我有以下代碼:MySQL和PHP - For Each?
SELECT q21, q21coding AS Description FROM `tresults_acme` WHERE q21 IS NOT NULL AND q21 <> '' ORDER BY q21coding
它帶回以下(節選):
Text Description
Lack of up to date equal pay cases&legislation - t... Content needs updating
The intranet could contain more "up to date traini... Content needs updating
Poorly set out. It is hard to find things. Difficulty in navigating/finding content
Only use the intranet as a necessity. Will ask my ... Difficulty in navigating/finding content
現在,我想在一個PHP頁面上表中顯示此但我有一些因爲我想顯示它的方式的問題,它需要如下:
Content needs updating
----------------------
[List all the comments relating to this description]
Difficulty in navigating/finding content
----------------------------------------
[List all the comments relating to this description]
等等。
現在我認爲這是一個PHP中的For Each循環,但是我的頭腦很糟糕 - 任何想法和建議都非常受歡迎!
感謝,
外觀上的jqGrid顯示此表+搜索和編輯開始看:http://www.trirand.net/demophp.aspx – 2010-10-11 09:25:51