MySQL 问题 - 警告 : mysql_num_rows() expects parameter 1 to be resource, bool 值给出

Question

需要了解错误，我需要修复错误，以便显示数据库中的车辆。这是 bool 值，为什么它不起作用这是 9 个月前网站的备份。如果有人可以帮助我，请告诉我

               </div>

<?php if(isset($_GET['dmsg']) && $_GET['dmsg']==1) { echo '<div align="center" style="color:green; font-weight:bold;">vehicles Deleted successfully..</div>';} ?>
<a href="insert-vehicles.php" style="  background: none repeat scroll 0 0 green;    color: white; float: right;  margin: 0 20px 0 0;  padding: 10px;">Insert new Vehicles</a>
            <div class="offer_area">

                <div class="offer_head row">

                    <form action="" method="get" name="form1" class="key_form">

                        <label>Keyword :(search by Make, Model, Year, Stock no)</label>

                        <input name="keyword" type="text"  value="<?php if($_GET['keyword']) {echo $_GET['keyword']; } ?>"/>

                        <input name="search" class="search" type="submit" value="search" />

                    </form>

                </div>
                <br />
                <br />
                <br>
<br>
<table width="90%" align="center" cellpadding="2" cellspacing="0">
  <tr>
    <th scope="col">IMAGE</th>
    <th scope="col">STOCK NUMBER</th>
    <th scope="col">MAKE</th>
    <th scope="col">MODEL</th>
    <th scope="col">YEAR</th>    
    <th scope="col">AUCTION DATE</th>
     <th scope="col">SALE DOCUMENT</th>
     <th scope="col">CURRENT BID PRICE</th>
     <th scope="col">BUY NOW PRICE</th>
     <th scope="col">ACTION</th>
  </tr>
  <?php
   $id=1;
   while ($GetUserQryRow = mysql_fetch_array($GetUserQryRs)) {
   ?>
  <tr>
    <td align="center"><img src="<?php echo $GetUserQryRow['defaultImg'];  ?>" width="96"/> </td>
    <td align="center"><?php echo $GetUserQryRow['StockNumber'];  ?> </td>
    <td align="center"><?php echo $GetUserQryRow['Make']; ?></td>
    <td align="center"><?php echo $GetUserQryRow['Model']; ?> </td>
    <td align="center"><?php echo $GetUserQryRow['Year'];  ?></td>
    <td align="center"><?php echo date('m, d Y H:i:s A',strtotime($GetUserQryRow['LiveAuctionDate']));  ?> </td>
    <td align="center"><?php echo $GetUserQryRow['SaleDocument'];  ?> </td>
    <td align="center"><?php  if(is_numeric($GetUserQryRow['CurrentBidPrice'])){ echo number_format($GetUserQryRow['CurrentBidPrice'],2); }  ?> </td> 
    <td align="center"><?php if(is_numeric($GetUserQryRow['ibuyCost'])){echo number_format($GetUserQryRow['ibuyCost'],2); } ?> </td>
    <td align="center"><a href="edit-vehicles.php?id=<?php echo $GetUserQryRow['StockNumber'];  ?>">Update</a>&nbsp;&nbsp;<a href="javascript:void(0);" onclick="confirmdel('<?php echo $GetUserQryRow['StockNumber'];  ?>');">Delete</a> </td>
  </tr>
    <?php $id++; } ?>
</table>




          </div>



  <?php include('sections/footer.php'); ?>

    </div>

        <p>&nbsp;</p>
</body>

</html>

Answer 1

问题在于，当您使用 mysql_query() 发出 sql 查询时，您的代码中绝对没有错误处理。您应该通过检查每个 mysql_query() 的返回值来添加错误处理，并记录/显示使用 mysql_error() 遇到的任何错误消息。

$res=mysql_query(...);
if($res) {
  //do your normal stuff
}
else {
  echo mysql_error();
  //decide to stop or render futher parts of the website
}

根据 mysql 错误消息，您可以解决您遇到的问题。然而，我的猜测是，与数据库的连接有缺陷。密码、用户名或主机名错误。

编辑:请注意 mysql_*() 函数已处于弃用状态，并且它们的支持将从 php7 中删除。转换为 mysqli 或 PDO。