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我試圖加載存儲在數據庫中的圖像,但不工作。我只能看到鏈接,但沒有圖像。我在存儲圖片的字段中使用longblob。不能從數據庫加載圖像
的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++;
}
$stmt->close();
echo'</table>';
?>
'$ stmt-> execute(); $ stmt-> store_result(); $ stmt-> bind_result($ Foto);'不應該在綁定它們之後存儲結果嗎? – hjpotter92 2013-04-08 10:19:08
這沒有看起來適合我的圖像路徑..'src =「images.php?id = =$id?>」' – bipen 2013-04-08 10:20:52
@bipen哦,但它是! – hjpotter92 2013-04-08 10:22:11