從SQL查詢更改PHP頁面上的顯示

Question

我有一個SQL查詢經過搜索並顯示數百列的記錄。但是兩列的顯示必須改變;生效日期和清除日期。目前它們都是這樣顯示的：（例如1991年8月29日）。我想將輸出更改爲1991年8月29日。我不確定我會怎麼做，因爲它將數據直接從數據庫中提取到表中。
請注意我在這個項目中使用MS-SQL和PHP。

這是我的SQL查詢：

$sql = "select SerialNum as [Serial Number],ts_sitename As Site,(case m.Scratched 
    when 0 then 'Live' 
    when 1 then 'Free' 
    END) as Status, Note as Comment, (case Destroyed when 0 then 'NO' 
    when 1 then 'YES' END) as [Destroyed], 
    SUBSTRING(cast(mg_effectivedate as char), 1, 12) AS [Effective Date], 
    SUBSTRING(cast(mg_effectivedate as char), 12, 8) AS [Effective Time], 
    SUBSTRING(cast(ScratchedDate as char), 1, 4) + '-' + 
    SUBSTRING(cast(ScratchedDate as char), 3, 2) + '-' + 
    SUBSTRING(cast(ScratchedDate as char), 5, 2) 
    AS [Scratched Date], 
    SUBSTRING(cast(ScratchedDate as char), 10, 2) + ':' + 
    SUBSTRING(cast(ScratchedDate as char), 12, 2) + ':' + 
    SUBSTRING(cast(ScratchedDate as char), 14, 2) 
    AS [Scratched Time], 
    SUBSTRING(cast(mg_scratchdate as char), 1, 12) AS [Purge Date], 
    (select fl_filename from TheFiles_tab where mg_filenum = fl_filenum) as [Dataset], 
    (select hs_hostname from TheHosts_tab where mg_hostnum = hs_hostnum) as [Host], 
    (select UserCode from [User] where mg_usernum = UserId) as [UserCode] 
    from ((Media m left join MediaGenT g on m.MediaId = g.mg_medianum) 
    join TheSites_tab s on m.SiteId = s.ts_sitenum) 
    join Note n on m.NoteId = n.NoteId 
    where SerialNum like '" . $userQuery . "%'"; 

這我的代碼顯示兩列：

echo '<th>Effective Date</th>'; 
    echo '<th>Purge Date</th>'; 

    foreach ($result as $r) 
    { 
    echo "<tr>"; 
    echo "<td>".$r['Serial Number'] . " </td>"; 
    echo "<td>".$r['Site'] . " </td>"; 
    echo "<td>".$r['Status'] . "</td> "; 
    echo "<td>".$r['Comment'] . " </td>"; 
    echo "<td>".$r['Destroyed'] . " </td>"; 
    echo "<td>".$r['Effective Date'] . " </td>"; 
    echo "<td>".$r['Effective Time'] . "</td> "; 

    echo "<td>"; 
    if ($r['Scratched Date'] == "") { 
    echo "NULL"; 
    } 
    else { 
    echo $r['Scratched Date']; 
    } echo "</td>"; 

    echo "<td>" ; 
    if ($r['Scratched Time'] == "") { 
    echo "NULL"; 
    } 
    else { 
    echo $r['Scratched Time']; 
    } echo "</td>"; 


    echo "<td>".$r['Purge Date'] . " </td>"; 
    echo "<td>".$r['Dataset'] . "</td> "; 
    echo "<td>".$r['Host'] . " </td>"; 
    echo "<td>".$r['UserCode'] . " </td>"; 
    echo "<td>".$r['NoteId'] . " </td>"; 
    echo "</tr>"; 
    } 

請注意，我是新來的SO社區。我一直使用這個網站進行研究，但第一次剛剛創建了一個帳戶。

Answer 1

如果我理解正確，你只是想替換連字符日期中的空格？

現在您的日期格式： MMM DD YYYY

你想：如果我沒有記錯 DD-MMM-YYYY

，在PHP中，你可以這樣做：

$effective_date = date("d-M-Y", strtotime($r['Effective Date'])); 

那將數據庫提取的值轉換爲時間戳，然後將該時間戳格式化爲新的日期符號。查看php的date（）函數，瞭解您有哪些其他選項：PHP: date - Manual

您也可以更改其他日期。

編輯

總PHP代碼：

echo '<th>Effective Date</th>'; 
echo '<th>Purge Date</th>'; 

foreach ($result as $r) 
{ 
    echo "<tr>"; 
    echo "<td>".$r['Serial Number'] . " </td>"; 
    echo "<td>".$r['Site'] . " </td>"; 
    echo "<td>".$r['Status'] . "</td> "; 
    echo "<td>".$r['Comment'] . " </td>"; 
    echo "<td>".$r['Destroyed'] . " </td>"; 
    echo "<td>".date("d-M-Y", strtotime($r['Effective Date'])) . " </td>"; 
    echo "<td>".$r['Effective Time'] . "</td> "; 

    echo "<td>"; 
    if ($r['Scratched Date'] == "") { 
    echo "NULL"; 
    } 
    else { 
    echo $r['Scratched Date']; 
    } echo "</td>"; 

    echo "<td>" ; 
    if ($r['Scratched Time'] == "") { 
    echo "NULL"; 
    } 
    else { 
    echo $r['Scratched Time']; 
    } echo "</td>"; 


    echo "<td>".date("d-M-Y", strtotime($r['Purge Date'])) . " </td>"; 
    echo "<td>".$r['Dataset'] . "</td> "; 
    echo "<td>".$r['Host'] . " </td>"; 
    echo "<td>".$r['UserCode'] . " </td>"; 
    echo "<td>".$r['NoteId'] . " </td>"; 
    echo "</tr>"; 
}