PHP - Blank Entries In The Database

Hi,
I want to be able to generate visitor statistics for a blog I'm creating. I'm going to be collecting numerous pieces of data when a post is viewed, including a time stamp of the visit.

I need to be able to select timestamps that were within the current day, the previous day, the day before that ect.. So that I can generate the statistics.
Show it for the current week (current day and 6 previous days).

So it would be the entries where the timestamp was made on the days: 11/1, 10/1, 9/1, 8/1, 7/1, 6/1, 5/1. For example.

Not quite sure how I could do this.

Thanks.

For some reason my for loop only seems to be storing some data in my fields, most of the time the data is not true but simply duplicating one of the entries. So instead of having a string of 1,2,3,4 for instance it would store all of them as 4. The entires are selected using fields in a multiple select box, in theory of someone choses 10 unique entries from the "Systems" category all 10 should be made into rows in the MySQL table.

Any idea why this is happening?

The form,

if ($type == "games") {

echo "<tr><td>".$name."</td><td><select name='".$name."_".$i."'[] multiple='multiple'><option value='' selected>--------</option>";

$sqla = mysql_query("SELECT * FROM ".$pre."games ORDER BY `name` ASC") or die(mysql_error());

while($row2a = mysql_fetch_array($sqla)) {
$system = mysql_fetch_array(mysql_query("SELECT * FROM ".$pre."systems WHERE id = '".$row2a[system]."'"));
echo "<option value='".$row2a[id]."'>".$row2a[name]." - ".$system[name]."</option>";
}
echo "</select></td></tr>";
}

if ($type == "system") {
echo "<tr><td>".$name."</td><td><select name='".$name."_".$i."'[] multiple='multiple'><option value='' selected>--------</option>";

$sqlb = mysql_query("SELECT * FROM ".$pre."systems ORDER BY `name` ASC") or die(mysql_error());
while($row2b = mysql_fetch_array($sqlb)) {
echo "<option value='".$row2b[id]."'>".$row2b[name]."</option>";
}
echo "</select></td></tr>";
}

The PHP code,

while($row = mysql_fetch_array($query)) {
$name = "$row[name]";

for ($i = 0; $i < count($_POST["systems_$i"]); $i++) {
   mysql_query("INSERT INTO ".$pre."fielddata VALUES (null, 'systems', '".$_POST["systems_$i"]."', '".$fetch[0]."', 'content')");
}

for ($i = 0; $i < count($_POST["games_$i"]); $i++) {
   mysql_query("INSERT INTO ".$pre."fielddata VALUES (null, 'games', '".$_POST["games_$i"]."', '".$fetch[0]."', 'content')");
}

I coded the following but some parts are not working (the "curveball" mentioned at the end).
How would YOU do this?

I have a page (A) displaying a form with a textarea field using a WYSIWYG interface.
The user will enter a list of unordered items and upon submit the string will look something like this:

This is my list of food items:
 <ul>
  <li> Lettuce </li>
  <li> Tomatoes </li>
  <li> Eggs </li>
 <ul>




Upon submit, I need to give each list item an unique id and store this id along with the text next to it in a separate table:

List_item_id   List_item_text  List_item_state
food_1             Lettuce
food_2             Tomatoes
food_3             Eggs





The user will then land in another page with a form displaying the list. The text is now in the page itself and not in a textarea.
Instead of the bullets a dropdown list appears, and the user can select "buy" or "sell" for each list item. 
This page (B) looks something like this:

<form>
This is my list of food items:
  <br><select name="food_1">
   <option value="buy">Buy</option>
   <option value="sell">Sell</option>
  </select>
  Lettuce

  <br><select name="food_2">
   <option value="buy">Buy</option>
   <option value="sell">Sell</option>
  </select>
  Tomatoes

  <br><select name="food_2">
   <option value="buy">Buy</option>
   <option value="sell">Sell</option>
  </select>
  Eggs

<input type=submit value="submit">
</form>




When the user hits "submit" the table with the list items will be updated with the values selected in the dropdown list:

List_item_id   List_item_text  List_item_state
food_1             Lettuce        Buy
food_2             Tomatoes       Buy
food_3             Eggs           Sell



The next time the user goes to Page B the list will remember the states.


Here's the curveball: At any point the user may click on "EDIT LIST" on Page B so they may add more items (at the begining, middle or end of the string). On edit mode the list items should appear as bullets again inside the WYSIWYG interface.

Keep in mind that some of the text in the string may not be a list item (ie, "this is my list of food...")

