将PHP变量转换为javascript变量时出现未定义的参考错误

Question

For an internet programming assignment we need to parse data from .json files using PHP. 对于Internet编程，我们需要使用PHP从.json文件中解析数据。

Here is the php code where I get the text from the files: 这是我从文件中获取文本的php代码：

$pointsA = json_decode(utf8_encode(file_get_contents("pointsA.json")), true);
$pointsB = json_decode(utf8_encode(file_get_contents("pointsB.json")), true);
$mapCenter = utf8_encode(file_get_contents("center.json"));

And here is the code where I attempt to pass it to javascript. 这是我尝试将其传递给javascript的代码。

    <script type="text/javascript">
        var mapCenter, aMatches, bMatches;
        mapCenter = <?php echo $mapCenter;?>;
        aMatches = <?php echo json_encode($aMatches);?>;
        bMatches = <?php echo json_encode($bMatches);?>;
    </script>

This last block is located in the of the html code, and it is located before any of the other code used. 最后一块位于html代码的中，并且位于所使用的任何其他代码之前。 SPecifically, the head looks like this: 特别是，头部看起来像这样：

<head>
    <title>Route Plotting</title>
    <link rel="stylesheet" type="text/css" href="style.css"/>
    <script type="text/javascript">
        var mapCenter, aMatches, bMatches;
        mapCenter = <?php echo $mapCenter;?>;
        aMatches = <?php echo json_encode($aMatches);?>;
        bMatches = <?php echo json_encode($bMatches);?>;
    </script>
    <script
        type="text/javascript"
        src="https://maps.googleapis.com/maps/api/js?key=AIzaSyAdt2pyYUsR8xq7n_0InalSC7mahTk6Hcg&amp;libraries=places">
    </script>
    <script
        type="text/javascript"
        src="googleMapsCalendar.js">
    </script>
</head>

When I render the webpage in Google Chrome I get the following console error message: 当我在Google Chrome浏览器中呈现网页时，出现以下控制台错误消息：

Uncaught SyntaxError: Unexpected token var index.php:12
Uncaught ReferenceError: mapCenter is not defined

The line number for the first error message is not correct, as it leads to a comment. 第一条错误消息的行号不正确，因为它会引起注释。 I believe it is talking about first line in the first tag that i posted. 我相信这是在谈论我发布的第一个标签中的第一行。

The second error message points to this piece of code: 第二条错误消息指向这段代码：

var mapCenterLoaction = new google.maps.LatLng(mapCenter['center']['lat'], mapCenter['center']['lat']);

Which is the fist place I attempt to use 'mapCenter' variable in an external javascript file. 这是我尝试在外部javascript文件中使用“ mapCenter”变量的第一位置。

Any help would be appreciated. 任何帮助，将不胜感激。

EDIT: 编辑：

here is the generated code from inspecting the first element in the head: 这是检查头部的第一个元素生成的代码：

        var mapCenter, aMatches, bMatches;
        mapCenter = var center=
{
    "center": { "lat" : "44.974", "long" : "-93.234" },
    "zoom": "15"
};
        aMatches = [0];
        bMatches = [0];

Does that mean that I don't need the 'var' keyword when defining these, but only to echo them in? 这是否意味着在定义它们时我不需要'var'关键字，而只是将它们回显？

If that's the case, how do I refer to these variables in an external javascript file? 在这种情况下，如何在外部javascript文件中引用这些变量？

SECOND EDIT: 第二编辑：

The issue was that the file that I was parsing which I thought was json was actually javascript. 问题是我解析的文件（我认为是json）实际上是javascript。 Thank you to all who contributed to solving my problem! 感谢所有为解决我的问题做出贡献的人！

Answer 1

Try this: 尝试这个：

mapCenter = '<?php echo $mapCenter;?>';

I think you get this error because $mapCenter is empty. 我认为您会收到此错误，因为$ mapCenter为空。

将PHP变量转换为javascript变量时出现未定义的参考错误

问题描述

1 个解决方案

解决方案1
0 2014-11-03 21:17:52

将PHP变量转换为javascript变量时出现未定义的参考错误

问题描述

1 个解决方案

解决方案1 0 2014-11-03 21:17:52

解决方案1
0 2014-11-03 21:17:52