Help with Displaying Data from database

**Feckie** · 04-09-2009, 04:29 PM

Hi I am a noob so be gentle,

I have got the database written, also the upload form, wich works....

What I want todo is to pull the information into a table (as Devined below)
to display all the entries in an array of tables...

This is where I am completly stuck, anyone can help I would be greatful..

Upload form:

Code:

<?php

# DATABASE CONFIGURE

$hostname = "localhost";
$db_user = "root";
$db_password = "root";
$database = "film";
$db_table = "film_info";


# STOP HERE

# THIS CODE IS USED TO CONNECT TO THE MYSQL DATABASE
$db = mysql_connect($hostname, $db_user, $db_password);
mysql_select_db($database,$db);
?>
<html>
<head>
<title>Upload Form</title>
</head>
<body>

<?php
if (isset($_REQUEST['Submit'])) {

$sql = "INSERT INTO $db_table(title,info,download,view,year,plus) values ('".mysql_real_escape_string(stripslashes($_REQUEST['title']))."','".mysql_real_escape_string(stripslashes($_REQUEST['info']))."','".mysql_real_escape_string(stripslashes($_REQUEST['download']))."','".mysql_real_escape_string(stripslashes($_REQUEST['view']))."','".mysql_real_escape_string(stripslashes($_REQUEST['year']))."','".mysql_real_escape_string(stripslashes($_REQUEST['plus']))."')";
if($result = mysql_query($sql ,$db)) {
echo '<h1>Thank you</h1>Your information has been entered into our database<br><br><h3><a href="upload.php">New Entry</a></h3>';
} else {
echo "ERROR: ".mysql_error();
}
} else {
?>
<h1>Films Update Form</h1><hr>
<form method="post" action="">
<b>Title:</b>   <br>
<input type="text" name="title" size="100">
<br>
<b>Info:</b>    <br>
<input type="text" name="info" size="100">
<br>
<b>Download:</b> <br>
<input type="text" name="download" size="100" value=".avi">
<br>
<b>View:</b>    <br>
<input type="text" name="view" size="100" value=".php">
<br>
<b>Year:</b>   <br>
<input type="text" name="year" size="100" value="2009">
<br>
<b>Plus:</b>  <br>
<input type="text" name="plus" size="100">
<br>
<br><br>
<input type="submit" name="Submit" value="Submit">
</form>
<?php
}
?>
</body>
</html>

Table Format:

Code:

<table border="1" width="301">
    <tr>
        <td width="291">
            <table width="292">
                <tr>
                <td width="282" height="40" align="center"> . $row['title'] . </td>
                </tr>
                <tr>
                    <td width="282" height="66"> . $row['info'] . </td>
                </tr>
            </table>
            <table width="292">
                <tr>
                    <td width="153" align="center" height="44"> . $row['year'] . </td>
                    <td width="64" height="44">
                        <p align="center"><a href="" . $row['view'] . "" target="detail" title="View Film"><img src="FILMSTRIp.GIF" border="0"></a></p>
                    </td>
                    <td width="61" height="44">
                        <p align="center"><a href="" . $row['download'] . "" title="Download Film"><img src="download_48.png" border="0"></a></p>
                    </td>
                </tr>
</table>

        </td>
    </tr>
</table>

**AdrielGreene** · 04-09-2009, 11:12 PM

Your form will need an action variable. You can have it send the submitted info back to the file your working on.

This is to update fields in the DB.

Code:

$a = "UPDATE db_name SET field='value' WHERE condition";	
	$b = mysqli_query ($dbinfo, $a) or trigger_error("Query: $a\n<br />MySQL Error: " . mysqli_error($dbinfo));

This is to call fields from the DB.

Code:

$a = "SELECT field FROM db_name WHERE condition";
$whatyouwantdisplayed = mysqli_query ($dbinfo, a) or trigger_error("Query: $a\n<br />MySQL Error: " . mysqli_error($dbinfo));

I too am a n00b so I hope this will help you. Goodluck.

**Feckie** · 04-10-2009, 07:18 AM

Wow that lost me completly...............

Originally Posted by AdrielGreene

Your form will need an action variable. You can have it send the submitted info back to the file your working on.

This is to update fields in the DB.

Code:

$a = "UPDATE db_name SET field='value' WHERE condition";	
	$b = mysqli_query ($dbinfo, $a) or trigger_error("Query: $a\n<br />MySQL Error: " . mysqli_error($dbinfo));

This is to call fields from the DB.

Code:

$a = "SELECT field FROM db_name WHERE condition";
$whatyouwantdisplayed = mysqli_query ($dbinfo, a) or trigger_error("Query: $a\n<br />MySQL Error: " . mysqli_error($dbinfo));

I too am a n00b so I hope this will help you. Goodluck.

**forum_amnesiac** · 04-14-2009, 12:19 PM

What I do is the basic PHP query:-

$result = mysql_query('SELECT * FROM `mytable` WHERE condition');

I then put the field values into variables:-

$name=mysql_result($result,$id,"Name");
$company=mysql_result($result,$id,"Enterprise");

and then in the HTML secion I use this to display the values from the table for a form input:-

<td>input type="text" name="FName" value="<? echo "$name"?>" /></td>

Hope this is what you were looking for.

Thread: Help with Displaying Data from database

Thread Tools

Search Thread

Help with Displaying Data from database

Bookmarks

Bookmarks

Posting Permissions