在PHP中导致错误

时间:2013-01-31 18:42:24

标签: php mysql datediff

我有sql语句在mysql中运行很好,我正在使用datediff。当我尝试在php中使用它时,我得到一个“mysql_fetch_arrary()期望参数1是一个资源,布尔值在”

中给出

这句话是......

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

我知道该语句适用于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 |

当我删除datediff( hwar , now())时,我的页面正常运行。我想将其用作字段

$datediff=$row['datediff'];

关于它为什么不起作用的任何线索?

3 个答案:

答案 0 :(得分:0)

尝试在查询中设置限制

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

this post

答案 1 :(得分:0)

试试这个:

替换这段sql语句:datediff(hwar,now())

使用:datediff(hwar,now())作为Diffdate

以这种方式恢复: $ DATEDIFF = $行[ 'Diffdate'];

Bye XD

答案 2 :(得分:0)

您应该为选择器添加别名:

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

然后您可以访问名为'diff'

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