小编Gor*_*oro的帖子

我正在尝试使用AmCharts制作图表,并在他们的网站上关注此示例:https://www.amcharts.com/kbase/using-data-loader-to-connect-charts-to-mysql-data-base/

我的原始代码是

var chart = AmCharts.makeChart("chart_1", {
        "pathToImages": App.getGlobalPluginsPath() + "amcharts/amcharts/images/",

        "dataProvider": [{
            "year": 2009,
            "income": 23.5,
            "expenses": 18.1
        }, {
            "year": 2010,
            "income": 26.2,
            "expenses": 22.8
        }],
            // bunch of graph stylings
    });

    $('#chart_1').closest('.portlet').find('.fullscreen').click(function() {
        chart.invalidateSize();
    });
}

我尝试从mysql加载数据

var chart = AmCharts.makeChart("chart_1", {
    "type": "serial",
    "dataProvider": {
        "url": "data.php"
    },
        "pathToImages": App.getGlobalPluginsPath() + "amcharts/amcharts/images/",

      // bunch on graphs style

    });

    $('#chart_1').closest('.portlet').find('.fullscreen').click(function() {
        chart.invalidateSize();
    });
}

包含硬编码数据的原始代码正在运行.下面用mysql不起作用.它是空白页面,不显示图表.控制台上也没有错误.

data.php 正在工作并返回json数组.

有人可以帮忙吗？

如果还需要php部分

$query = "
       SELECT *
       FROM …

我正在尝试在处理表单时插入日期,但是在MySQL中插入就像

0000-00-00

我不希望用户插入日期.这是我的表格

<form action="upload.php" method="post" enctype="multipart/form-data">
<textarea type="text" name="text" cols="40" rows="10"></textarea><br />
<input type="hidden" name="dt"/>
<input type="submit" name="upload" id="upload" value="???????" />

这是upload.php

<?php

if (isset($_POST['upload'])) {

$text = $_POST['text'];
$dt = date('Y-m-d', strtotime($_POST['Date']));

        $db = new mysqli("localhost", "*****", "", "****");

        if (mysqli_connect_error()) {
            printf("Connect failed: ", mysqli_connect_error());
        }
        mysqli_set_charset($db, "UTF8");

        $query = "INSERT INTO table (text, date) VALUES (?, ?)";
        $conn = $db->prepare($query);
        if ($conn == TRUE) {
            $conn->bind_param("si",$text, $dt );
            if (!$conn->execute()) {
                echo 'error insert';
            } else …