将变量添加到json

Question

所以这里有很多关于这个的问题，我尝试了很多但没有成功。

我正在尝试将一个 php 变量添加到一个 javascript 变量，然后将其放入 json 中。

<?php $symbol_json =  $row['symbol_json']; echo $symbol_json; ?> <!--This is OK!-->

</div>

</center>

<center>

<!-- Widget BEGIN -->

<div class="widget-container">

  <div class="widget-container__widget"></div>

  <div class="widget-copyright"><a href="https://aaaaaaa.com/symbols/<?php echo $row['symbol_html']; ?>/" rel="noopener" target="_blank"><span class="blue-text"><?php echo $row['symbol']?> Symbol Info</span></a></div>

  <script>

    var myVar = '<?php echo $symbol_json ?>'; alert(myVar); //This works ok!

    var myObject = JSON.parse('<?php echo json_encode($symbol_json) ?>'); alert(myObject); //Alert works ok!

  </script>

  <script type="text/javascript" src="https://aaaaaaa.com/external-embedding/symbol-info.js" async>

  {

  //"symbol": "SDT", //This how it is at present and it works. I need to change it to a variable like below.

  "symbol": myObject, //Whether I use myObject or myVar (which both contain SDT) the widget loads a different symbol. 

  "width": 1000,

  "locale": "uk",

  "colorTheme": "light",

  "isTransparent": false

}

  </script>

</div>

<!-- Widget END -->

Answer 1

你为什么不把你的变量放在同一个脚本标签中？作为一般性建议，请不要将 PHP 与 Javascript 或 HTML 混淆。这是让你的代码有味道的最好方法。您必须解析 JSON 才能将其用作变量，因此&nbsp;var&nbsp;myObject&nbsp;=&nbsp;'<?php&nbsp;echo&nbsp;json_encode($symbol_json)&nbsp;?>';成为&nbsp;var&nbsp;myObject&nbsp;=&nbsp;JSON.parse('<?php&nbsp;echo&nbsp;json_encode($symbol_json)&nbsp;?>');