問題在PHP

Question

頭看看下面的代碼

<?php 

$username = "root"; 
$password = ""; 
$host = "localhost"; 
$database = "binaries"; 

@mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error()); 

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

$id = 5; 

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

$query = mysql_query("SELECT * FROM tbl_images WHERE id='".$id."'"); 
$row = mysql_fetch_array($query); 
$content = $row['imag']; 
header('Content-type: image/jpg'); 
echo '<table><tr><td height="700" width="700">';// Line X 
print $content; 

echo '</td></tr></table>';//Line Y 

} 

?> 

當我評論了線的X和Y的圖像被顯示出來，否則not.What可能是可能的原因是什麼？

編輯：以下馬特的意見後。

show.php

echo '<table><tr><td> 
    <img src="image.php"/> 
    </td></tr></table>'; 

image.php

$query = mysql_query("SELECT * FROM tbl_images WHERE id='".$id."'"); 
$row = mysql_fetch_array($query); 
$content = $row['imag']; 
header('Content-type: image/jpg'); 

print $content; 

即使這樣做之後，我沒有得到預期的結果。

編輯：「image.php」的 代碼：

<?php 
    $username = "root"; 
    $password = ""; 
    $host = "localhost"; 
    $database = "binaries"; 

    @mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error()); 

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




$query = mysql_query("SELECT * FROM tbl_images WHERE id=5"); 
$row = mysql_fetch_array($query); 
$content = $row['imag']; 
header('Content-type: image/jpg'); 
echo $content; 

?> 

Answer 1

因爲當瀏覽器被告知的MIME類型爲image/jpg，它期待看到的最後一件事是<table ...

當你設置MIME類型，你告訴瀏覽器「我給你發送一個圖像。」但是，HTML標記肯定不是圖像數據，所以瀏覽器不知道如何呈現它。