php: use variable name to display specific image on diffrent page - php

Question

Php beginner here. I have php loop on page to display images (basically a gallery) and each image has a title displayed below. Those titles are links to second page where only one image is shown.
The problem is that every link has a different name (value of name is specific /ID of image).
<form action="image_link.php" method="post">
<?php>
$sql_select = "SELECT * FROM images ORDER BY data DESC";
$sql_table_selected = mysql_query($sql_select);
while ($table_row = mysql_fetch_array($sql_table_selected)){
echo "<div class=\"galery-1-3\"><figure>";
echo "<img src=\"" . $table_row['file_path'] . "\" />";
echo "<figcaption>" . $table_row['tytul'] . "</figcaption></figure></div>";
}
?>
I'm getting a single image displayed but no matter what title I click, I always get last image from the table. How can I catch a specific ID on image_link.php, or should I use some flexible kind of address like /image_link.php?id=34. I don't know where to start.
image_link.php contains basically the same code without the loop:
$sql_select = "SELECT * FROM images WHERE ID = $SomehowGetID";
$sql_table_selected = mysql_query($sql_select);
$table_row = mysql_fetch_array($sql_table_selected);
echo "<div class=\"galery-big\"><figure>";
echo "<img src=\"" . $table_row['file_path'] . "\" />";
echo "<figcaption>" . $table_row['tytul'] . "</figcaption></figure></div>";
Thanks in advance for help.

Answer 1

No need for a form
first script :
<?php
$sql_select = "SELECT * FROM images ORDER BY data DESC";
$sql_table_selected = mysql_query($sql_select);
while ($table_row = mysql_fetch_array($sql_table_selected)){
echo "<div class=\"galery-1-3\"><figure>";
echo "<img src=\"" . $table_row['file_path'] . "\" />";
echo "<figcaption><a href=\"image_link.php?id=" . $table_row['ID']
. "\" name=\"" . $table_row['ID']
. "\">" . $table_row['tytul'] . "</a></figcaption></figure></div>";
}
?>
second script
$SomehowGetID=$_GET['id'];
$sql_select = "SELECT * FROM images WHERE ID = $SomehowGetID";
$sql_table_selected = mysql_query($sql_select);
$table_row = mysql_fetch_array($sql_table_selected);
echo "<div class=\"galery-big\"><figure>";
echo "<img src=\"" . $table_row['file_path'] . "\" />";
echo "<figcaption>" . $table_row['tytul']
. "</figcaption></figure> </div>";
I'd advice you though to use prepared statement instead of simple query constructions.

Answer 2

Change query to:
$sql_select = "SELECT * FROM images WHERE ID =" $_GET['id'];

Answer 3

echo
"<figcaption>
<a href='image_link.php?id=$table_row[ID]' name='$table_row[ID]'>"
.$table_row['tytul'] ."
</a>
</figcaption>
</figure></div>";
You need to pass the id as a parameter in your href with $table_row[ID] as value.
In image_link.php
Use:
$sql_select = "SELECT * FROM images WHERE ID = '$_GET[ID]'";

Answer 4

You had the right idea by saying that you want to pass the id of the image.
your code should looks like this :
<form action="image_link.php" method="post">
<?php>
$sql_select = "SELECT * FROM images ORDER BY data DESC";
$sql_table_selected = mysql_query($sql_select);
while ($table_row = mysql_fetch_array($sql_table_selected)){
echo "<div class=\"galery-1-3\"><figure>";
echo "<img src=\"" . $table_row['file_path'] . "\" />";
echo "<figcaption>" . $table_row['tytul'] . "</figcaption></figure></div>";
}
?>
image_link.php:
$sql_select = "SELECT * FROM images WHERE ID =" . mysql_real_escape_string($_GET['id']);
$sql_table_selected = mysql_query($sql_select);
$table_row = mysql_fetch_array($sql_table_selected);
echo "<div class=\"galery-big\"><figure>";
echo "<img src=\"" . $table_row['file_path'] . "\" />";
echo "<figcaption>" . $table_row['tytul'] . "</figcaption></figure></div>";
Please note that the use of mysql_* function is highly deprecated, you should use mysqli extention instead