Hello,

I am using the following code to display images managed by a MySQL database.  Basically another program manages a bunch of images, but this script displays certain ones (ones with INCLUDE = 1 in the database) on my main page.  My question is, is there an easy way to limit the number of images it displays, say to 5?  I'm not too concerned which images actually display (ascending or descending)... or better yet, random!  Most importantly, I only want five to display.  Each image will be linked to the full page, which displays all the images.

Any ideas?

Thanks!


Code: [Select]
<?php
$username="XXXXXXX";
$password="XXXXXXX";
$database="XXXXXXX";

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM ft_form_12 WHERE col_24='1'";
// $query="SELECT * FROM ft_form_12";
// SELECT * FROM ft_form_12 WHERE col_24='1'
$result=mysql_query($query);

$num=mysql_numrows($result);

mysql_close();
?>

<?php
$i=0;
while ($i < $num) {

$f20=mysql_result($result,$i,"col_23"); //Photo file name
$f21=mysql_result($result,$i,"col_24"); //INCLUDE
?>

<a href="http://www.domain.com/display_whole_page.shtml"><img src="http://www.domain.com/the_file/pictures/<?php echo $f20; ?>" height="50" border="0"></a>&nbsp;&nbsp;&nbsp;
               
<?php
$i++;
}
?>

Hello. Many thanks for your help.

I am writing a PHP/MySQL dating-site and have hit a programming impass.

I have a database full of members and a search form consisting of checkboxes.

So to search, a member ticks say...gender: female; age: 21,22,23,24,25,26; height: 5'4",5'5",5'6",5'7"; county: cornwall,devon,somerset

How can a run a check on the database selecting all entries that fall into the selected criteria. For example a 23 year old female of 5'5" living in Cornwall and a 26 year old female of 5'4" living in Somerset?

The key index of my database is 'id' and the fields a age,height,county
The names of the form checkboxes a
Gender: male, female; Age: 21,22,23,24 etc; Height: 5_4,5_5,5_6 etc; county: cornwall,devon etc

Im trying to create a website where users login, and then when they add a new entry to the database there name is put as the author. This is how my tables are set up. One table is named job and has the columns id, jobtext, jobdate, and authorid. Another table is called author. This table contains the columns id, username, password, and name. Authorid from the job table matches with id from the author table. When a user logins in this code is used to register the name...session_start();
$_SESSION['myusername'] = $_POST['myusername'];
$_SESSION['mypassword'] = $_POST['mypassword'];
header("location: index.php");
}
else {
echo "Wrong Username or Password";
}

This is the form users use to add a new entry...
if (isset($_GET['add']))
{
$pagetitle = 'New Job';
$action = 'addform';
$text = '';
$authorid = '';
$id = '';
$button = 'Add job';

include $_SERVER['DOCUMENT_ROOT'] . '/jobs/includes/db.inc.php';

// Build the list of authors
$sql = "SELECT id, name FROM author";
$result = mysqli_query($link, $sql);
if (!$result)
{
$error = 'Error fetching list of authors.';
include 'error.html.php';
exit();
}

while ($row = mysqli_fetch_array($result))
{
$authors[] = array('id' => $row['id'], 'name' => $row['name']);
}

// Build the list of categories
$sql = "SELECT id, name FROM category";
$result = mysqli_query($link, $sql);
if (!$result)
{
$error = 'Error fetching list of categories.';
include 'error.html.php';
exit();
}

while ($row = mysqli_fetch_array($result))
{
$categories[] = array(
'id' => $row['id'],
'name' => $row['name'],
'selected' => FALSE);
}

include 'form.html.php';
exit();
}

if (isset($_GET['addform']))
{
include $_SERVER['DOCUMENT_ROOT'] . '/includes/db.inc.php';

$text = mysqli_real_escape_string($link, $_POST['text']);
$author = mysqli_real_escape_string($link, $_POST['author']);

if ($author == '')
{
$error = 'You must choose an author for this job.
Click &lsquo;back&rsquo; and try again.';
include 'error.html.php';
exit();
}

$sql = "INSERT INTO job SET
jobtext='$text',
jobdate=CURDATE(),
authorid='$author'";
if (!mysqli_query($link, $sql))
{
$error = 'Error adding submitted job.';
include 'error.html.php';
exit();
}

