未捕获的引用错误:.. 未定义 .. 但我可以在错误字符串中看到该变量 [英] Uncaught reference error: .. is not defined .. but I can see the variable in the error string

问题描述

这行代码 var $sTest = <?php echo $a = $_GET['ID']?> 返回未捕获的引用错误:.. 未定义."

This line of code var $sTest = <?php echo $a = $_GET['ID']?> returns a "Uncaught reference error: .. is not defined."

对于这个问题，如果我们假设我传递的是ABC"的 ID.当我在 Chrome 中使用开发人员工具并检查资源时，我可以看到 $_GET 已 起作用，因为它向我显示了变量值.这种情况下的错误消息是未捕获的引用错误:未定义 ABC."

For this question, if we assume I am passing an ID of "ABC." When I use the Developer Tools in Chrome and inspect the Resources, I can see that the $_GET has worked because it shows me the variable value. The error message in this case is "Uncaught reference error: ABC is not defined."

我不知道如何将 php 变量分配给脚本变量.感谢帮助.

I can't work out how to assign the php variable to a script var. Help appreciated.

推荐答案

这是因为你必须引用你的 php 脚本的输出:

It's because you have to quote your php script's output:

var $sTest = <?php echo $a = $_GET['ID']?>

将成为

var $sTest = something;

在您的 Javascript 环境中，something 未定义.如果你想保留尽可能多的类型信息，你应该使用 json_encode 在 php 端是这样的:

And in your Javascript enviroment the something is not defined. If you want to preserve as much type information as you can, you should use json_encode on the php side like this:

var $sTest = <?php print json_encode($_GET['ID']);?>;

这篇关于未捕获的引用错误:.. 未定义 .. 但我可以在错误字符串中看到该变量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！