并非所有行都输出PHP Mysql

Question

in this code I am trying to output all images with the tag I searched. I created 2 images with the same tag 'hi'.

The program is only outputting one of the images though. But when I echo the $rows number it equals 2. When I ran the query in MYSQL it also returned 2 rows. I do not see why ALL the images are being shown.

 $input = ($_GET['input']);
$query = "SELECT * FROM `photo`.`photo` WHERE `tags` LIKE '%$input%'";
$result = mysql_query($query);
$data = mysql_fetch_array($result) or die (mysql_error());
$rows = mysql_num_rows($result);
$image = $data['image'];
     header('Content-type: image/jpeg');
     echo $image;

This isn't all my code but it is all that I think is necessary. Here is what is being shown in another page.

echo "<img src='photolarge.php?input=$input'>"

When I echo $rows it outputs the number 2 so I know the query is working.

Answer 1

You need to use a loop to iterate through the result array.

$query = "SELECT * FROM `photo`.`photo` WHERE `tags` LIKE '%$input%'";
$result = mysql_query($query);
while($rows = mysql_fetch_array($result)) {
    echo $rows[image];
}

As I mentioned in the comments, please don't use mysql_ functions. You also need to sanitize your parameter as your code is vulnerable to SQL injection.

Answer 2

您需要获取第4行中所有返回的行，而不仅仅是第一行。

Answer 3

try this

   $query = "SELECT * FROM `photo`.`photo` WHERE `tags` LIKE '%$input%'";
   $result = mysql_query($query);
   while($row = mysql_fetch_array($result)) {
   echo $row['image'];
   }

tips: try to change to MYSQLI or PDO ,MYSQL is no longer maintained