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问在变量名称中使用php和mysql获取值[value1，value2，value2]
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Stack Overflow用户

提问于 2012-12-21 10:24:34

回答 3查看 783关注 0票数 0

如何在php和mysql中从变量s1，s2，s3插入数组值：

$(function () {

    var s1 = [100, 200, 300]; //How to Get Value from mysql database
    var s2 = [30, 80, 90]; //How to Get Value from mysql database
    var s3 = [120, 90, 80]; //How to Get Value from mysql database

    // Can specify a custom tick Array.
    // Ticks should match up one for each y value (category) in the series.

    var ticks = ['2010', '2011', '2012'];
    var plot1 = $.jqplot('chart3', [s1, s2, s3], {

        // The "seriesDefaults" option is an options object that will
        // be applied to all series in the chart.

        seriesDefaults: {
            shadow: true, // show shadow or not.
            renderer: $.jqplot.BarRenderer,
            rendererOptions: {
                fillToZero: true
            }
        },

        // Custom labels for the series are specified with the "label"
        // option on the series option.  Here a series option object
        // is specified for each series.

        series: [
            {label: 'Hotel'},
            {label: 'Event Regristration'},
            {label: 'Airfare'}
        ],

        // Show the legend and put it outside the grid, but inside the
        // plot container, shrinking the grid to accomodate the legend.
        // A value of "outside" would not shrink the grid and allow
        // the legend to overflow the container.

        legend: {
            show: true,
            placement: 'outsideGrid'
        },
        axes: {
            // Use a category axis on the x axis and use our custom ticks.
            xaxis: {
                renderer: $.jqplot.CategoryAxisRenderer,
                ticks: ticks
            },
            // Pad the y axis just a little so bars can get close to, but
            // not touch, the grid boundaries.  1.2 is the default padding.
            yaxis: {
                pad: 1.05,
                tickOptions: {
                    formatString: '$%d'
                }
            }
        },
        grid: {
            borderColor: '#000', // CSS color spec for border around grid.
            borderWidth: 2.0, // pixel width of border around grid.
            shadow: true // draw a shadow for grid.
        }
    });
    // Bind a listener to the "jqplotDataClick" event.  Here, simply change
    // the text of the info3 element to show what series and ponit were
    // clicked along with the data for that point.
    $('#chart3').bind('jqplotDataClick',

    function (ev, seriesIndex, pointIndex, data) {
        $('#info3').html('series: ' + seriesIndex + ', point: ' + pointIndex + ', data: ' + data);
    });
});

php

jquery

jqplot

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回答 3

Stack Overflow用户

回答已采纳

发布于 2012-12-21 10:44:04

2种方式：

Ajax

用法：$.getJSON ( http://api.jquery.com/jQuery.getJSON/ )

var ses = {};

$.getJSON('page_adress.php', {variable_you_want_to_pass1: 'its value', variable_you_want_to_pass2: 'var 2 value'}, function(data) {
    ses = data;
});

在你的PHP中：

<?php
$passed_var_1 = $_REQUEST['variable_you_want_to_pass1'];
//.... etc


//Here you get your data from mysql, cast it into array

header('Content-type: application/json');
echo json_encode($dbdata);
?>

因此，基本上在请求完成后，您将拥有在JavaScript中传输的精确数组。请记住，此技术使用AJAX。如果你想避免这种情况，你将不得不使用第二种技术。

动态创建JS

让PHP生成你的javascript。在这种情况下，您将在主页中使用

<script src="js_data.js.php" type="text/javascript"></script>

在js_data.js.php文件中：

<?php
header("content-type: application/x-javascript");

$s1 = array(100,200,300);
//....

var s1 = [<?=implode(', ', $s1)?>],
    s2 = [<?=implode(', ', $s2)?>],
    s3 = [<?=implode(', ', $s3)?>];

?>

票数 0

Stack Overflow用户

发布于 2012-12-21 10:53:41

第一个方法(不带ajax和json)(Untidy Way)

首先从数据库中获取值，并将其放入PHP变量中。

然后将html元素放入页面中，并为其赋值。

然后使用document.getElement方法在javascript中使用它。

//假设您已经从$valueFrmDB中的数据库中获取了值。

$valueFrmDB;

现在，以html元素为例(您可能需要获取多个元素)

<input type="hidden" id="something" name="something" value="echo value of $valueFrmDB here" />;

然后，在javascript中

var vfd = document.getElementById('something').value;

将字符串转换为数组

第二种方法(使用ajax和json) (简单且正确，但必须了解ajax和json)

使用ajax从数据库获取值

然后使用json将这些值传递给javascript。

票数 0

Stack Overflow用户

发布于 2013-02-06 09:55:56

简单地说，您可以通过以下方式完成此操作：

<?php
   $query = mysql_query("SELECT * FROM attendence");
   $results = array(array());

   while($line = mysql_fetch_array($query)){
    $results[] = $line;
   }
?>

Javascript

  <script type="text/javascript">
     $(document).ready(function(){
     var data = <?php echo json_encode($results); ?>; //array uses here
     var plot1 = jQuery.jqplot ('chart1', [data], 
       { 
       seriesDefaults: {
       renderer: jQuery.jqplot.PieRenderer, 
        rendererOptions: {
        showDataLabels: true}
       }, 
      legend: { show:true, location: 'e' }
      });
     });
  </script>