在表

时间:2018-04-24 10:46:28

标签: php blob

我知道在桌子中存储图像是不受欢迎的,但你能告诉我为什么这不起作用吗?我试图在桌子的最后一栏显示一个小图像,没有任何乐趣。

header("content-type:image/jpeg");

// Retrieve the boat name from the URL.    
$angler = $_GET['angler'];

$sql = "SELECT DateCaught, Angler, Species, Notes, PhotoName, Photo FROM SpeciesHuntAnglers WHERE Angler='$angler' ORDER BY DateCaught ASC";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
     echo "<br><br><table border=1 bordercolor=#EBF4F9 cellspacing=0 cellpadding=3 style=color:#3D6594><tr><th>Date Caught</th><th>Angler</th><th>Species</th><th>Notes</th><th>PhotoName</th><th>Photo</th></tr>";
    // output data of each row
    while($row = $result->fetch_assoc()) {
  echo "<tr><td>" . date('F Y', strtotime($row["DateCaught"])). "</td><td>" . $row["Angler"]. "</td><td>" . $row["Species"]. "</td><td>" . $row["Notes"]. "</td><td>" . $row["PhotoName"]. "</td><td><img src="data:image/jpeg; base64,'.base64_encode($row['Photo']).'"/></td></tr>";
    }
    echo "</table><br>";
  } 

1 个答案:

答案 0 :(得分:0)

Rajnish - 谢谢。它应该是:

<img src='data:image/jpeg;base64,".base64_encode($row['Photo'])."'/>