有时我必须将一些变量从 PHP 传递到 JS 脚本.现在我是这样做的:
From time to time I have to pass some variables from PHP to JS script. For now I did it like this:
var js-variable = "<?php echo $php-variable; ?>";
但这非常难看,我无法在 .js 文件中隐藏我的 JS 脚本,因为它必须由 PHP 解析.处理这个问题的最佳解决方案是什么?
But this is very ugly and I can't hide my JS script in .js file because it has to be parsed by PHP. What is the best solution to handle this?
如果您不想使用 PHP 来生成您的 javascript(并且不介意额外调用您的网络服务器),请使用 AJAX 来获取数据.
If you don't want to use PHP to generate your javascript (and don't mind the extra call to your webserver), use AJAX to fetch the data.
如果您确实想使用 PHP,请在输出前始终使用 json_encode 进行编码.
If you do want to use PHP, always encode with json_encode before outputting.
<script>
var myvar = <?php echo json_encode($myVarValue); ?>;
</script>
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