$jobid = mysqli_insert_id($link);

if (isset($_POST['categories']))
{
foreach ($_POST['categories'] as $category)
{
$categoryid = mysqli_real_escape_string($link, $category);
$sql = "INSERT INTO jobcategory SET
jobid='$jobid',
categoryid='$categoryid'";
if (!mysqli_query($link, $sql))
{
$error = 'Error inserting job into selected category.';
include 'error.html.php';
exit();
}
}
}

header('Location: .');
exit();
}


Form.html.php =
<?php include_once $_SERVER['DOCUMENT_ROOT'] .
'/includes/helpers.inc.php'; ?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<title><?php htmlout($pagetitle); ?></title>
<meta http-equiv="content-type"
content="text/html; charset=utf-8"/>
<style type="text/css">
textarea {
display: block;
width: 100%;
}
</style>
</head>
<body>
<?php session_start(); ?>

<h1><?php htmlout($pagetitle); ?></h1>
<form action="?<?php htmlout($action); ?>" method="post">
<div>
<label for="text">Type your job he </label>
<textarea id="text" name="text" rows="3" cols="40"><?php
htmlout($text); ?></textarea>
</div>
<div>
<label for="author">Author:</label>
<select name="author" id="author">
<option value="">Select one</option>
<?php foreach ($authors as $author):?>
<option value="<?php htmlout($author['id']); ?>"<?php
if ($author['id'] == $authorid)
echo ' selected="selected"';
?>><?php htmlout($author['name']); ?></option>
<?php endforeach; ?>
</select>
</div>
<fieldset>
<legend>Categories:</legend>
<?php foreach ($categories as $category): ?>
<div><label for="category<?php htmlout($category['id']);
?>"><input type="checkbox" name="categories[]"
id="category<?php htmlout($category['id']); ?>"
value="<?php htmlout($category['id']); ?>"<?php
if ($category['selected'])
{
echo ' checked="checked"';
}
?>/><?php htmlout($category['name']); ?></label></div>
<?php endforeach; ?>
</fieldset>
<div>
<input type="hidden" name="id" value="<?php
htmlout($id); ?>"/>
<input type="submit" value="<?php htmlout($button); ?>"/>
</div>
</form>
</body>
</html>


Right now, under authors, it displays all the authors in the database. I want it to just show/submit the authorid of the logged in user.

Maybe some of the great coders here can help this noob out.
So here is what I have so far:

Code: [Select]
//set age criteria for deletion
$age = 60;

//get current date
$datenow  = date("Y-m-d");

//set the range we want to delete
$delete_range = $datenow - $age;

//get old user_id from users table
$oldusers_users = mysql_query ("SELECT user_id FROM users WHERE lastvisit < $delete_range ");

//get user_id from images table that correspond to users table
$oldusers_images = mysql_query ("SELECT user_id FROM images WHERE $oldusers_users=user_id.images ");

//find folders that correspond to the usernames and delete
$oldusers_files = mysql_query ("SELECT username FROM users WHERE lastvisit < $delete_range ");
//print out username
//$result = mysql_($oldusers_files);
$foldername = mysql_result($oldusers_files, 0);

$sigspath = "sigs/";
unlink($sigspath . $foldername . ".gif/index.php");
rmdir($sigspath . $foldername . ".gif");
//now delete user_id's from database
mysql_query("DELETE * FROM users WHERE user_id=$oldusers_users");
mysql_query("DELETE * FROM images WHERE user_id=$oldusers_images");
unset($oldusers_users, $oldusers_images, $oldusers_files, $foldername, $sigspath );
mysql_close($link);
Right now it is only returning the first entry, not the entire list that meets the criteria.  It does delete that one file though but does not remover the rows from the database.

I am sure the database stuff is jacked up I am really new to that part.  What am I doing wrong here or is there a better way to do this perhaps

Probably a silly question, but say you have a value in a mysql database that (naturally) can be either be blank or have characters in it.

If in my PHP I am doing a check to see if that returned field from the database HAS or DOES NOT HAVE a value entered in it, what is the proper way to do it?

For example, you could do either of the following below (and I'm sure there are probably 10 other ways to do it as well).  Is one of these quicker/more efficient than the other? 

I  know it probably does not matter in the grand scheme of things because it's such a quick check either way, but if one way is more "proper" than the other, please let me know...
Code: [Select]
if ($mysqlvalue!=="")
OR
if (strlen($mysqlvalue>0))
OR
 

Hey all. I'm a noob so bare with me. Just started studying php and have an assignment due. At this phase of the project I have to create a form where a user can register data and I should retrieve that data and store it in my database. Now I have already created the form (student_reg.php) and a separate file with php code (demo.php)

My database is name "registration" and I am trying to put this info in the "student" table. It has about 15 fields but for now I am only trying trying to insert the 1 field of data to test if it works. Dont know if that may cause problems?

