Question

我正在尝试根据MYSQL中的数据在PHP中生成JavaScript图表不确定此代码有什么问题，因为所有标签都已正确放置。以下是我的代码。任何帮助将不胜感激。感谢

<?php

include("includes/fusioncharts.php");

$hostdb = "localhost";  // MySQl host
$userdb = "root";  // MySQL username
$passdb = "";  // MySQL password
$namedb = "fusioncharts_phpsample";  // MySQL database name

$dbhandle = new mysqli($hostdb, $userdb, $passdb, $namedb);

if ($dbhandle->connect_error) {
exit("There was an error with your connection: ".$dbhandle->connect_error);
}


$strQuery = "SELECT Name, Population FROM Country ORDER BY Population DESC LIMIT 10";


$result = $dbhandle->query($strQuery) or exit("Error code ({$dbhandle->errno}): {$dbhandle->error}");


if ($result) {

$arrData = array(
  "chart" => array
  (
    "caption" => "Top 10 Most Populous Countries",
    "paletteColors" => "#0075c2",
    "bgColor" => "#ffffff",
    "borderAlpha"=> "20",
    "canvasBorderAlpha"=> "0",
    "usePlotGradientColor"=> "0",
    "plotBorderAlpha"=> "10",
    "showXAxisLine"=> "1",
    "xAxisLineColor" => "#999999",
    "showValues" => "0",
    "divlineColor" => "#999999",
    "divLineIsDashed" => "1",
    "showAlternateHGridColor" => "0"
  )

);

$arrData["data"] = array();

while($row = mysqli_fetch_array($result)) {
array_push($arrData["data"], array(
  "label" => $row["Name"],
  "value" => $row["Population"]
    )
  );
}

$jsonEncodedData = json_encode($arrData);

$columnChart = new FusionCharts("column2D", "myFirstChart" , 600, 300, "chart-1", "json", $jsonEncodedData);
$columnChart->render();
$dbhandle->close(); 

?>

Answer 1

您错过了}行的结束if ($result) {。

Answer 2

if ($result) {条件不会在任何地方结束，因为}缺失。

在最后一行添加}。

解析错误：意外的文件结束

2 个答案: