Php 谷歌可视化Api+SQL
我想在MySql数据库中创建一些数据的GViz,但我遇到了一些问题 以下是迄今为止的资料来源:Php 谷歌可视化Api+SQL,php,mysql,api,Php,Mysql,Api,我想在MySql数据库中创建一些数据的GViz,但我遇到了一些问题 以下是迄今为止的资料来源: <?php $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); mysql_select_db($dbname); $int_y_pos = -1; $int_y_step_small = 1; $sql = "SELECT * from table')"; $
<?php
$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error
connecting to mysql');
mysql_select_db($dbname);
$int_y_pos = -1;
$int_y_step_small = 1;
$sql = "SELECT * from table')";
$sql = mysql_query($sql);
$rownum = mysql_num_rows($sql);
?>
<html>
<head>
<script type="text/javascript" src="http://www.google.com/jsapi"></
script>
<script type="text/javascript">
google.load("visualization", "1", {packages:
["table"]});
google.setOnLoadCallback(drawData);
function drawTable() {
var data = new google.visualization.DataTable();
data.addColumn('string', 'column1');
data.addColumn('string', 'column2');
<?php
echo " data.addRows($rownum);\n";
while($row = mysql_fetch_assoc($sql)) {
$int_y_pos += $int_y_step_small;
echo " data.setValue(" . $int_y_pos . ", 0, new column1(" .
$row['column1'] . "));\n";
echo " data.setValue(" . $int_y_pos . ", 0, new column2(" .
$row['column2'] . "));\n";
}
?>
var table = new google.visualization.Table(document.getElementById('table_div'));
table.draw(data, {showRowNumber: true});
google.visualization.events.addListener(table, 'select', function() {
var row = table.getSelection()[0].row;
alert('You selected ' + data.getValue(row, 0));
});
}
</script>
</head>
<body>
<div id="table_div" style="width: 940px; height: 240px;"></div>
</body>
</html>
以前集成的任何帮助或示例都将非常有用
谢谢
Gareth谢谢你的精彩剧本,不管它对我有什么影响 我把它改成这样
$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');
mysql_select_db($dbname);
$int_y_pos = -1;
$int_y_step_small = 1;
$sql = "SELECT * from cache";
$sql = mysql_query($sql);
$rownum = mysql_num_rows($sql);
?>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="content-type" content="text/html; charset=utf-8"/>
<title>
Google Visualization API Sample
</title>
<script type="text/javascript" src="http://www.google.com/jsapi"></script>
<script type="text/javascript">
google.load('visualization', '1', {packages: ['table']});
</script>
<script type="text/javascript">
function drawVisualization() {
// Create and populate the data table.
var data = new google.visualization.DataTable();
data.addColumn('string', 'col1');
data.addColumn('string', 'col2');
<?php
echo " data.addRows($rownum);\n";
while($row = mysql_fetch_assoc($sql)) {
$int_y_pos += $int_y_step_small;
echo " data.setCell(" . $int_y_pos . ", 0,'". $row['col1'] . "');\n";
echo " data.setCell(" . $int_y_pos . ", 1,'" . $row['col2'] . "');\n";
}
?>
// Create and draw the visualization.
visualization = new google.visualization.Table(document.getElementById('table'));
visualization.draw(data, null);
}
google.setOnLoadCallback(drawVisualization);
</script>
</head>
<body style="font-family: Arial;border: 0 none;">
<div id="table"></div>
</body>
</html>
这对我很有效:
1.设置表的标题
<?php
$dataTable = array(
'cols' => array(
// each column needs an entry here, like this:
array('type' => 'string', 'label' => 'Col1'),
array('type' => 'string', 'label' => 'Col2'),
array('type' => 'string', 'label' => 'Col3')
)
);
编码为JSON:
$json = json_encode($dataTable);?>
然后装载容器并将其显示为表格:
function drawTable() {
var data = new google.visualization.DataTable(<?php echo $json; ?>);
var table = new google.visualization.Table(document.getElementById('your-div-id'));
table.draw(data);
}
function drawTable() {
var data = new google.visualization.DataTable(<?php echo $json; ?>);
var table = new google.visualization.Table(document.getElementById('your-div-id'));
table.draw(data);
}