php – SQL命令的类型转换警告

我现在已经坚持了一段时间,我知道这是初学者,但是找不到任何类似的问题.

我试图显示我的最后一个主题细节,但我收到了警告.

*Warning: pg_exec() [<a href='function.pg-exec'>function.pg-exec</a>]:
Query Failed: 
ERROR: operator does not exist: character varying = integer LINE 4: WHERE
t_cat = 3 ^
HINT: No operator matches the given name and argument type(s). You
might need to add explicit type casts.*

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$topicsearh = pg_exec($db, 
     "SELECT t_id, t_subject, t_date, t_cat
      WHERE t_cat = " . $row['s_id'] . "
      ORDER BY t_date DESC LIMIT 1"
);      
if(!$topicsearh){
          echo 'Last topic Could not be displayed.';
}
else{
      while($trow = pg_fetch_assoc($topicsearh))
      echo '<a href="topicview.PHP?id=' . $trow['t_id'] . '">' . $trow['t_subject'] .   
          '</a> at ' . date('d-m-Y', strtotime($trow['t_date']));
}

解决方法:

您需要定义FROM表.

例

SELECT t_id, t_subject, t_date, t_cat FROM TABLE_NAME WHERE...
                            ----------^^^^^^^^^^^^^^^-----

并且还如下所述.

WHERE t_cat = '". $row['s_id'] ."'