Help integrating php/mysql with existing html page.

April 9, 2012

<p>Hello, I know little about web design so I hope this makes sense.

I am trying to integrate a very basic php site into an existing html website to create a blog. I have one page where I can enter info which is then published onto a page called blogentries.php which works fine on it's own. I want the info to be published into the blogentries div on my website that says "I WANT BLOG ENTRIES HERE" close to the bottom.

I thought it would be a simple case of making the connection to the database in the header and put the rest of the php into the div, but obviously there is more to it than that.

I was wondering if someone could look at the code for the two pages and explain how it should be done?

Thank you

blogentries.php


<?php
mysql_connect("localhost:8889","root","root");
mysql_select_db("Alexweb");
echo "<html><head><link rel='stylesheet' href='style.css' type='text/css'></head><body>";
$query = "select * from content";
echo "<p>Query: ".$query;
$result = mysql_query($query);

$rows = mysql_num_rows($result);
for($i=0;$i<$rows;$i++) {
$row = mysql_fetch_array($result);
echo "<div class='".$row['class']."'>";
echo "<p> Date: ".$row['date']."</p>";
echo "<p><h3>".$row['title']."<?h3></p>";
echo "<p><img src='".$row['image']."'></p></div><p></p>";
echo "<p>".$row['text']."</p>";

}

echo "</body></html>";
?>
[/CODE]

[b]blogpage in html[/b]

[CODE]
<!doctype html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>AJWatson - Blog</title>
<meta name="keywords" content=" art design, digital media design, graphic design, logo design, ">
<link rel="shortcut icon" href="favicon.ico" />

<link href="css/style.css" rel="stylesheet" type="text/css" />

<link rel="stylesheet" href="http://fonts.googleapis.com/css?family=Droid+Serif:regular,italic,bold,bolditalic" type="text/css" />
<link rel="stylesheet" href="http://fonts.googleapis.com/css?family=Droid+Sans:regular,bold" type="text/css" />
</head>

<body>


<div id="topcontrol" style="position: fixed; bottom: 5px; left: 960px; opacity: 1; cursor: pointer;" title="Go to Top"></div>
<div id="header-wrapper">
<div id="header">
<div id="logo"><a href="index.html"><img src="images/logo.png" width="100" height="80" alt="logo" /></a></div>
<div id="header-text">
<h4>Blog</h4>
<h6><a href="index.html">Home</a> ? Blog</h6>
</div>
</div>
</div>



<div id="menu-wrapper">
<div id="main-menu">
<ul>
<li><a href="index.html">Home</a></li>
<li><a href="about.html">About</a></li>
<li><a href="portfolio.html">Portfolio</a></li>
<li><a class="selected" href="blog.html">Blog</a></li>
<li><a href="contact.html">Contact</a></li>
</ul>
</div>


<div id="footer">
<div id="social-box">
<ul>
<li>
<div class="facebook"><a href=""></a></div>
</li>
<li>
<div class="twitter"><a href=""></a></div>
</li>
<li>
<div class="linkedin"><a href=""></a></div>
</li>
</ul>
</div>
<h6></h6>
</div>


</div>



<div id="content">



<div id="blogentries">

I WANT BLOG ENTRIES HERE

</div>


</div>


</div>
</body>
</html>
[/CODE]

April 18, 2012

What about settting the image src to "./blogimages/"?

Thanks, I tried that but still have the same issue.

April 18, 2012

Thanks, I tried that but still have the same issue.

Ok, sorry that it didn't work. I just figured it wasn't finding it because the path wasn't starting from the parent directory. I apologize that I couldn't be more helpful.

April 18, 2012

Ok, sorry that it didn't work. I just figured it wasn't finding it because the path wasn't starting from the parent directory. I apologize that I couldn't be more helpful.

No worries. It's all worth trying for me :)

April 18, 2012

i'm not entirely sure what you mean. i don't want to keep bugging people with basic questions though so i'll do a bit of googling. thanks again for your help :)

I think this is what he means:


if(isset($_FILES['photo']))
{
	$photo = $_FILES['photo'];
}
[/CODE]

No worries. It's all worth trying for me :)

True. It is a learning experience. :)

April 18, 2012

I think this is what he means:
if(isset($_FILES['photo']))
{
	$photo = $_FILES['photo'];
}
[/CODE]
Just to add to this, if you're checking if the photo has a name, or if it's FALSE, default to false. So:

[CODE]
$photo = FALSE;
if(isset($_FILES['photo']))
{
$photo = $_FILES['photo']['name'];
}
[/CODE]

For further explanation: PHP is a bit wonky on what it returns/type checking. It's best to look up in the php manual what a function returns, although in this case since you're dealing with an associative array, the manual doesn't actually say. In that case, you have to check what it's returning, assume it could return a whole bunch of strange things, or check for FALSE on something further up.