So far my demo.php connects to the database successfully. The problem I am having is the information I submit does not showing in the database field. However, I have an id field in my case a student number field which is set to auto increment so it automatically updates (1,2,3,4) with each new sumbitted info. When I submit the info the id field gets updated each time with a new row ( so i assume some information is somehow going through ) however the info I entered will remain blank in its field. :banghead:

Heres my form (student_reg.php)

Code: [Select]
<div id="apdiv3">

<FORM action = "demo.php" method ="post ">

<p>Course name:</p>
<INPUT TYPE = "text"  name="input"/>
<INPUT TYPE = "Submit" VALUE = "Submit"/>
</div>
</FORM>

and heres my php code to retrieve the form info

Code: [Select]
<?PHP


include 'includes/config.php';

//connect to database
$link=mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);



//error check
if (!$link)
{
die('could not connect: ' . mysql_error());
}

$db_selected=mysql_select_db(DB_NAME, $link);



// error check
if (!$db_selected)
{
die('can\t use ' . DB_NAME . ': ' . mysql_error());
}
// Connected to database



// retrieve form data
$value=(isset($_POST['input']));

$sql = "INSERT INTO student (sname) VALUES ('$value')";


if (!mysql_query($sql))
{
die ('Error: ' . mysql_error());
}
mysql_close();


?>

I also recieved a notice "undefined index" or something like that from this part of the code "$value=(isset($_POST['input']));" before I inserted "isset" to the code. I'm not sure if what I did fixed the problem or only hid it, or whether that notice was related to the problem I'm having of not retrieving the info.

Haha sorry guys know this is probably childs play for you but I just started out and really need to fix this so I can catch up with this project. Feel free to add tips and comments on other areas. Thanks :mrgreen:

I have 2 files; Newfault.php and thankyou.php

1) Data is entered into a form in Newfault.php
2) The data from this form is retieved in thankyou.php and is then inserted into a table called "Calls"

Problem:

My entered record is being added to my database OK, but it is also adding a blank record for some reason and I can't work out why. Can anyone help? This is for my uni assignment.

Thanks,

Ladykudos

Hi there,

I am using this code to send the users email address to the database. That works fine, but i keep getting blank info added to the database. Does anyone know how i can stop this?

<?php
$con = mysql_connect("*","*","*");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("ogs_mailinglist1", $con);

$sql="INSERT INTO mailinglist (email)
VALUES
('$_POST[rec_email]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }

mysql_close($con);
?>



Thanks,

hi all i am having a big problem that i have been trying to find out what is going on for weeks. i have a echo script that echos data that is in my database, and if i have to refresh the page it will add blank data in my database and the top echo info is blank as well. how can i fix this, and i was wanting to know how do i echo out my info in a textarea. i added a jepg to show you what i mean.

here is my code

echoforms.php

<?php

error_reporting(0);


require_once('demo.php');
/*Open the connection to our database use the info from the config file.*/
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);

$sql = "SELECT company_name, contact_name, address, street_number, postcode, contact_number, contact_email, budget, description FROM 3dartactforms";

$results = mysql_query($sql);
      
if (!$results) {
   die('Invalid query: ' . mysql_error());
}



while($result = mysql_fetch_array( $results )){
   echo '<div style="border: 1px solid #e4e4e4; padding: 15px; margin-bottom: 10px;">';
   echo date("d/m/y");
   echo '<p>Company Name: ' . $_POST['company_name'] . '</p>';
   echo '<p>Contact Name: ' . $_POST['contact_name'] . '</p>';
   echo '<p>Address: ' . $_POST['address'] . '</p>';
   echo '<p>Street Number: ' . $_POST['street_number'] . '</p>';
   echo '<p>Postcode: ' . $_POST['postcode'] . '</p>';
   echo '<p>Contact Number: ' . $_POST['contact_number'] . '</p>';
   echo '<p>Contact Email: ' . $_POST['contact_email'] . '</p>';
   echo '<p>Budget: ' . $_POST['budget'] . '</p>';
   echo '<p>Description: ' . $_POST['description'] . '</p>';
   echo '</div>';
}


?>

Below is the part of my code that creates the html table.  How do I add pagination every 30 entries?



