php for语句来获取mysql数据

I am storing the names and types of images in my database. Now I want to select the newest images from that database and display them to the user. I tried to use the following code but I am getting the same image name and type, but I want to get different images.

$new_image_count = mysql_query("SELECT COUNT(id) FROM newest");
$result = mysql_result($new_image_count, 0);

    $select_images = mysql_query("SELECT * FROM newest");

        $fetch = mysql_fetch_assoc($select_images);
        for($result; $result > 0; $result--){

        $name = $fetch['image_name'];
        $type = $fetch['image_type'];
        $name_dot_type = $name.".".$type;
        echo '<img src="main_images/'.$name_dot_type.'" width="300">';
        }

Any ideas about how to fix this problem ?

Remove the query to count rows (you don't need it)

and change the main part of the code to something like

$select_images = mysql_query("SELECT * FROM newest");
while ($fetch = mysql_fetch_assoc($select_images)) {
    // echo data from $fetch
}

PS: you'll recently get a comments about using PDO instead of mysql_

PPS: you get the same results because you fetch the row just once, and after that you output the same values in the loop

First off, you shouldn't be using mysql_*() functions anymore, and should switch to mysqli or PDO

Secondly, just move the $fetch = mysql_fetch_assoc($select_images); inside the for loop:

for($result; $result > 0; $result--){
    $fetch = mysql_fetch_assoc($select_images);
    $name = $fetch['image_name'];
    $type = $fetch['image_type'];
    $name_dot_type = $name.".".$type;
    echo '<img src="main_images/'.$name_dot_type.'" width="300">';
}