Question

我正在创建一个函数，在插入用户名后，将从数据库中检索用户显示图片。但是我的代码遇到了一些问题。请检查我ya ^^ 非常感谢你。

<?php
include("connection.php");

$name = $_SESSION['login_username']; // login_username is test123, this is $_SESSION from another .php file
$_SESSION['name'] = $storename;


echo '<img src="display2.php"width="90" height="90"/>'; //this is how i display my picture
?>

Display.php（不工作）

mysql_select_db($database) or die("Can not select the database: ".mysql_error());

$storename = $_SESSION['name']; // is there an error with my $_session statement?
$name = (string)$storename; 

if(!isset($name) || empty($name)){
     die("Please select your image!");

}else{


$query = mysql_query("SELECT * FROM customerdetail WHERE customer_username='$name'");
$num_row = mysql_fetch_array($query);
$content = $num_row['image'];

header('Content-type: image/jpg');
echo $content;

 }
}

Display.php的（工作）

    $storename = "test123"; // it worked if i store the id in string but not passing from another page.
    $name = (string)$storename;

if(!isset($name) || empty($name)){
     die("Please select your image!");
}else{

$query = mysql_query("SELECT * FROM customerdetail WHERE customer_username='$name'");
$row = mysql_fetch_array($query);
$content = $row['image'];

header('Content-type: image/jpg');
         echo $content;

}
?>

任何人都可以帮我弄清楚出了什么问题吗？提前谢谢。

Answer 1

header('Content-type: image/jpg');声明输出将是jpg图像，因为代码从<img src="">

运行

无法将文字显示为iamges

Answer 2

在你的php文件的最顶层（两者都是），把这行代码放在<?php标签之后，当然）：

session_start();

图片无法在php中显示

2 个答案: