I have two issues:-
Issue No :1 (PHP related)
I can't display an array of images on a PHP page.Here is the code i am trying..
getUser.php
<?php
$q=$_GET["q"];
$con = mysql_connect('localhost', 'sulen', '123');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("holidayNet", $con);
$sql="SELECT * FROM image WHERE id = '".$q."'";
$result = mysql_query($sql);
echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Age</th>
<th>Hometown</th>
<th>Picture</th>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "<td>" . $row['Age'] . "</td>";
echo "<td>" . $row['Hometown'] . "</td>";
$dir = 'images';
$file_display = array('jpg','jpeg','png','gif');
if (file_exists ($dir) == false) {
echo 'Directory \'', $dir, '\' not found!';
}
else{
$dir_contents = scandir($dir);
foreach($dir_contents as $file) {
$file_type = strtolower(end(explode('.', $file)));
If($file !== '.' && $file !== '..' && in_array($file_type, $file_display) ==
true) {
echo "<td>" . $row ['<img src="', $dir, '/', $file, '" alt="', $file, '" />'] . "</td>";
}
}
}
/*echo "<td>" . $row['Job'] . "</td>";*/
echo "</tr>";
}
echo "</table>";
mysql_close($con);
?>
<br />
one.html
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html>
<head>
<script type="text/javascript">
function showUser(str)
{
if (str=="")
{
document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","getuser.php?q="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>
<form>
<select name="users" onchange="showUser(this.value)">
<option value="">Select a person:</option>
<option value="1">Sulman qb</option>
<option value="2">asd asd</option>
<option value="3">Glenn Quagmire</option>
<option value="4">Joseph Swanson</option>
</select>
</form>
<br />
<div id="txtHint"><b>Person info will be listed here.</b></div>
</body>
</html>
Issue No 2 (AJAX Related)
You might have seen in the above code that while using AJAX i am retrieving image directory path from database. Now, how i am going to give that directory path to php file getUser.php or
$dir = 'images'(need that path here !);
this line is wrong:
echo "<td>" . $row ['<img src="', $dir, '/', $file, '" alt="', $file, '" />'] . "</td>";
why not just, also why are you using $row[] to display the image, $row is meant for retrieving the values in your database corresponding to the table field. $row['table field'];
echo '<td>';
echo '<img src =';
echo $dir."/".$file;
echo '/>';
echo '</td>';
also try
$dir = "images/";