Two other things, you're going to want to do something to all of those text entries you have coming in, throwing $_POST['text'] into a MySQL database is a disaster waiting to happen. Also, it's easier (IMO), to use NOW() for the MySQL time entry, then convert it using date() when you're pulling it out.

Feel free to ask questions here, I'm learning myself, I find it helpful to answer things like this. :)

April 18, 2012

thanks for the replies guys and the extra php advice.

so should the check for whether or not $photo is False be in the php that gathers the info from, or the php that outputs the image and text the form? I thought it would be in the php that gathers the info from the form but I've tried putting it in there with no luck. This is what I tried...


//This gets all the other information from the form
	    $title=$_POST['title'];
	    $date= date("d F Y H:i:s");
$photo = FALSE;
   if(isset($_FILES['photo']))
   {
    $photo = $_FILES['photo']['name'];
   }  
	    $text=$_POST['text'];

[/CODE]

April 18, 2012

You should use is_uploaded_file() rather than isset when checking if the file is there.

Also I'd set $photo to an empty string, else the DB may store "FALSE" literally.

$photo = '';
if(is_uploaded_file($_FILES['photo']['tmp_name']))
{
	$photo = $_FILES['photo']['name'];
}

And you really need to escape your input texts with mysql_real_escape_string()

//This gets all the other information from the form
$title = mysql_real_escape_string($_POST['title']);
$date = date("d F Y H:i:s");
$photo = '';
if(is_uploaded_file($_FILES['photo']['tmp_name']))
{
	$photo = $_FILES['photo']['name'];
}
$text = mysql_real_escape_string($_POST['text']);

And the <b> problem from my last post is in some other lines as well, you should check.

April 18, 2012

You check in addblog.php... I just realized this is probably an issue with how MySQL is storing "FALSE". So when printing the results, use this instead:


if(!empty($info['photo']))
{
			    echo "<p><img src=/blogimages/" . $info['photo'] . " /></p><br />";
}
else
{
			    echo "<p><img src='/blogimages/noimage.png' /></p><br />";
}
[/CODE]

isset checks for boolean values, empty will check for NULL, FALSE, and empty strings.

April 18, 2012

Thanks so much guys! I managed to get it working with is_uploaded_file() and then the !empty command.

I will look over my code and tidy it up, I know it's a bit of a mess. I will also check out the mysql_real_escape_string() in more depth too, I had a quick look but didn't quite get it.

Thanks again, really appreciate it.

April 18, 2012

Long story short, just put mysql_real_escape_string() around text you're going to insert into a database. Like:


$post = mysql_real_escape_string($_POST['post_text']);
[/CODE]

Again though, before this hits the web you should have an understanding of mysqli, and preferably prepared statements. Although it might seem like a lot to grasp right now, it will save you a lot of trouble in the long run.

April 18, 2012

Long story short, just put mysql_real_escape_string() around text you're going to insert into a database. Like:
$post = mysql_real_escape_string($_POST['post_text']);
[/CODE]
Again though, before this hits the web you should have an understanding of mysqli, and preferably prepared statements. Although it might seem like a lot to grasp right now, it will save you a lot of trouble in the long run.

Thanks for the tip, I will make sure I check it out. My brain is about to explode with php for now so I'll take a look tomorrow with fresh eyes. :)

April 19, 2012

Thanks so much guys! I managed to get it working with is_uploaded_file() and then the !empty command.

I will look over my code and tidy it up, I know it's a bit of a mess. I will also check out the mysql_real_escape_string() in more depth too, I had a quick look but didn't quite get it.

Thanks again, really appreciate it.

No problem! I am glad that I was able to help. Also, helping others learn different programming languages and techniques is a nice challenge, keeps the concepts fresh in my mind, and I enjoy going to look up things that I don't know or remember (in an attempt to help out) so that I can learn or remember them as well. It is a win-win situation for everyone involved! :)

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