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 - js jQuery, always have a number display at least two digits 00 - Stack Overflow

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    javascript - js jQuery, always have a number display at least two digits 00 - Stack Overflow

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    I'm using the following to add one to number:

    <div id="count">00</div>
<div id="count">03</div>
<div id="count">08</div>
<div id="count">12</div>

$('#count').text(function(i,txt) { return parseInt(txt, 10) + 1; });

    I always want there two be 2 places, 00 even if the number is under 10. How can I get the func above, with JS, to always return the 2 00 places? So if the number putes to 3, it injects 03 into #count?

    Thanks

    I'm using the following to add one to number:

    <div id="count">00</div>
<div id="count">03</div>
<div id="count">08</div>
<div id="count">12</div>

$('#count').text(function(i,txt) { return parseInt(txt, 10) + 1; });

    I always want there two be 2 places, 00 even if the number is under 10. How can I get the func above, with JS, to always return the 2 00 places? So if the number putes to 3, it injects 03 into #count?

    Thanks

    • javascript
    • jquery
    Share Improve this question edited Nov 23, 2010 at 20:36 user113716 323k64 gold badges453 silver badges441 bronze badges asked Nov 23, 2010 at 19:41 AnApprenticeAnApprentice 111k202 gold badges636 silver badges1k bronze badges 1
    • 1 FYI it is invalid to have more than a single element with the same ID. This will not valid correctly and JavaScript may behave unexpectedly on different browsers. – Hamish Commented Nov 23, 2010 at 19:56
    Add a ment  | 

    4 Answers 4

    Reset to default 5 
    $('#count').text(function(i,txt) { var c = parseInt(txt, 10) + 1; return (c<10) ? "0"+c : c; });

    EDIT: But having multiple elements with the same ID is gonna cause problems somewhere.

    Here's a little different approach from the others. Doesn't use a conditional operator.

    $('#count').text(function(i, txt) {
    return ("0" + (+txt + 1)).slice(-2);
});

    It just assumes it will need the extra 0, then returns a slice of the last two characters in the string.

    You can add a "0" if it's less than 10, like this:

    $('#count').text(function(i,txt) { 
  var num = parseInt(txt, 10) + 1; 
  return num < 10 ? "0" + num : num;
});

    I think it's just an example, but if it's not note that IDs have to be unique.

    Try this one:

    $('#count').text(function(i,txt) { return txt.length == 1 ? '0'+txt : txt; });

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