如何转义 PHP 中回显的 Javascript 代码
我有这段代码是从 php 页面的 jquery 数据对象中捕获的。
echo "
var $d = $('<div/>', {
id: 'hi' + $('#textResp').children().length,
class: 'eventdiv',
html: 'hello'
}).hide().fadeIn(3000);
$('#textResp').append($d)
";
问题是， ' 没有被转义。我尝试过使用 /' 来转义，但出现错误。我确信我做错了，有谁知道在哪里放置 /' 而不是 '？
I have this code that is captured in the jquery Data object from a php page.
echo "
var $d = $('<div/>', {
id: 'hi' + $('#textResp').children().length,
class: 'eventdiv',
html: 'hello'
}).hide().fadeIn(3000);
$('#textResp').append($d)
";
Problem is, the 's are not escaped. I have tried using /' to escape, but it comes up with an error. I am sure I am doing this wrong, does anyone know where to put the /' instead of '?
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您可以使用 php nowdoc< /a> 而不是引号，这会简化事情：
然后在里面使用你想要的任何内容（quote 或 dquote）。当然，这是未解析的，因此如果 $d 实际上引用了 php var 那么你就会遇到问题。
You could use a php nowdoc instead of quotes at all which would simplify things:
then use whatever you want inside (quote or dquote). This is, of course, unparsed so if $d was actually referring to a php var then you would have problems.
你的撇号实际上看起来很好。但是，在双引号字符串中，PHP 会将任何以美元符号开头的字符串作为变量进行计算，并且不会产生所需的结果。尝试用
\$替换$的 jquery 相关实例。像这样：Your apostrophes actually look fine. But, within a double quoted string, PHP will evaluate any string beginning with a dollar sign as a variable and not produce the desired result. Try replace the jquery related instances of
$with\$. Like this:在php中使用
json_encode函数，它的行为类似于rails中的escape_javascript函数。只需将字符串参数传递给 json_encode 函数，它就会为您返回转义字符串，请参阅下面的示例代码：
use
json_encodefunction in php, it behaves like theescape_javascriptfunction in rails.just pass a string argument to the
json_encodefunction, and it return the escaped string for you, see the sample code below:您需要在所有
'之前使用\。然而，这令人费解，为什么你觉得需要转义字符？看来您只是回显此输出，如果这是在
You will need to use
\before all's.However, this is puzzling, why do you feel you need escape characters? It appears you are simply echoing this output, if this is between
<script />tags, you should be fine.PHP 将尝试扩展出现在双引号字符串中的变量
$name。由于$d对于 PHP 解释器来说看起来像一个变量，因此它会尝试用变量的值替换它。假设您没有在任何地方定义
$d，这将产生一个空白空间，并且可能会产生一个通知（如果您使用错误级别 E_NOTICE）。为了防止这种情况发生，请使用反斜杠转义美元符号（将
$替换为\$）PHP will attempt to expand variables,
$name, that occur in strings wrapped in double quotes. Since$dlooks like a variable to the PHP interpreter, it will try to replace it with the variable's value.Assuming that you don't have
$ddefined anywhere, that will produce an empty space and, possibly, a notice (if you are using error level E_NOTICE).To prevent that from happening, escape dollar signs with a backslash (replace
$with\$)使用单引号构建字符串。仅当您明确包含要评估的变量时才使用双引号。 PHP 正在尝试评估其中的所有 $ 引用。通过使用单引号，您将避免许多转义问题。
Use single quotes for your string construction. Only use double quotes when you specifically are including variables that you want evaluated. PHP is trying to evaluate all of those $ references you have in there. By using single quotes, you will avoid many escaping problems.