MySQL查询表-不显示第一个结果 [英] MySQL querying table - Doesn't show first result

问题描述

我有一个具有以下结构的数据库(四边形):

I have a DB (quartos) created with the following structure:

id_quarto  tipo_quarto  vista_quarto |
a      Single   Mar | 
b      Single   Mar |
c      Single   Mar | 
d      Single   Serra |

当tipo_quarto = Single和vista_quarto = Mar哪个值来自表单时，我希望它返回id_quarto的结果.

I want it to return the results of id_quarto, when tipo_quarto=Single and vista_quarto=Mar which values come from a form.

所以我写了以下内容:

$strSQL = "
SELECT id_quarto 
FROM quartos 
WHERE tipo_quarto='". $_POST['tipo_quarto'] ."' 
    AND vista_quarto='". $_POST['vista_quarto'] ."'";
$rs = mysql_query($strSQL);
$row = mysql_fetch_array($rs);

然后我将其循环并按如下所示写入表中:

Then i loop it and write in a table as followed:

while($row = mysql_fetch_array($rs)) {
<?
<table border="1">
     <tr align="left">
         <td width="75"><?php echo $row['id_quarto']; ?></td>
     </tr>
</table>

这里的问题是它不返回id_quarto = a，仅返回b和c.为什么会这样，我该怎么解决?

The problem here is that it does not return the id_quarto=a , only b and c. Why is that and what can i do to fix it?

谢谢.

推荐答案

在mysql_query($strSQL);之后，您还有多余的$row = mysql_fetch_array($rs);.然后在while循环中，再次读取$row(结果集中的第二行). 这样您的代码就会看起来

You have extra $row = mysql_fetch_array($rs); just after mysql_query($strSQL);. Then in while loop you read $row again (second row in resultset). So your code will look

$strSQL = "SELECT id_quarto FROM quartos 
  WHERE tipo_quarto='". $_POST['tipo_quarto'] ."' 
  AND vista_quarto='". $_POST['vista_quarto'] ."'";
$rs = mysql_query($strSQL);
// $row = mysql_fetch_array($rs); Don't need this line!!!
while($row = mysql_fetch_array($rs)) 
{
   // output ....
}

此外，添加用于处理mysql错误的代码总是很有意义.

Also, it always makes sense to add code for handling mysql errors.

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