<b>how to display image by echo() in existing html code?</b>

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mianmajidali
Forum Commoner
Posts: 30
Joined: Tue Dec 01, 2009 8:05 pm

<b>how to display image by echo() in existing html code?</b>

Post by mianmajidali »

hi to all,
i am using php code in existing html file with extenssion ".php" , it is not showing the images and css working. here i m putting some code for example,
Please guide me, how to display images and use css in php ??
---------------------
file name is data.php

<html>
<head>
<link rel="stylesheet" type="text/css" href="global.css">
note: CSS not woring
</head>
<body bgcolor="#DF6804">
<table align="center" border="0" cellpadding="0" cellspacing="0" width="1000">
<tr align="center">
<td colspan="7">
<img src="Images/header-new.jpg" alt="os-co header" />
note: image not displaying
</td>
</tr>
<tr class="top-nev-row" align="center">

<td width="14%" height="30"><a href="index.html">Home</a></td>

<td width="14%" height="30"><a href="mission.html">Mission Statement</a></td>

<td width="14%" height="30"><a href="processing.htm">Processing</a></td>

<td width="14%" height="30"><a href="products.htm">Products</a></td>

<td width="14%" height="30"><a href="">Contact us</a></td>

<td width="14%" height="30"><a href="">Query</a></td>

<td width="14%" height="30"><a href="">Login</a></td>
</tr>
<tr>

<td colspan="7" class="cont-content" valign="top" height="400">
<table align="center" width="100%" border="1" cellspacing="3" >
<tr bgcolor="#cccccc">
<td align="center"><strong>Date</strong></td>
<td align="center"><strong>Name</strong></td>
<td align="center"><strong>Email</strong></td>
<td align="center"><strong>Company</strong></td>
<td align="center"><strong>Country</strong></td>
<td align="center"><strong>Telephone</strong></td>
<td align="center"><strong>Fax</strong></td>
<td align="center"><strong>Mobile</strong></td>
<td align="center"><strong>Contact via</strong></td>
<td align="center"><strong>Query Detail</strong></td>
</tr>
<tr>

<?php

$con = mysql_connect("localhost","username","password");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("my_db", $con);

$result = mysql_query("SELECT * FROM query");
$name= mysql_query("SELECT name FROM members");
$pass= mysql_query("SELECT pass FROM members");
if("'$_POST[name]'==$name" && "'$_POST[pass]'==$pass")
{
while($row = mysql_fetch_array($result))
{

echo "<tr>";

echo "<td>" . $row['date'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['email'] . "</td>";
echo "<td>" . $row['company'] . "</td>";
echo "<td>" . $row['country'] . "</td>";
echo "<td>" . $row['tel'] . "</td>";
echo "<td>" . $row['fax'] . "</td>";
echo "<td>" . $row['mob'] . "</td>";
echo "<td>" . $row['box'] . "</td>";

echo "<td>" . $row['query']."</td>";

echo "</tr>";
}

}/*if close*/
else
echo "Wrong Username or Password";


mysql_close($con);
?>
</tr>
</table>

</td>
</tr>
<tr align="center">
<td colspan="7" class="footer" align="center" height="30">
<p>Copyright &copy; 2009-2012 All Rights Reserved.</p>
</td>
</tr>
</table>
</body>
</html>
-------------------------------------
Thanks
User avatar
John Cartwright
Site Admin
Posts: 11470
Joined: Tue Dec 23, 2003 2:10 am
Location: Toronto
Contact:

Re: <b>how to display image by echo() in existing html code?</b>

Post by John Cartwright »

The images and style sheets have nothing to do with PHP (assuming PHP renders them ok, but thats not the case here). Check to make sure the paths are setup correctly and the files exist on the said path.
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