如何将 PHP 字符串传递到 Javascript 函数调用中?

声明:本页面是StackOverFlow热门问题的中英对照翻译,遵循CC BY-SA 4.0协议,如果您需要使用它,必须同样遵循CC BY-SA许可,注明原文地址和作者信息,同时你必须将它归于原作者(不是我):StackOverFlow 原文地址: http://stackoverflow.com/questions/13905313/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me): StackOverFlow

提示:将鼠标放在中文语句上可以显示对应的英文。显示中英文
时间:2020-10-26 20:07:35  来源:igfitidea点击:

How do I pass a PHP string into a Javascript function call?

phpjavascriptstringparameters

提问by Brecon

Possible Duplicate:
Pass a PHP string to a Javascript variable (and escape newlines)

可能的重复:
将 PHP 字符串传递给 Javascript 变量(并转义换行符)

So, essentially I am trying to pass a string from a PHP page as an argument for a javascript function. The PHP is included in the page that the script is on, but they are in two separate files.

所以,本质上我试图从 PHP 页面传递一个字符串作为 javascript 函数的参数。PHP 包含在脚本所在的页面中,但它们位于两个单独的文件中。

However, whenever I try to pass a string, it stops the javascript from running. I can pass PHP integers, but not strings. Is it possible to fix this? I've been looking around the internet for what might be causing my error, but I can't find anything. Any help would be greatly appreciated.

但是,每当我尝试传递字符串时,它都会停止运行 javascript。我可以传递 PHP 整数,但不能传递字符串。有没有可能解决这个问题?我一直在互联网上寻找可能导致我出错的原因,但我找不到任何东西。任何帮助将不胜感激。

This is the PHP and HTML. The function runs when submit is clicked.

这是 PHP 和 HTML。该函数在单击提交时运行。

$comment = "7b";
echo"
    <table>
         <p><input type='text' id='newPostComment' value='' placeholder='WRITE YOUR COMMENT HERE...' size='35'/><input type='submit' value='UPLOAD' onclick='uploadPostComment($parentID, $comment);fetchComment()'/></p>
    </table>

This is the javascript function.

这是javascript函数。

function uploadPostComment(PostCID, Comment){

      var PostCommentData = "fsPostID="+PostCID;
      PostCommentData += "&fsPostComment="+Comment;

      var xmlHttp = new XMLHttpRequest();
      xmlHttp.open("post", "uploadPostComment.php", true);

      xmlHttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
      xmlHttp.setRequestHeader("Content-length", PostCommentData.length);
      xmlHttp.setRequestHeader("Connection", "close");

      xmlHttp.onreadystatechange = function()
         {
             if(xmlHttp.readyState == 4)
             {
                if(xmlHttp.status == 200)
                {
                    alert("Message Posted");
                }
                else
                {
                    alert("Oh no, an error occured2"); 
                }
             }
         };
         xmlHttp.send(PostCommentData);}

回答by cryptic ツ

echo"
    <table>
         <p><input type='text' id='newPostComment' value='' placeholder='WRITE YOUR COMMENT HERE...' size='35'/><input type='submit' value='UPLOAD' onclick='uploadPostComment(\"$parentID\", \"" . str_replace('"', '\"', $comment) . "\");fetchComment()'/></p>
    </table>

You need to quote the fields. If your field is an integer it does not need to be quoted, if it is a string it needs to be. I quoted both because I don't know if parentID is a string id or a numeric id. Do what your application needs of course.

您需要引用字段。如果您的字段是整数,则不需要引用,如果是字符串,则需要引用。我引用了两者,因为我不知道 parentID 是字符串 id 还是数字 id。做你的应用程序当然需要什么。

Update: in regards to Michael's comment about breaking if comment contains a quote. We now escape all double quotes in $comment to prevent that.

更新:关于迈克尔关于打破如果评论包含引用的评论。我们现在转义 $comment 中的所有双引号以防止这种情况发生。

回答by Eugen Rieck

<input type='submit' value='UPLOAD' onclick='uploadPostComment($parentID, $comment);fetchComment()'/>

is your culprit, especially uploadPostComment($parentID, $comment);- if $comment is a string, you need to put quotes in place: uploadPostComment($parentID, \"$comment\");

是你的罪魁祸首,尤其是uploadPostComment($parentID, $comment);- 如果 $comment 是一个字符串,你需要把引号放在适当的位置:uploadPostComment($parentID, \"$comment\");

回答by Viral Patel

Wrap the string in escaped double quotes like \"$comment\"

将字符串包裹在转义的双引号中,例如 \"$comment\"

$comment = "7b";
echo"
    <table>
         <p><input type='text' id='newPostComment' value='' placeholder='WRITE YOUR COMMENT HERE...' size='35'/><input type='submit' value='UPLOAD' onclick='uploadPostComment(\"$parentID\", \"$comment\");fetchComment()'/></p>
    </table>