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我使用Dygraph来显示图形。我把它静态这样工作dygraph php动态生成数据库中的内容
<div id="graphdiv"></div>
<script type="text/javascript">
g = new Dygraph(
// containing div
document.getElementById("graphdiv"),
// CSV
"Date,High,Low\n" +
"2015-05-07,75,40\n" +
"2015-05-08,70,50\n" +
"2015-05-09,80,60\n" +
"2015-05-10,60,40\n" +
"2015-05-11,50,30\n" +
"2015-05-12,0,0\n"
);
</script>
现在我想用PHP动态生成从数据库中的内容。我试图这样做。我创建了关联数组来从数据库中获取信息,然后添加PHP绑定
<?php
//CREATE SQL STATEMENT
$sql_temperatures = "SELECT * FROM tbltemperatures";
//CONNECT TO MYSQL SERVER
require('inc-conndygraph.php');
//EXECUTE SQL STATEMENT
$rs_temperatures = mysqli_query($vconndygraph, $sql_temperatures);
//CREATE AN ASSOCIATIVE ARRAY
$rs_temperatures_rows = mysqli_fetch_assoc($rs_temperatures);
?>
<!doctype html>
<html>
<head>
<!-- LINK TO THE DYGRAPH LIBRARY -->
<script type="text/javascript" src="dygraph-combined-dev.js"></script>
</head>
<body>
<!-- CONTAINER HOLDING GRAPH -->
<div id="graphdiv"></div>
<script type="text/javascript">
g = new Dygraph(
// containing div
document.getElementById("graphdiv"),
"Date,High,Low\n" +
// CSV
<?php do { ?>
"<?php echo json_encode($rs_temperatures_rows['tdate']); ?>,<?php echo json_encode($rs_temperatures_rows['thigh']); ?>,<?php echo json_encode($rs_temperatures_rows['tlow']); ?>\n"
<?php } while ($rs_temperatures_rows = mysqli_fetch_assoc($rs_temperatures)); ?>
);
</script>
从理论上讲,这应该在功能正常工作,但是当我尝试它不会显示任何浏览器来查看它。我错过了一些愚蠢的东西吗?
我在想我可能连接不正确,它显示只是不返回?我不擅长JavaScript和任何帮助,将不胜感激。