3

我正在尝试加载存储在数据库中的图像,但无法正常工作。我只能看到链接,但看不到图像。我在存储图片的字段中使用 longblob。

JavaScript:

  <script>

    $(document).ready(function(){
         $.ajax({type: "POST",
                 url: "cargaImg.php",
                 success:function(data) {
                    $('#pinta').html(data);
                    }

         });  
    });
  </script> 

PHP

(images.php)

 <? require('conecta.php');
   $stmt=$oConni->prepare("SELECT PIC FROM FOTOS WHERE ID_PIC=?");
   $stmt->bind_param('i',$_GET['id']);
   $stmt->execute();
   $stmt->store_result();
   $stmt->bind_result($Foto);
   while ($stmt->fetch()) {
     header('Content-Type: image/jpeg');
     print $Foto;
   }
 ?>

(cargaImg.php)

    <?php
        require('conecta.php');
        ini_set('display_errors',1); error_reporting(E_ALL);  

        $cSQL="SELECT ID_PIC, PIC, NOMBRE FROM FOTOS";

        $stmt=$oConni->prepare($cSQL) or die($oConni->error);
        //$stmt->bind_param('i',$_POST['local']);
        $stmt->execute();                               
        $stmt->bind_result($id, $pic, $nombre);
        //$i=0;
        echo '<table cellspacing="0">';
        while ($stmt->fetch()) {

            if (!empty($pic)){ ?>
            <tr><td><img class="sifoto" src="images.php?id=<?=$id?>" width="60" height="60"  /></td></tr>
            <?}
            echo'<tr><td value="'.$id.'"><a target="_blank" href="'.$nombre.'">Enlace</a></td></tr>'; 
            //$i++;
        }   
        $stmt->close();
        echo'</table>';

     ?>
4

1 回答 1

1

已解决:我需要这个:$stmt->store_result();

    <?php
        require('conecta.php');
        ini_set('display_errors',1); error_reporting(E_ALL);  
        $cSQL="SELECT ID_PIC, PIC, NOMBRE FROM FOTOS";
        $stmt=$oConni->prepare($cSQL) or die($oConni->error);
        //$stmt->bind_param('i',$_POST['local']);
        $stmt->execute(); 
        $stmt->store_result();
        $stmt->bind_result($id, $pic, $nombre);
        //$i=0;
        echo '<table cellspacing="0">';
        while ($stmt->fetch()) {

            if (!empty($pic)){ 
            echo'<tr><td><img class="sifoto" src="images.php? id='.$id.'" width="100" height="100"  /></td></tr>';
            }
            echo'<tr><td value="'.$id.'"><a target="_blank" href="'.$nombre.'">Enlace</a></td></tr>'; 
            //$i++;
        }   
        $stmt->close();
        echo'</table>';

?>
于 2013-04-08T11:57:23.027 回答