Question

Python 3.6.3 Django的== 1.11.3 bootstrap4

我正在尝试创建一个条形图，其中包含从数据表中检索的数据。如何将数据（标签，数据）值映射到图表js文件中的数据表。假设硬编码数据作为数据表存在，您如何编写js文件？

请告诉我怎么做。

HTML

<div class="col-lg-6">
  <div class="card bar-chart-example">
    <div class="card-header d-flex align-items-center">
      <h4>Bar Chart Example</h4>
    </div>
    <div class="card-body">
      <canvas id="barChartExample"></canvas>
    </div>
  </div>
</div>

图表js

var barChartExample = new Chart(BARCHARTEXMPLE, {
        type: 'bar',
        data: {
            labels: ["January", "February", "March", "April", "May", "June", "July"],
            datasets: [
                {
                    label: "Data Set 1",
                    backgroundColor: [
                        'rgba(51, 179, 90, 0.6)',
                        'rgba(51, 179, 90, 0.6)',
                        'rgba(51, 179, 90, 0.6)',
                        'rgba(51, 179, 90, 0.6)',
                        'rgba(51, 179, 90, 0.6)',
                        'rgba(51, 179, 90, 0.6)',
                        'rgba(51, 179, 90, 0.6)'
                    ],
                    borderColor: [
                        'rgba(51, 179, 90, 1)',
                        'rgba(51, 179, 90, 1)',
                        'rgba(51, 179, 90, 1)',
                        'rgba(51, 179, 90, 1)',
                        'rgba(51, 179, 90, 1)',
                        'rgba(51, 179, 90, 1)',
                        'rgba(51, 179, 90, 1)'
                    ],
                    borderWidth: 1,
                    data: [65, 59, 80, 81, 56, 55, 40],
                }
            ]
        }
    });

Answer 1

查看DataTables API

从DataTable后面获取数据：

<?php

if (empty($_FILES) || $_FILES['file']['error']) {
    die('{"OK": 0, "info": "Failed to move uploaded file."}');
}

$chunk = isset($_REQUEST["chunk"]) ? intval($_REQUEST["chunk"]) : 0;
$chunks = isset($_REQUEST["chunks"]) ? intval($_REQUEST["chunks"]) : 0;

$fileName = isset($_REQUEST["name"]) ? $_REQUEST["name"] : $_FILES["file"]["name"];
$filePath = "uploads/$fileName";


// Open temp file
$out = @fopen("{$filePath}.part", $chunk == 0 ? "wb" : "ab");
if ($out) {
    // Read binary input stream and append it to temp file
    $in = @fopen($_FILES['file']['tmp_name'], "rb");

    if ($in) {
        while ($buff = fread($in, 4096))
            fwrite($out, $buff);
    } else
        die('{"OK": 0, "info": "Failed to open input stream."}');

    @fclose($in);
    @fclose($out);

    @unlink($_FILES['file']['tmp_name']);
} else
    die('{"OK": 0, "info": "Failed to open output stream."}');


// Check if file has been uploaded
if (!$chunks || $chunk == $chunks - 1) {
    // Strip the temp .part suffix off 
    rename("{$filePath}.part", $filePath);
}

die('{"OK": 1, "info": "Upload successful."}');

如何用数据值替换数据值（js）？

1 个答案: