将PHP变量传递给Jquery弹出窗口 [英] Pass PHP Variable into Jquery popup

问题描述

我有以下设置：
JQuery：

I have the following setup: JQuery:

$('.btn-setting').click(function(e){
        e.preventDefault();
        $('#myModal').modal('show');
    });

这会打开一个HTML文件作为对话框，问题是我无法将任何PHP变量传递到该页面供参考：

Which opens a HTML file as a Dialog box, the problem is that I cannot pass any PHP variables onto the page for reference:

<a href="<?= echo $UserID; ?>" class="btn btn-info btn-setting">Click for dialog</a>

有人建议让我这样做的方法吗？

Anyone suggest a method to enable me to do so?

好的。问题是现在，我正在查看页面的来源..然而，它显示了我所期待的，现在当我点击按钮调出对话框时，它显示的是不同的数据......

Alright. The problem is now, i'm viewing the source of the page.. Yet, it's showing what i'm expecting, now When I click the button to bring up the dialog box, it's showing different data...

不推荐我知道，但是现场直播：

Not recommended I know, but heres a live:

92.236.219.136/Admin/

92.236.219.136/Admin/

用户：堆栈

通过：堆栈

推荐答案

这是最强大的方式，特别是如果你想传递的不仅仅是一个字符串：

This is the most robust way, especially if you want to pass more than just one string:

<?php

// Call this function to pass an object to the JS code
function php_vars_to_js($id, $obj) {
  echo "<script id='$id' type='text/php_data'>";
  echo htmlspecialchars( json_encode( obj ) );
  echo '</script>';
}

// for examlpe
php_vars_to_js( "php_vars", array("a_string" => "foo", "a_num" => 12 ) );

然后以这种方式阅读：

var php_vars = $.jsonDecode( $("php_vars").text() );

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