Not all rows outputting 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.

^^^^as true as it is, its getting a little tedious for regular S.O users to see it 10+ times a day — user557846
– user557846, Commented Feb 7, 2013 at 23:24
regardless of house many times its seen, it still needs to be said :/ — Ascherer
– Ascherer, Commented Feb 7, 2013 at 23:27
Oh come on "don't use mysql_" doesn't need posting on every question that mentions it. Let's be realistic, until every copy of the php mysql extension is deleted from all servers in all of the world, it will still work. Obsessive is an understatement. — Popnoodles
– Popnoodles, Commented Feb 7, 2013 at 23:34

Kermit · Accepted Answer · 2013-02-07 23:24:57Z

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.

answered Feb 7, 2013 at 23:24

Kermit

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3 Comments

Nick D Over a year ago

Ok well I know what you are doing but I am also re-sizing the image in the same file. so I tried codewhile($data = mysql_fetch_array($result)) { $image = $data['image']; code And that still only would output 1 image

Nick D Over a year ago

I need a variable so I can re-size it.

Nick D Over a year ago

Ok I tried what you did and it still did not work ob_start(); $input = $_GET['input']; $query = "SELECT * FROM photo.photo` WHERE tags LIKE '%$input%'"; $result = mysql_query($query); while($rows = mysql_fetch_array($result)) { header('Content-type: image/jpeg'); echo $rows['image'];`

DYN · Accepted Answer · 2013-02-07 23:25:10Z

0

You need to fetch all returned rows in line 4, not only the first one.

answered Feb 7, 2013 at 23:25

DYN

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Comments

echo_Me · Accepted Answer · 2013-02-07 23:27:02Z

0

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

answered Feb 7, 2013 at 23:27

echo_Me

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