Question

我几天前发帖 Pass MySQL medium blob into IOS NSData\UIImage 但是还没有设法找到解决方案。我的具体问题是，当我从数据库，blob或文本中检索图像时，它返回NULL，其中没有变量的行只返回空白

while ($row = mysql_fetch_array($result)) {

 $add = array(); 
 $add["Id"] = $row["Id"]; 
 $add["Mail"] = $row["Mail"];
 $add["Category"] = $row["Category"]; 
 $add["Phone"] = $row["Phone"]; 
 $add["Msgs"] = $row["Msgs"];
 //image from blob
 $add["Picture"] = $row["Picture"];

 // push single add into final response array 
 array_push($response["adds"], $add); 
}

还有其他方法可以解决blob \ text中的图像吗？！我试过https://stackoverflow.com/a/6273415/1333294，但对我没什么用。

Answer 1

最后!!

    $img = $row["Picture"];
    $b64img = base64_encode ($img);
    $b64img = mysql_real_escape_string($b64img);
    $add["Picture"] = $b64img;

这很简单！和xcode得到它作为base64字符串和坚果NULL。谢谢你的建议...

Answer 2

添加以下代码

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

echo $row['picture'];

// Display image with img html tag

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

echo 'Hello world.';

请查找示例代码

$response["adds"] = array();
$add = array(); 
$add["Id"] = 1; 
$add["Mail"] = "test@mm.com";
$add["Category"] = "category"; 
$add["Phone"] = "1234567"; 
$add["Msgs"] = "ths s txt msg";
//image from blob
$add["Picture"] = "pciture";

// push single add into final response array 
array_push($response["adds"], $add); 
print_r($response["adds"]);
header("Content-type: image/jpeg");
print $response["adds"][0]['Picture'];

来自MySQL的blob \ text图像返回NULL

2 个答案: