get.php(特别注意WHERE idclient=GET variableID)

  include('lib/connection.php');

 if($_GET) {
$query = "select DATE( data ), COUNT(*)
            from smarty_pedidos  **WHERE idclient=GET variableID**
            group by DATE( data )";
$result = mysql_query( $query );
$rows = array();
while( $row = mysql_fetch_array( $result ) ) {
    $rows[] = array( '0' => $row['0'] , '1' => $row['1'] );

}

//print json_encode($rows);
print json_encode($rows, JSON_NUMERIC_CHECK);
 }

Answer 1

在 ajax 调用中，将变量添加到 url...

jQuery.ajax({
  url: 'get.php?variableID=variableVALUE',
  ...  

然后在 php 代码中，使用 $_GET['variableID']

$query = "select DATE( data ), COUNT(*)
  from smarty_pedidos  WHERE idclient=$_GET['variableID']
  group by DATE( data )";