Have Geek Brain

How do I use Php inside Php?

Can someone tell / show me how to use PHP inside PHP. I am trying to change the url of an image depending on what the MySQL database is. Here is an example of what I am trying to do. Be aware that $ idx already has a value from the page url.

<?php
$query  = "SELECT * FROM comment WHERE uname='$idx'";
$result = mysql_query($query);
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo "<img src='' name='comm' width='75px' height='60px' id='mainimage' />";
}
?>

How do I initialize an image from another table?

+2


a source to share


4 answers


In the loop, while

you would execute another SQL query. I love how you put it, "Php inside Php", that's pretty much what you do.

while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
  $image_query = "SELECT image_url FROM your_table";
  $image_result = mysql_query($image_query);
  $image = mysql_fetch_assoc($image_result);

  echo "<img src='" . $image['image_url'] . "' name='comm' width='75px' height='60px' id='mainimage' />";
}


Make sure the variable names for your query and result are different from the original query, because you are still using the variable $result

from the original query for each iteration of the loop. So here I am prefixing them with "image_".

+5


a source


You can combine data from multiple tables in one SQL query. See: http://www.w3schools.com/sql/sql_join.asp

Example:



SELECT column_name(s)
FROM table_name1
JOIN table_name2
ON table_name1.column_name=table_name2.column_name
+7


a source


Assuming under the random assumption that the other table might be called accounts

and uname

is its primary key, you can do JOIN

to grab data from both tables in one go. If you can match what you want to do in a single query, that is usually much faster than doing an additional query in a row

<?php
    $result = mysql_query(
        "SELECT comment.*, account.url AS imgurl ".
        "FROM comment JOIN account ON comment.uname=account.uname ".
        "WHERE comment.uname='".mysql_real_escape_string($idx)."'"
    );
?>
<?php while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { ?>
    <img src="<?php echo htmlspecialchars($row['imgurl']) ?>" alt="" class="mainimage" />
<?php } ?>

Notes:

  • you need to avoid the text that you insert into the SQL query, or you have serious security issues. Use mysql_real_escape_string

    , or let parameterized queries be processed;

  • Likewise, when text fits into an HTML string, you need htmlspecialchars

    or have other (less serious, but still good) security concerns;

  • you shouldn't use the same id

    for an element more than once (which will be in a loop). The same can be said for name

    on <img>

    , although you have few reasons why you would really want to use img name

    ;

  • width="75px"

    will not work on the attribute, since px

    -general is CSS, not HTML. Omit the block, or better, set a rule in CSS ( .mainimage { width: 75px; height: 60px; }

    ) to do it all at once.

+1


a source


How do I initialize an image from another table?

You just need to ask for another table. You can ask by creating a subquery (or joining) or by creating a second query inside the loop.

0


a source






More articles:


All Articles