如何让DISCUZ发贴"禁用 URL 识别" 默认是选中状态?

在网上找到一个解决办法,就是全局禁用url识别,方法如下:

打开：include\javascript\post.js

查找并删除之：

CODE:

function parseurl(str, mode) {
        str= str.replace(/\s*\[code\]([\s\S]+?)\[\/code\]\s*/ig, function($1, $2) {return codetag($2);});
        str = str.replace(/([^>=\]"'\/]|^)((((https?|ftp):\/\/)|www\.)([\w\-]+\.)*[\w\-\u4e00-\u9fa5]+\.([\.a-zA-Z0-9]+|\u4E2D\u56FD|\u7F51\u7EDC|\u516C\u53F8)((\?|\/|:)+[\w\.\/=\?%\-&~`@':+!]*)+\.(jpg|gif|png|bmp))/ig, mode == 'html' ? '$1<img src="$2" border="0">' : '$1');
        str = str.replace(/([^>=\]"'\/@]|^)((((https?|ftp|gopher|news|telnet|rtsp|mms|callto|bctp|ed2k|thunder|synacast):\/\/))([\w\-]+\.)*[:\.@\-\w\u4e00-\u9fa5]+\.([\.a-zA-Z0-9]+|\u4E2D\u56FD|\u7F51\u7EDC|\u516C\u53F8)((\?|\/|:)+[\w\.\/=\?%\-&~`@':+!#]*)*)/ig, mode == 'html' ? '$1<a href="$2" target="_blank">$2</a>' : '$1$2');
        str = str.replace(/([^\w>=\]"'\/@]|^)((www\.)([\w\-]+\.)*[:\.@\-\w\u4e00-\u9fa5]+\.([\.a-zA-Z0-9]+|\u4E2D\u56FD|\u7F51\u7EDC|\u516C\u53F8)((\?|\/|:)+[\w\.\/=\?%\-&~`@':+!#]*)*)/ig, mode == 'html' ? '$1<a href="$2" target="_blank">$2</a>' : '$1$2');
        str = str.replace(/([^\w->=\]:"'\.\/]|^)(([\-\.\w]+@[\.\-\w]+(\.\w+)+))/ig, mode == 'html' ? '$1<a href="mailto:$2">$2</a>' : '$1[email]$2[/email]');
        for(var i = 0; i <= codecount; i++) {
                str = str.replace("[\tDISCUZ_CODE_" + i + "\t]", codehtml[i]);
        }
        return str;
}

更好的解决办法是在viewthread.htm, forumdisplay.htm, post_editor.htm, viewthread_fastreply.htm中找到

CODE:

"parseurloff" value="1"

后面都添加：

CODE:

checked

即可.