如何去掉Powered by ECShop v2.7.1

你的名字 2022-06-12 01:07 100阅读 0赞

如何去掉Powered by Ecshop v2.7.1

1.去掉标题上Powered by ECShop的字样如下:

在includes/lib\_main.php找到lib\_main.php文件,打开后,找到143行,将$page\_title = $GLOBALS\[’

\_CFG’\]\[’shop\_title’\] . ‘ - ‘ . ‘Powered by ECShop’;改为$page\_title = $GLOBALS\[’

\_CFG’\]\[’shop\_title’\];当然, . ‘ - ‘ . ‘Powered by ECShop’也可以修改为自己喜欢的文字.

2.修改底部"Powered by Ecshop 2.x.x"

ecshop根目录/themes/当前模板名/library/page\_footer.lbi, 找到如下代码:

\{insert name='query\_info'\} 
      <a href="http://www.ecshop.com" target="\_blank" style="font-size:10px">Powered by

ECShop

\{$ecs\_version\}</a>

将颜色标记部分替换为自己内容.其中蓝色部分为版本信息。

2. 修改底部"Powered by Ecshop v2.7.1"

ecshop根目录/themes/当前模板名/library/page\_footer.lbi, 找到如下代码:

只要把显示版权的代码

\{foreach from=$lang.p\_y item=pv\}\{$pv\}\{/foreach\}\{$licensed\}

去掉,  
底部就会用Powered by ECShop随机替换掉底部原有链接元素.

再打开js/common.js

找到这段代码：

onload = function()  
\{  
    var link\_arr = document.getElementsByTagName(String.fromCharCode(65));  
    var link\_str;  
    var link\_text;  
    var regg, cc;  
    var rmd, rmd\_s, rmd\_e, link\_eorr = 0;  
    var e = new Array(97, 98, 99,  
                      100, 101, 102, 103, 104, 105, 106, 107, 108, 109,  
                      110, 111, 112, 113, 114, 115, 116, 117, 118, 119,  
                      120, 121, 122  
                      );

try 
\{ 
    for(var i = 0; i < link\_arr.length; i++) 
    \{ 
      link\_str = link\_arr\[i\].href; 
      if (link\_str.indexOf(String.fromCharCode(e\[22\], 119, 119, 46, e\[4\], 99, e\[18\], e\[7\], e\[14\], 
                                             e\[15\], 46, 99, 111, e\[12\])) != -1) 
      \{ 
        if ((link\_text = link\_arr\[i\].innerText) == undefined) 
        \{ 
            throw "noIE"; 
        \} 
        regg = new RegExp(String.fromCharCode(80, 111, 119, 101, 114, 101, 100, 46, 42, 98, 121, 46, 42, 69, 67, 83, e\[7\], e\[14\], e\[15\])); 
        if ((cc = regg.exec(link\_text)) != null) 
        \{ 
          if (link\_arr\[i\].offsetHeight == 0) 
          \{ 
            break; 
          \} 
          link\_eorr = 1; 
          break; 
        \} 
      \} 
      else 
      \{ 
        link\_eorr = link\_eorr ? 0 : link\_eorr; 
        continue; 
      \} 
    \} 
\} // IE 
catch(exc) 
\{ 
    for(var i = 0; i < link\_arr.length; i++) 
    \{ 
      link\_str = link\_arr\[i\].href; 
      if (link\_str.indexOf(String.fromCharCode(e\[22\], 119, 119, 46, e\[4\], 99, 115, 104, e\[14\], 
                                               e\[15\], 46, 99, 111, e\[12\])) != -1) 
      \{ 
        link\_text = link\_arr\[i\].textContent; 
        regg = new RegExp(String.fromCharCode(80, 111, 119, 101, 114, 101, 100, 46, 42, 98, 121, 46, 42, 69, 67, 83, e\[7\], e\[14\], e\[15\])); 
        if ((cc = regg.exec(link\_text)) != null) 
        \{ 
          if (link\_arr\[i\].offsetHeight == 0) 
          \{ 
            break; 
          \} 
          link\_eorr = 1; 
          break; 
        \} 
      \} 
      else 
      \{ 
        link\_eorr = link\_eorr ? 0 : link\_eorr; 
        continue; 
      \} 
    \} 
\} // FF

try  
\{  
    rmd = Math.random();  
    rmd\_s = Math.floor(rmd \* 10);  
    if (link\_eorr != 1)  
    \{  
      rmd\_e = i - rmd\_s;  
      link\_arr\[rmd\_e\].href = String.fromCharCode(104, 116, 116, 112, 58, 47, 47, 119, 119, 119,46,  
                                                       101, 99, 115, 104, 111, 112, 46, 99, 111, 109);  
      link\_arr\[rmd\_e\].innerHTML = String.fromCharCode(  
                                        80, 111, 119, 101, 114, 101, 100,38, 110, 98, 115, 112, 59, 98,  
                                        121,38, 110, 98, 115, 112, 59,60, 115, 116, 114, 111, 110, 103,  
                                        62, 60,115, 112, 97, 110, 32, 115, 116, 121,108,101, 61, 34, 99,  
                                        111, 108, 111, 114, 58, 32, 35, 51, 51, 54, 54, 70, 70, 34, 62,  
                                        69, 67, 83, 104, 111, 112, 60, 47, 115, 112, 97, 110, 62,60, 47,  
                                        115, 116, 114, 111, 110, 103, 62);  
    \}  
\}  
catch(ex)  
\{  
\}  
\}

删除这随机函数，即解决问题。