用PHP替换HTML替换函数 [英] HTML replaceWith function with PHP

问题描述

  $（'＃content'）。replaceWith（< div>+<？php echo $ tableau_Text？> +< / div>）; 
  

但是不适合我。它对我没有任何意义。当我将 PHP代码更改为静态类型时：

  $（ '#content'）。replaceWith（< div> Hello World< / div>）; 
  

它可以完美地与此相符。

我的html＃内容定义是： -

 < div id =content> 
<？php echo $ show_menu;？> 
< / div> 
  

有什么想法吗？非常感谢！

更新：

var $ tableau_text 看起来像这样：

  $ tableau_Text =< ul class ='object_list'id ='object_list'data -role ='listview'data-inset ='true'data-theme ='c'> $ TableauText< / ul>; 
  

其中 $ TableauText 是相同的格式。

  $ TableauText =< li class ='menuA'id ='menuA'data-role ='list-divider'data -icon ='false'> River Flow< / li> 
< li class ='menuB'id ='menuB'data-icon ='arrow-r'>< a href ='＃'target ='foo'> Rossiniere< / a>< /立GT; 
< li class ='menuB'id ='menuB'data-icon ='arrow-r'>< a href ='＃'target ='foo'> Montsalvens< / a>< /立GT; 
< li class ='menuA'id ='menuA'data-role ='list-divider'data-icon ='false'> Lake Level< / li> 
< li class ='menuB'id ='menuB'data-icon ='arrow-r'>< a href ='＃'target ='foo'> Rossiniere< / a>< /立GT; 
< li class ='menuB'id ='menuB'data-icon ='arrow-r'>< a href ='＃'target ='foo'> Montsalvens< / a>< /立GT; 
< li class ='menuB'id ='menuB'data-icon ='arrow-r'>< a href ='＃'target ='foo'> Lessoc< / a>< /立GT; 
< li class ='menuB'id ='menuB'data-icon ='arrow-r'>< a href ='＃'target ='foo'> Rossens< / a>< /立GT; 
< li class ='menuB'id ='menuB'data-icon ='arrow-r'>< a href ='＃'target ='foo'> Maigrauge< / a>< /立GT; 
< li class ='menuB'id ='menuB'data-icon ='arrow-r'>< a href ='＃'target ='foo'> Schiffenen< / a>< /立GT; 
< li class ='menuB'id ='menuB'data-icon ='arrow-r'>< a href ='＃'target ='foo'> Maigrauge< / a>< /立GT; 
  

解决方案：我粘贴了几个具有相同ID的列表元素，在这里，这让我的代码变得疯狂......

解决方案

试试这个：

  $（'＃content'）。replaceWith（< div> +<？php echo json_encode（$ tableau_Text）;？> +< / div> ;）; 
  

鉴于：

  $ TableauText ='hello world'; 
 
 $ tableau_Text =
< ul class ='object_list'id ='object_list'data-role ='listview'data-inset ='true'data-theme ='c' > 
 $ TableauText 
< / ul>; 
  

浏览器将会显示：

<$ p $数据 - 角色='列表视图'数据

$（'＃content'）。replaceWith（< div> +\\\
< ul class ='object_list'id ='object_list' -inset ='true'data-theme ='c'> \\\
hello world \\\\\
< \ / ul>+< / div>）


其中呈现为

  hello world
  

I am trying to execute this code in javascript:

 $('#content').replaceWith("<div>" + <?php echo $tableau_Text?> + "</div>");

But is not working for me. It shows nothing to me. When I change the PHP code for something static like:

 $('#content').replaceWith("<div>Hello World</div>");

it works perfectly with that.

My html #content definition is:--

<div id="content">
    <?php echo $show_menu;?>
</div>

Any idea? Thank you very much!

UPDATE:

The var $tableau_text looks like this:

$tableau_Text = "<ul class='object_list' id='object_list' data-role='listview' data-inset='true' data-theme='c'>$TableauText</ul>";

Where $TableauText are the list items in the same format.

$TableauText = <li class='menuA' id='menuA' data-role='list-divider' data-icon='false'>River Flow</li>
<li class='menuB' id='menuB' data-icon='arrow-r'><a href='#' target='foo'>Rossiniere</a></li>
<li class='menuB' id='menuB' data-icon='arrow-r'><a href='#' target='foo'>Montsalvens</a></li>
<li class='menuA' id='menuA' data-role='list-divider' data-icon='false'>Lake Level</li>
<li class='menuB' id='menuB' data-icon='arrow-r'><a href='#' target='foo'>Rossiniere</a></li>
<li class='menuB' id='menuB' data-icon='arrow-r'><a href='#' target='foo'>Montsalvens</a></li>
<li class='menuB' id='menuB' data-icon='arrow-r'><a href='#' target='foo'>Lessoc</a></li>
<li class='menuB' id='menuB' data-icon='arrow-r'><a href='#' target='foo'>Rossens</a></li>
<li class='menuB' id='menuB' data-icon='arrow-r'><a href='#' target='foo'>Maigrauge</a></li>
<li class='menuB' id='menuB' data-icon='arrow-r'><a href='#' target='foo'>Schiffenen</a></li>
<li class='menuB' id='menuB' data-icon='arrow-r'><a href='#' target='foo'>Maigrauge</a></li>

Solution: I had several list elements with the same id which I didn't realized after I paste it here and that was making my code crazy...

解决方案

Try this:

$('#content').replaceWith("<div>" + <?php echo json_encode($tableau_Text); ?> + "</div>");

Given:

$TableauText = 'hello world ""';

$tableau_Text = "
        <ul class='object_list' id='object_list' data-role='listview' data-inset='true' data-theme='c'>
        $TableauText
        </ul>";

The browser will see:

$('#content').replaceWith("<div>" + "\n        <ul class='object_list' id='object_list' data-role='listview' data-inset='true' data-theme='c'>\n        hello world \"\"\n        <\/ul>" + "</div>")​

Which renders as

hello world ""

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