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javascript - 如何将 MySQL 查询数据传递给 JavaScript 变量？

问题描述

这是从mysql获取数据..

if (!empty($result1)) {
    while ($row1 = mysqli_fetch_array($result1)) {
        $caseno = $row1['cases'];
        echo "<b>" . $caseno . "<br>";
    }
}

我想将那里的数据传递$caseno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>

标签： javascriptphp

解决方案

我更仔细地分析了您的代码，并注意到它gaugevalue必须是一个整数数组，而您正试图将一个 DOM 元素传递给它。所以你的完整代码应该是这样的：

<?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>

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