te')); return $arr; } /* 遍历用户所有主题 * @param $uid 用户ID * @param int $page 页数 * @param int $pagesize 每页记录条数 * @param bool $desc 排序方式 TRUE降序 FALSE升序 * @param string $key 返回的数组用那一列的值作为 key * @param array $col 查询哪些列 */ function thread_tid_find_by_uid($uid, $page = 1, $pagesize = 1000, $desc = TRUE, $key = 'tid', $col = array()) { if (empty($uid)) return array(); $orderby = TRUE == $desc ? -1 : 1; $arr = thread_tid__find($cond = array('uid' => $uid), array('tid' => $orderby), $page, $pagesize, $key, $col); return $arr; } // 遍历栏目下tid 支持数组 $fid = array(1,2,3) function thread_tid_find_by_fid($fid, $page = 1, $pagesize = 1000, $desc = TRUE) { if (empty($fid)) return array(); $orderby = TRUE == $desc ? -1 : 1; $arr = thread_tid__find($cond = array('fid' => $fid), array('tid' => $orderby), $page, $pagesize, 'tid', array('tid', 'verify_date')); return $arr; } function thread_tid_delete($tid) { if (empty($tid)) return FALSE; $r = thread_tid__delete(array('tid' => $tid)); return $r; } function thread_tid_count() { $n = thread_tid__count(); return $n; } // 统计用户主题数 大数量下严谨使用非主键统计 function thread_uid_count($uid) { $n = thread_tid__count(array('uid' => $uid)); return $n; } // 统计栏目主题数 大数量下严谨使用非主键统计 function thread_fid_count($fid) { $n = thread_tid__count(array('fid' => $fid)); return $n; } ?>javascript - How to debug only one js file - Stack Overflow

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    I have lot of my JS files in my projects. Some of them are external libraries. I want to know what JS code is executing for current user interaction. I have set up break point in chrome as describe here. But there are already lots of JS files and lots are added by Visual Studio of his own. So it bees difficult for me to get to exact code. So in this case, I need something that will enable me to debug only xyz.js file.

    I have lot of my JS files in my projects. Some of them are external libraries. I want to know what JS code is executing for current user interaction. I have set up break point in chrome as describe here. But there are already lots of JS files and lots are added by Visual Studio of his own. So it bees difficult for me to get to exact code. So in this case, I need something that will enable me to debug only xyz.js file.

    • javascript
    • google-chrome
    • debugging
    • developer-tools
    • javascript-debugger
    Share Improve this question edited May 23, 2017 at 10:33 CommunityBot 11 silver badge asked Apr 29, 2016 at 12:26 ImadImad 7,49014 gold badges62 silver badges122 bronze badges
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    2 Answers 2

    Reset to default 9

    Open up settings in chrome dev tools, click on "Blackboxing" or possibly "Manage framework blackboxing" and set up file names or regex patterns. Any scripts matching this will not be debugged when you are stepping through your own code and reach a point at which you would enter that script.

    Based on adamtickner's explanation here is a detailed description. Let's assume the file you want to debug is "src/app.js"

    In Chrome open the debugger using F12 and then click settings (gear icon ⚙ , top right window corner). Select the "Ignore List" menu item.

    So far so easy. The tricky part is to add the right regular expression. We need an expression saying "everything, except...". This can be done by a technique called "negative look-arounds"

    Our positive expression would be:

    src\/app\.js

    We just escaped the slash and the dot as these are reserved regex characters.

    Now, using negative look-arounds, our expression should look like this:

    ^((?!src\/app\.js).)*$

    This expression will find anything which does NOT contain our file path. You can test it with regex101. Add this pattern to your Ignore List and you are done! Chrome will maybe show a misleading message when you debug in this file like "This script is on the debugger's ignore list" - just ignore this message, you can still debug into the ignored files.

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