How do i make images appear when making a search from a MYSQL database?

[usman07]

I currently have a MYSQL database which works and is connected, at the moment is brings up information from the database but beside the information from the database I also want images to appear. How do i do this?

 

Any help is appreciated.

[AyKay47]

Where are the images coming from?

[usman07]

I dont really know how to set it up? I dont know where to put the images etc, are they suppose to be somehow in the MYSQL database or could they be in a directory e.g. cutouts/Homepage.

[AyKay47]

Typically, the images are stored in a directory and the file paths are stored in the database.

[usman07]

ahh right ok, how do i go about setting file paths in the database?

[AyKay47]

create a varchar type field named whatever and insert the file paths.

So if you have an image at /some/file/path/image.jpg, that is what will go into that field, along with any other corresponding image data that you want.

[Jessica]

Are you trying to have a specific image come up based on the search results, or the same image next to each result? your question is kind of vague.

[usman07]

Yeah i'm trying to have a specific image come up based on the search results

[Jessica]

Okay then continue with the above method.

[usman07]

Thanks

[usman07]

Ok I have done that, what do i do now? this is a image of mysql table of 1 of my entries, is this right?

[AyKay47]

looks right.

Now you need to grab that data from the db using a select statement.

Have you attempted a query yet?

[usman07]

No i haven't yet, dunno how to really. php is kinda new to me.

[AyKay47]

Here is some Pseudo code to get you going:

$sql = 'SELECT * from table';
$query = mysql_query($sql) or die(mysql_error() . '<br />' . $sql);
while($row = mysql_fetch_assoc($query))
{
    ?>
        <img src="<?php echo $row['images']; ?>" />
    <?php
}

 

this will simply display each image in the table, you will need to tweak it to suit your needs.

[usman07]

thank you very much for your help, i really do appreciate it. where abouts would i be placing that code in my php?

 

heres my php code:

<?php

$server = "mysql10.000webhost.com";      // Enter your MYSQL server name/address between quotes
$username = "";    // Your MYSQL username between quotes
$password = "";    // Your MYSQL password between quotes
$database = "";    // Your MYSQL database between quotes

$con = mysql_connect($server, $username, $password);       // Connect to the database
if(!$con) { die('Could not connect: ' . mysql_error()); }  // If connection failed, stop and display error
mysql_select_db($database, $con);  // Select database to use
// Query database
$result = mysql_query("SELECT * FROM Properties");

if (!$result)
{
    echo "Error running query:<br>";
    trigger_error(mysql_error());
}
elseif(!mysql_num_rows($result))
{
    // no records found by query.
    echo "No records found";
}
else
{
    $i = 0;
    echo '<div style="font-family:helvetica; font-size:15px; padding-left:15px; padding-top:20px;">';
    while($row = mysql_fetch_array($result)) {     // Loop through results
        $i++;
        echo "Displaying record $i<br>\n";
        echo "<b>" . $row['id'] . "</b><br>\n";      // Where 'id' is the column/field title in the database
        echo $row['Location'] . "<br>\n";            // Where 'location' is the column/field title in the database
        echo $row['Property_type'] . "<br>\n";       // as above
        echo $row['Number_of_bedrooms'] . "<br>\n";  // ..
        echo $row['Purchase_type'] . "<br>\n";       // ..
        echo $row['Price_range'] . "<br>\n";         // ..
    }
echo '</div>';
}

mysql_close($con);  // Close the connection to the database after results, not before.
?>

[AyKay47]

again, this code will need to be tweaked so it looks right:

 

while($row = mysql_fetch_array($result)) {     // Loop through results
        $i++;
        echo "Displaying record $i<br>\n";
        echo "<b>" . $row['id'] . "</b><br>\n";      // Where 'id' is the column/field title in the database
        echo $row['Location'] . "<br>\n";            // Where 'location' is the column/field title in the database
        echo $row['Property_type'] . "<br>\n";       // as above
        echo $row['Number_of_bedrooms'] . "<br>\n";  // ..
        echo $row['Purchase_type'] . "<br>\n";       // ..
        echo $row['Price_range'] . "<br>\n";         // ..
        echo '<img src="'. $row['images'] .'" />';   //image
    }

[usman07]

thats brilliant, it appears, how would i get it to appear beside the text rather than below?

[AyKay47]

remove the <br>\n from the previous line.

You can add CSS to the image also.

[usman07]

How would i apply css to the image, would I give it an ID then apply the css in my style.css or do i apply it within the php code?

[AyKay47]

the first statement, add either a class or an id (depending upon how many images you want the css to affect) and style it in an external css page.

[deragoku]

Simply each image is displayed in the table, you will need to be adjusted to suit your needs.