0

大家好,我想通过 GET 方法将两个查询中的两个值发送到另一个页面

一个值发送正确但第二个值失败意味着我只收到一个值

但不是第二个,我需要在新查询的其他页面中使用这两个值,并且

执行一个动作

请问有人吗??????

这是我的代码:

    <?php
    require_once('db.php');
    if(isset($_POST['property']))
    {
    $property=$_POST['property'];
    $propertyquery = "SELECT PropertyID, PropertyImageID, PropertyName, PropertyStatus, PropertyLocation, PropertyRegion FROM properties WHERE PropertyImageID =$property";
     $propertyquery_run= mysql_query($propertyquery);
    if (mysql_num_rows($propertyquery_run) > 0) 
    {
        while ($propertyrow = mysql_fetch_array($propertyquery_run)) 
          {
        $imagequery ="SELECT PropertyImageID, ImagePath FROM propertyimages WHERE PropertyImageID='".$propertyrow['PropertyImageID']."' LIMIT 1";
         $imagequery_run=mysql_query($imagequery);        
    ?>
            <table style="margin-left:348px;">
    <tr><td>PropertyID:</td><td><span style="color:#0F0; font-weight:bold;"><?php echo $propertyrow['PropertyID'] ?></span></td></tr><tr><td>PropertyName:</td><td><span style="color:#0F0; font-weight:bold;"><?php echo $propertyrow['PropertyName'] ?></span></td></tr>
    <tr><td>PropertyStatus:</td><td><span style="color:#0F0; font-weight:bold;"><?php echo $propertyrow['PropertyStatus'] ?></span></td></tr>
    <tr><td>PropertyLocation:</td><td><span style="color:#0F0; font-weight:bold;"><?php echo $propertyrow['PropertyLocation'] ?></span></td></tr>
    <tr><td>PropertyRegion:</td><td><span style="color:#0F0; font-weight:bold;"><?php echo $propertyrow['PropertyRegion'] ?></span></td></tr>
            </table>
        <center>
    <?php }
                if(mysql_num_rows($imagequery_run) > 0)  
                {
                    while ($imagerow = mysql_fetch_array($imagequery_run))  
                    {
                    ?>
                       <div style="border:solid 2px #9F0; border-radius:5px; height:222px; width:544px;">

<!-- sending value started -->  

    <a href="DisplayImages.php?PropertyImageID=<?php echo $imagerow['PropertyImageID']     ?>&PropertyID=<?php echo $propertyrow['PropertyID'] ?>"> 



                       <img src="<?php echo $imagerow['ImagePath'];  ?>" width="544" height="222" ></a>    <!-- sending value Ended -->

                       </div><br />    
                    <?php
                    }

                }
            }
    }
    else
    {
        echo 'Go Back And Set Your Session.Thanks';
    }
    ?>
4

1 回答 1

-1

可能是这个工作检查

<a href="DisplayImages.php?PropertyImageID=<?php echo $imagerow['PropertyImageID']; ?>&PropertyID=<?php echo $propertyrow['PropertyID']; ?>"> 

在上面的代码中,你错过了两个回声';'

于 2013-04-23T05:19:19.163 回答