嘿,我在使用php顯示圖像時遇到問題。這些圖像被存儲在mysql中的表格'images'中。還有另一個餐桌需要取出這些圖像並根據restid顯示各自的圖像。但是,它在獲取圖像並不顯示它們時面臨着一個問題。請幫忙! 這是imageupload.php:在php中顯示圖像時出錯mysql mysql
<?php
require 'connect.inc.php';
?>
<html>
<head>
<title>Uploading image</title>
</head>
<body>
<?php
echo "<form action='imageupload.php' method='POST' enctype='multipart/form-data'>
Upload: <input type='file' name='image'><input type='submit' value='Upload' >
</form>";
if(isset($_FILES['image']['tmp_name']))
{
$image = addslashes(file_get_contents($_FILES['image']['tmp_name']));
$image_name = addslashes($_FILES['image']['name']);
$image_size = getimagesize($_FILES['image']['tmp_name']);
if($image_size==FALSE)
echo "That's not an image";
else
{
$query = "INSERT INTO images VALUES ('','$image_name','$image','22')";
$result = mysqli_query($con, $query);
if(!$result)
{
echo "Problem uploading";
}
else
{
echo "Image uploaded ";
$query2 = "SELECT * FROM images WHERE restid = '22'";
$result2 = mysqli_query($con,$query2);
while($info = mysqli_fetch_array($result2))
{
header("Content-type: image/jpeg");
echo $info['image'];
}
}
}
}
else
{
"Please upload a file";
}
?>
</body></html>
這是getimage.php(它取該圖像,並將其顯示):
<?php
require 'connect.inc.php';
$id = $_REQUEST['id'];
$image = "SELECT * FROM images WHERE imgid = $id" ;
$image = mysqli_query($con, $image);
$image = mysqli_fetch_assoc($image);
$image = $image['image'];
header("Content-type: image/jpeg");
echo $image;
?>
connect.inc.php是連接到數據庫中的文件。我提到其他鏈接,但沒有得到任何堅實的幫助。請提供幫助。
爲什麼你neet改變標題顯示圖像?保存在圖像名稱中的名稱是否正常?爲什麼喲不使用'
'? –
是的是我試過
,但它顯示圖像的加密形式,即以文本形式。 –
user2758453
您保存圖像,原始名稱或base64的自定義圖像嗎? –