Question

以下代码将通过url：GETDATA.PHP显示来自数据库的数据，并在工作中显示我的数据。但我的问题是，如何传递$ iddoc = $ _GET [＆＃39; id＆＃39;];到我的GETDATA.PHP..so它可以根据获取文档ID创建图表..

Histogram.php

<?php
session_start();

if(isset($_SESSION['idmember'])){
    $idmembersession = $_SESSION['idmember'];
    $iddoc = $_GET['id'];

include'configure.php';
}
  ?>

  <html xmlns="http://www.w3.org/1999/xhtml">
  <head>
<script src="http://code.jquery.com/jquery-1.11.0.min.js"></script>
    <meta http-equiv="content-type" content="text/html; charset=utf-8"/>
  <title>
   HISTOGRAM SAM
  </title>
  <script type="text/javascript" src="http://www.google.com/jsapi"></script>
  <script type="text/javascript">

  google.load("visualization", "1", {packages:["corechart"]});


  function drawVisualization() {

    var jsonData=null;
    var json = $.ajax

    ({

      url:"getdata.php", // make this url point to the data file
      dataType: "json",
      async: false,
      success: (
    function(data) {
        jsonData = data;  
       var data = new google.visualization.DataTable(jsonData);


      var options = {
      title: 'SAM Histogram Results',
      hAxis: {title: 'Sustainability Percentage'}, 
      titleTextStyle: {color: 'black'},

hAxis: { ticks:[1,2,3,4,5,6,7,8,9,10,11,12,      

           13,14,15,16,17,18,19,20,21,22,23,24] },

    }

     var chart = new google.visualization.ColumnChart
                  (document.getElementById('visualization')).
     draw(data,options);  

     })
      }).responseText;
      }     

    google.setOnLoadCallback(drawVisualization);
    </script>
    </head>
    <body>
     <div id="visualization" style="width: 1300px; height: 800px;"></div>
   </body>
  </html>

访问getdata.php

<?php
session_start();

if(isset($_SESSION['idmember'])){
    $idmembersession = $_SESSION['idmember'];
    $iddoc = $_GET['id'];

             include 'configure.php';

  $query = "SELECT * FROM criteria_match
            where document = '$iddoc' and idmember ='$idmembersession'";

   $sql_query = mysql_query($query) or die('Error 3 :'.mysql_error());

   $table = array();
    $table['cols'] = array(

    array('label' => 'criteria', 'type' => 'number'),
    array('label' => 'percentage', 'type' => 'number')


    );

    $rows = array();
    while($r = mysql_fetch_assoc($sql_query)) {
     $temp = array();

    $temp[] = array('v' => (int)$r['criteria']);
    $temp[] = array('v' => (float)$r['percentage']); 




     $rows[] = array('c' => $temp); 
     }
     $table['rows'] = $rows;

     $jsonTable = json_encode($table);


      header('Cache-Control: no-cache, must-revalidate');
     header('Content-type: application/json');



      echo $jsonTable;
 } 
 ?>

Answer 1

将该变量作为ajax中的数据值回显：

...
url:"getdata.php", // make this url point to the data file
type: "POST"  // or GET, depends on your choice
data {
   iddoc: '<?php echo htmlspecialchars($iddoc); ?>'
},
dataType: "json",
async: false,
...

在Getdata.php中，您将获得$_POST["iddoc"]。

我假设您正确地获得了$_GET['id']。

谷歌图表：如何保持并将_GET ID传递给网址：

1 个答案: