datediff在php中導致錯誤

我有sql語句，在mysql中使用很好，我正在使用datediff。當我嘗試在PHP中使用它，我得到一個「mysql_fetch_arrary（）預計參數1是在給定的資源，布爾」datediff在php中導致錯誤

這聲明是...

$result = mysql_query("select hname, hsn, hmodel, hmake, htype, hwar, datediff(`hwar`, now()) from host where stype='physical';",$db);

我知道的聲明當我刪除datediff( hwar , now())，我的網頁作品在MySQL

mysql> select hname, hsn, hmodel, hmake, htype, hwar, datediff(`hwar`, now()) from servers.host where stype='physical'; 
+--------------+---------+--------+-----------+-------+------------+-------------------------+ 
| hname  | hsn  | hmodel | hmake  | htype | hwar  | datediff(`hwar`, now()) | 
+--------------+---------+--------+-----------+-------+------------+-------------------------+ 
| moscow  | XXXXXXX | Dell | PowerEdge | R710 | 2013-09-13 |      225 | 
| sydney  | XXXXXXX | Dell | PowerEdge | R710 | 2013-09-15 |      227 |

工作。我想使用它作爲字段

$datediff=$row['datediff'];

任何線索，爲什麼它不起作用？

來源

2013-01-31 user2030147

的'mysql'擴展是邪惡的訪問列。改爲使用'PDO_MySQL'或'MySQLi'。請參閱http://php.net/en/mysql-connect – KingCrunch

嘗試，使限制在查詢

$result = mysql_query("select hname, hsn, hmodel, hmake, htype, hwar, datediff(`hwar`, now()) as dif from host where stype='physical' LIMIT 0,10 ;",$db);

看this post

來源

2013-01-31 18:52:01

試試這個：

更換這塊sql語句：DATEDIFF（hwar，NOW（））

With：datediff（hwar，now（））as Diffdate

並以這種方式恢復： $ datediff = $ row ['Diffdate'];

再見XD

來源

2013-01-31 18:53:58 Hackerman

你應該給選擇一個別名：

select hname, hsn, hmodel, hmake, htype, hwar, datediff(`hwar`, now()) AS diff from host where stype='physical';

然後你可以名爲「差異」

來源

2013-01-31 18:54:11 richjhall18

datediff在php中導致錯誤

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