Google可视化API php ajax
我正在使用这个实现Google Visualization图表使用ajax示例重新加载，以便从Mysql数据库动态检索数据，并且得到“Uncaught SyntaxError：意外的标记<”错误。 这是我的HTML文件
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="content-type" content="text/html; charset=utf-8"/>
<script type="text/javascript" src="http://www.google.com/jsapi"></script>
<script type="text/javascript">
google.load("jquery", "1.6.1");
google.load('visualization', '1', {packages: ['table']});
</script>
<script type="text/javascript">
function drawVisualization(dataFromAjax) {
var data = new google.visualization.DataTable();
data.addColumn('number', 'InvoiceNo');
data.addColumn('string', 'B/L No');
data.addColumn('date', 'Date');
data.addColumn('string', 'Customer Name');
data.addColumn('number', 'Amount');
data.addRows(dataFromAjax);
var table = new google.visualization.Table(document.getElementById('table'));
table.draw(data);
}
function makeAjaxCall() {
$.ajax({url:'test.php',
data: {},
success: function(responseData) {
var arrayForGviz = eval("(" + responseData + ")");
drawVisualization(responseData);
}
});
}
</script>
</head>
<body>
<input type="button" onclick="makeAjaxCall();return false;" value="Click to get data"></input>
<div id="table"></div>
</body>
</html>
这是我的PHP文件test.php
$con = mysql_connect("localhost","userName","password");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("DB_NAME", $con);
$result = mysql_query("call cargosys.rpt_salesByDate('2013/03/05','2013/03/10')");
$output = array();
while($row = mysql_fetch_array($result)) {
// create a temp array to hold the data
$temp = array();
// add the data
$temp[] = $row['inv_no'];
$temp[] = ''' . $row['bl_no'] . ''';
$temp[] = ''' . $row['inv_date'] . ''';
$temp[] = ''' . $row['cust_name'] . ''';
$temp[] = $row['Amount'];
// implode the temp array into a comma-separated list and add to the output array
$output[] = '[' . implode(', ', $temp) . ']';
}
// implode the output into a comma-newline separated list and echo
echo implode(",n", $output);
//echo json_encode($output);
mysql_close($con);
在成功功能里面呢：
console.log(responseData);
然后将responseData复制并粘贴到jsonlint.com的JSON验证器中
我猜测这是一个JSON语法错误，验证者会指出你的错误。
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