声明 chart.js 无法识别的变量时，我无法显示图表

Question

我想在我的图表中显示 2021 年每个月获得的患者总数。但是图表上没有显示任何信息，而是图表消失了。 执行代码并检查页面时，它告诉我未定义变量 labeldeMes 和 datosdeMes。

你能帮我解决这个错误吗？ 我不知道我做错了什么

index.php

//HTML document

<?php 

$querymeses = 'SELECT Monthname(fecha_nacimiento) AS mes, count(*) AS Total FROM pacientes GROUP BY mes' WHERE YEAR(fecha_nacimiento) = '2021';
$resmeses =mysqli_query($conexion,$querymeses);
$rowmeses=mysqli_fetch_assoc($resmeses);

$labelMeses="";
$datosMeses="";

while($rowmeses=mysqli_fetch_assoc($resmeses)){
  $labelMeses=$labelMeses.$rowmeses['mes'].",";
  $datosMeses=$datosMeses.$rowmeses['Total'].",";
}
$labelMeses=rtrim($labelMeses,",");
$datosMeses=rtrim($datosMeses,",");


?>
 
<script type="text/javascript">
  var labeldeMes=[<?php echo $labelMeses; ?>]; //Here I declare the variable labeldeMes
  var datosdeMes=[<?php echo $datosMeses; ?>]; //Here I declare the variable datosdeMes
</script>

<section class="panelControl">
    <div class="container">
        
       <div class="container">
        <div class="content">
            <div class="row">
                <div class="col-lg-4">
                <div class="card  custom-card">
                <div class="card-header">
                    <h5 class="card-category">Pacientes</h5>
                </div>

             <!--MY CAHRT-->

            <div class="card-body">
                    <div class="chart-area">
                    <canvas id="chartLinePurple"></canvas>
                    </div>
                </div>
                </div>
            </div>
        </div>
        
        
            </section>
<?php

 include('plantillas/footer.php');

?>

餐桌

CREATE TABLE `pacientes` (
  `id` int(11) NOT NULL,
  `nombre` varchar(30) COLLATE utf8_spanish_ci NOT NULL,
  `apellidos` varchar(30) COLLATE utf8_spanish_ci NOT NULL,
  `nombre_documento` varchar(20) COLLATE utf8_spanish_ci NOT NULL,
  `num_documento` int(12) NOT NULL,
  `telf` int(16) NOT NULL,
  `sexo` varchar(10) COLLATE utf8_spanish_ci NOT NULL,
  `embarazo` varchar(2) COLLATE utf8_spanish_ci NOT NULL,
  `fecha_nacimiento` date NOT NULL
  
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_spanish_ci;

INSERT INTO `pacientes` (`id`, `nombre`, `apellidos`, `nombre_documento`, `num_documento`, `telf`, `sexo`, `embarazo`, `fecha_nacimiento`) VALUES
(6, 'marta', 'mangue', 'dip', 655, 222987665, 'Femenino', '', '2021-12-28'),
(7, 'Mayo', 'Esono', 'dip', 5769, 551233234, 'Masculino', '', '2021-10-25'),
(40, 'Celso', 'NUESTRA', 'dip', 4334, 4343, 'Masculino', '', '2021-11-19'),
(41, 'Nathan', 'Narby Esimi', 'dip', 1111, 654890, 'Masculino', '', '2021-12-28');

图表.js

    var ctx = document.getElementById("chartLinePurple").getContext("2d");

    var gradientStroke = ctx.createLinearGradient(0, 230, 0, 50);

    gradientStroke.addColorStop(1, 'rgba(72,72,176,0.2)');
    gradientStroke.addColorStop(0.2, 'rgba(72,72,176,0.0)');
    gradientStroke.addColorStop(0, 'rgba(119,52,169,0)'); //purple colors

    var data = {

      
      labels: labeldeMes, //THIS IS WHERE I CALLED MY VARIABLE to show me the months

      
      datasets: [{
        label: "Months",
        data: datosdeMes, //THIS IS WHERE I CALLED MY VARIABLE datosdeMes
      }]
    };

    var myChart = new Chart(ctx, {
      type: 'line',
      data: data,
      options: gradientChartOptionsConfigurationWithTooltipPurple
    });
    }
}

这是footer.php文件

<!--LIBRARIES USED . -->


<script src="librerias/charts/charts.js"></script>

<script>

      
    $(document).ready(function() {
      // Initializing the graphics
      demo.initDashboardPageCharts();

    });


 
  </script>

Answer 1

当需要调试某些东西或进行一般开发时，调高 error_reporting 级别并确保显示错误是个好主意。 将以下两行添加到 index.php（或随处包含的另一个 PHP 文件）的顶部 -

<?php

error_reporting(E_ALL);
ini_set('display_errors', 1);

您对$querymeses的字符串分配无效 -

$querymeses = 'SELECT Monthname(fecha_nacimiento) AS mes, count(*) AS Total FROM pacientes GROUP BY mes' WHERE YEAR(fecha_nacimiento) = '2021';

你应该看到一个错误——

Parse error: syntax error, unexpected identifier "WHERE" in /tmp/preview on line 5

WHERE 子句需要位于 GROUP BY 之前，并且由于YEAR(fecha_nacimiento) = 2021条件，查询 不可搜索。 我喜欢使用heredoc / nowdoc ，因为我认为它使它更具可读性 -

$querymeses = <<<'SQL'
    SELECT Monthname(fecha_nacimiento) AS mes, count(*) AS Total
    FROM pacientes
    WHERE fecha_nacimiento BETWEEN '2021-01-01' AND '2021-12-31'
    GROUP BY mes
    SQL;

你有$rowmeses=mysqli_fetch_assoc($resmeses); 在你的 while 循环之上，这意味着你没有使用结果集的第一行。 当您将标签回显到您的 JS 中时，需要引用字符串。

把它们放在一起你最终会得到 -

<?php

error_reporting(E_ALL);
ini_set('display_errors', 1);

$querymeses = <<<'SQL'
    SELECT Monthname(fecha_nacimiento) AS mes, count(*) AS Total
    FROM pacientes
    WHERE fecha_nacimiento BETWEEN '2021-01-01' AND '2021-12-31'
    GROUP BY mes
    SQL;
$resmeses = mysqli_query($conexion, $querymeses);

$labelMeses = [];
$datosMeses = [];

foreach ($resmeses as $row) {
    $labelMeses[] = $row['mes'];
    $datosMeses[] = $row['Total'];
}

?>
 
<script type="text/javascript">
    var labeldeMes = ['<?php echo implode("','", $labelMeses); ?>']; //Here I declare the variable labeldeMes
    var datosdeMes = [<?php echo implode(',', $datosMeses); ?>]; //Here I declare the variable datosdeMes
</script>