我试图用下面的SQL语句计算年龄
round(datediff(now() - dateofbirth / 365))
给出以下错误,
1582 - 调用本机函数'datediff'时参数计数不正确
如何以正确的方式放置复选框的名称,然后它可以使用javascript运行
<?php $i=1; ?>
<?php do { $i++; ?>
<tr>
<td><input name="<?php echo 'checkbox[$i]' ; ?> " type="checkbox" id="<?php echo $row_RsActivitynoteMem['id']; ?>" value="<?php echo $row_RsActivitynoteMem['id']; ?>" />
<label for="checkbox"></label></td>
<td><?php echo $row_RsActivitynoteMem['Sname']; ?> <?php echo $row_RsActivitynoteMem['Ssurname']; ?></td>
<td width="20"><?php echo $row_RsActivitynoteMem['idactivity']; ?></td>
<td><?php echo $row_RsActivitynoteMem['kname']; ?></td>
<td> </td>
</tr>
<?php } while ($row_RsActivitynoteMem = mysql_fetch_assoc($RsActivitynoteMem)); ?>
<tr>
<td> </td>
<td> </td>
<td width="20"> </td>
<td> </td>
<td> </td>
</tr>
</table>
<input type="button" name="CheckAll" value="Check All"
onClick="checkAll(document.form1.checkbox)">
<input type="button" name="UnCheckAll" value="Uncheck All"
onClick="uncheckAll(document.form1.checkbox)">
<br>
</form>
</body>
</html> …这段代码不适用于mysql:
update member
set member.xy = memba.surba
from memba
where member.id =memba.id
低于0的代码有效:
update member inner join memba on member.id =memba.id
set member.xy = memba.surba
你能解释第一个代码有什么问题吗？