I am trying to display image on web page where image path is stored in MySQL table (table column name-location). but I can't able to display the image.
I have a dynamic drop down list box where it populated all the image name. I want to display the image once the user click on the image name from this drop down list. If I place the full path to the img src, then I can able to see my image from the HTML table. Below is my code so far i tried to get output.
Your advice will help me to complete my task. Need your help to update my knowledge.
$(function () {
$("#Code").change(function () {
$("#image").load("image.php?choice=" + $("#Code").val());
});
});
index.php
<?php
mysql_connect('890.23.89.100', 'root', '');
mysql_select_db('abc');
$sql = "SELECT Code,location FROM Product_List ORDER BY Code ASC";
$result = mysql_query($sql);
echo "<select id='Code' name='Code' style='width: 120px'>";
while ($row = mysql_fetch_array($result))
{
echo "<option value='" . $row['Code'] . "'>" . $row['Code'] . "</option>";
}
echo "</select>";
?>
<img
src="../label_image/6015.jpg (??? how to get the image path here. i put thos manually and can display the image on webpage)"
width="100%" height="100%">
image.php
<?php
$username = "root";
$password = "";
$hostname = "890.23.89.100";
$dbhandle = mysql_connect($hostname, $username, $password) or die("Unable to connect to MySQL");
$selected = mysql_select_db("abc", $dbhandle) or die("Could not select examples");
$choice = mysql_real_escape_string($_GET['choice']);
$query = "SELECT location FROM Product_List WHERE Code='$choice'";
$result = mysql_query($query);
while ($row = mysql_fetch_array($result))
{
echo "<option>" . $row{'location'} . "</option>";
}
?>