Question

我试图在浏览器上显示存储在mysql数据库中的3个图像，但我发现只显示loop WHILE上第一个迭代的第一个图像。如何在浏览器中显示3个图像？

<?php
$con = mysqli_connect("localhost","root","","othmane") or die(mysqli_error($con));
  
if($_SERVER['REQUEST_METHOD']=='GET'){
//$id = $_GET['id']; 
$sql = "SELECT image,image_type FROM images where id between 2 and 6";

$r = mysqli_query($con,$sql) or die(mysqli_error($con));;
  
$result=mysqli_fetch_array($r);
header('Content-Type:image/jpeg');
$ss=$result['image_type'];

while($row = $r->fetch_assoc()){
if ($ss == 'php') {
            echo ( $result['image']);
} else if ($ss == 'android') {
            echo base64_decode( $result['image'] );      
}
}
}else{
mysqli_close($con);
}
?>

此代码：

<?php
$con = mysqli_connect("localhost","root","","othmane") or die(mysqli_error($con));
  
if($_SERVER['REQUEST_METHOD']=='GET'){
//$id = $_GET['id']; 
$sql = "SELECT image,image_type FROM images where id between 1 and 3";

$r = mysqli_query($con,$sql) or die(mysqli_error($con));;
  
//$result=mysqli_fetch_array($r);
$result=var_dump(mysqli_fetch_all($r));
//header('Content-Type:image/jpeg');
while($row = mysqli_fetch_assoc($r))
{
	var_dump($row) ;
	}

/*while($row = $r->fetch_assoc()){
	$ss=$row['image_type'];
if ($ss == 'php') {
            echo ( $row['image']);
} else if ($ss == 'android') {
            echo base64_decode( $row['image'] );      
}
	}*/
}
else{
mysqli_close($con);
}
?>

，结果如下：

Answer 1

修改。正如我已经看过这个问题，你应该能够做到：

<img src="data:image/jpeg;base64,
 <?php echo base64_encode( $image_data ); ?>
" />

显示所有图片，因此下面应该有效：

while($row = mysqli_fetch_assoc($r)){
                $ss = $row['image_type'];
                if ($ss == 'php') {
<img src="data:image/jpeg;base64, <?php echo ( $row['image']); ?>" />
                } else if ($ss == 'android') {
<img src="data:image/jpeg;base64, <?php echo base64_decode( $row['image'] ); ?> " />     
                }
            }

当您执行 $ r-＆gt; fetch_assoc（）时（只要您的查询获得结果，您的While循环将继续），您将获取一个结果行数据到关联数组中，因此您可以将数据放在 $ row 变量中。您必须使用该变量来检查 [＆＃39; image_type＆＃39;] 并获取 [＆＃39;图像＆＃39;] 。

在您的代码中， $ result [＆＃39; image_type＆＃39;] 和 $ result [＆＃39; image＆＃39;] 绑定到您使用 $ result = mysqli_fetch_array（$ r）获取的第一条记录（作为文档说明：mysqli_result :: fetch_array - mysqli_fetch_array - 将结果行作为关联，数字数组或两者获取）

if($_SERVER['REQUEST_METHOD']=='GET'){

    $sql = "SELECT image,image_type FROM images where id BETWEEN 2 and 6";

    $r = mysqli_query($con,$sql) or die(mysqli_error($con));

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

    while($row = mysqli_fetch_assoc($r)){
        $ss = $row['image_type'];
        if ($ss == 'php') {
                    echo ( $row['image']);
        } else if ($ss == 'android') {
                    echo base64_decode( $row['image'] );      
        }
    }
}else{
    mysqli_close($con);
}

当你使用 procedular 样式时，一直使用它，如（来自php.net的例子）：

$link = mysqli_connect("localhost", "my_user", "my_password", "world");

$query = "SELECT Name, CountryCode FROM City ORDER by ID DESC LIMIT 50,5";

$result = mysqli_query($link, $query);

while ($row = mysqli_fetch_assoc($result)) {
    printf ("%s (%s)\n", $row["Name"], $row["CountryCode"]);
}

或只使用面向对象样式：

$mysqli = new mysqli("localhost", "my_user", "my_password", "world");

$query = "SELECT Name, CountryCode FROM City ORDER by ID DESC LIMIT 50,5";

$result = $mysqli->query($query);

while ($row = $result->fetch_assoc()) {
    printf ("%s (%s)\n", $row["Name"], $row["CountryCode"]);
}

更多信息，示例：mysqli_result::fetch_assoc

使用php显示浏览器上的图像列表

1 个答案: