php - 无法显示图像路径存储在数据库中的图像

Question

我正在尝试在网页上显示图像，其中图像路径存储在数据库中，图像存储在服务器中。但是我无法使用以下代码显示这些图像，所以请有人帮助我解决这个问题，..

<form method="post"  enctype="multipart/form-data" action="file_upload.php">
<table>

<?php

$dbhost = 'xxxxxxxx';
$dbuser = 'xxxxxxxxx';
$dbpass = 'xxxxxxxxxx';
$db_name = 'xxxxxxxxxx';
$tbl_name = 'xxxxxxxxxxx';

$conn = mysql_connect($dbhost, $dbuser, $dbpass);

if(! $conn )
{
  die('Could not connect: ' . mysql_error());
}
mysql_select_db("$db_name")or die("cannot select DB");

$query1 = mysql_query("select * from '$tbl_name' where id='1'");
$rows1 = mysql_fetch_array($query1);
$path1 = $rows1['image'];

$query2 = mysql_query("select * from '$tbl_name' where id='2'");
$rows2 = mysql_fetch_array($query2);
$path2 = $rows2['image'];

$query3 = mysql_query("select * from '$tbl_name' where id='3'");
$rows3 = mysql_fetch_array($query3);
$path3 = $rows3['image'];

echo '<tr><td><img src="$path1"></td>' ;
echo '<td><img src="$path2"></td>' ;
echo '<td><img src="$path3"></td></tr>' ;

?>

</form>
</table>

Answer 1

看一下这段代码:

$b = "aaaaaaa";

echo '"$b"';  // will displayt "$b"
echo "'$b'";  // will displayt 'aaaaaaa'
echo "\"$b\"";// will displayt "aaaaaaa"

enter code here

因此，对于您的问题来说，这只是因为如果您希望在字符串中计算变量，则必须将其放在“”而不是“”之间。

<table>
<?php 
	$path1 = "/upload/image1.png";
	$path2 = "/upload/image2.png";
	$path3 = "/upload/image3.png";
	
	echo "<tr><td><img src='$path1'></td>" ;
	echo "<td><img src='$path2'></td>" ;
	echo "<td><img src='$path3'></td></tr>" ;
?>
</table>

或者

<table>
<?php 
	$path1 = "/upload/image1.png";
	$path2 = "/upload/image2.png";
	$path3 = "/upload/image3.png";
	
	echo '<tr><td><img src="'.$path1.'"></td>' ;
	echo '<td><img src="'.$path2.'"></td>' ;
	echo '<td><img src="'.$path3.'"></td></tr>' ;
?>
</table>

阿纳斯