这里从mysql获取数据..
if (!empty($result1)) {
while ($row1 = mysqli_fetch_array($result1)) {
$caseno = $row1['cases'];
echo "<b>" . $caseno . "<br>";
}
}
我想将 $caseno
中的数据传递给我下面的 JavaScript..
<script type="text/javascript">
var gaugevalue = document.getElementById("$caseno");
var myConfig2 = {
"type": "gauge",
"scale-r": {
"aperture": 200, //Scale Range
"values": "0:50:10" //and minimum, maximum, and step scale values.
},
"series": [{"values": [gaugevalue]}]
//"series":[{"values":[40]}]
};
zingchart.render({
id: 'myChart',
data: myConfig2,
height: "90%",
width: "90%"
});
</script>
最佳答案
我更仔细地分析了您的代码,并注意到 gaugevalue
必须是一个整数数组,而您试图将 DOM 元素传递给它。所以你的完整代码应该如下所示:
<?php
$gauge_values = [];
if (!empty($result1)) {
while ($row1 = mysqli_fetch_array($result1)) {
$gauge_values[] = $row1['cases'];
}
}
?>
<script type="text/javascript">
var myConfig2 = {
"type": "gauge",
"scale-r": {
"aperture": 200, //Scale Range
"values": "0:50:10" //and minimum, maximum, and step scale values.
},
"series": [{"values": <?php echo json_encode($gauge_values); ?>}]
};
zingchart.render({
id: 'myChart',
data: myConfig2,
height: "90%",
width: "90%"
});
</script>
关于javascript - 如何将 MySQL 查询数据传递给 JavaScript 变量?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53789992/