$db = mysql_connect("localhost","trend_learnu","asdfasdf");
mysql_select_db("trend_learningdb",$db);

$result = mysql_query("SELECT * FROM Peoples ORDER BY number ASC");

$table = '<table width="608" border="0" align="center" cellpadding="10" cellspacing="0">
  <tr>
    <td style="color:#fff;" width="207" bgcolor="#425d74"><strong>Firstname</strong></td>
    <td style="color:#fff;" width="204" bgcolor="#425d74"><strong>Lastname</strong></td>
    <td style="color:#fff;" width="197" bgcolor="#425d74"><strong>Date</strong></td>
<td style="color:#fff;" width="197" bgcolor="#425d74"><strong>Weight</strong></td>
  </tr>';

while($row=mysql_fetch_array($result))
{
  $firstname = $row['Firstname'];
  $lastname = $row['Lastname'];
  $weight = $row['Weight'];
  $date = $row['Date'];

   $table.= '<tr>';
   $table.= '<td bgcolor="#e8f4ff">'.$firstname.'</td>';
   $table.= '<td bgcolor="#e8f4ff">'.$lastname.'</td>';
   $table.= '<td bgcolor="#e8f4ff">'.$date.'</td>';
   $table.= '<td bgcolor="#e0ffe6"><strong>'.$weight.'</strong></td>';
   $table.= '</tr>';
}
$table .= '</table>';

echo $table;



mysql_close($con)

the problem is that when I add an item to be displayed in the blog, shows three instead, which two are empty.
It should not create three entries when I just add one entry. How can I fix this?

<?PHP
/* define the blog content file name */
$filename = "myBlogContent.txt";
?>




<?php
/* check to see if the file exists */
if (!file_exists($filename)) {

echo "The Blog Is Empty";

}else{

/* get the file lines into an array */

$BlogArray = file($filename);

/* count the number of blog entries */
$count = count($BlogArray);

$i=0;

while($i<$count) {

$new_array = explode("|", $BlogArray[$i]);

echo "Posted by: " . $new_array[1] . "<br>";

echo "Posted on: " .  date("m/d/y h:iA", time($new_array[0])) . "<br>"
echo 

"Title: " . $new_array[2] . "<br>";


echo $new_array[3] . "<hr>";


$i ++;


}
}
?>





<?PHP
/* obtain the form data */
$who = $_POST['who'];
$title = $_POST['title'];
$content = $_POST['content'];
$content = str_replace(array("\r\n", "\r", "\n"), "<br>", $content); 

/* create timestamp variable for current date and time */
$when_ts = time(); 

/* define the blog content file name */
$filename = "myBlogContent.txt";

/* prepare the variables for adding to the file */

$new_line_content = $when_ts . "|" . $who . "|" . $title . "|" . $content . "\n";

/* open the file in the APPEND MODE */
$fh = fopen($filename, 'a+') or die("can't open file");

/* add the new content */
fwrite($fh, $new_line_content); 

/* close the file */
fclose($fh); 

 
//exit; // Closes further script execution . 
?>

Hi guys 

Im trying to delete multiple images from  a page with the code below, I have retirieved the images successfully but i can not delete the with checkbox,

wht it does is just refereshing the page and thats it, what im trying to do is to delete the image from the database but it wont.

can u help me please?

<?php
session_start();

include ("../../global.php");
//welcome messaage
$username=$_SESSION['username'];
echo "$username";

$query=mysql_query("SELECT id FROM users WHERE username='$username'");
while($row = mysql_fetch_assoc($query))
{
$user_id = $row['id'];
}

$ref=$_GET['reference'];


$images=mysql_query("SELECT * FROM img WHERE refimage='$ref'");

while($row = mysql_fetch_array($images))
{
$image=$row['image'];
$thumb=$row['thumb'];





?>

<table width="400" border="0" cellspacing="1" cellpadding="0">
<tr>
<td><form name="" method="post" action=''>
<tr>
<td align="center" bgcolor="#FFFFFF"><input name="checkbox[]" type="checkbox" id="checkbox[]" value="<? echo $rows['id']; ?>"></td>
<td bgcolor="#FFFFFF"><? echo "<a href='$image' rel='lightbox[roadtrip]'><img src= '$thumb' width='60' height='40' alt='$title'>";?></td>
<?
}
?>


<tr>
<td colspan="5" align="center" bgcolor="#FFFFFF"><input name="delete" type="submit" id="delete" value="Delete"></td>
</tr>


<?php

// Check if delete button active, start this
$delete = $_REQUEST['delete'];
if( $delete != '' ){
$checkbox = $_REQUEST['checkbox'];
$count = count($_REQUEST['checkbox']);
for($i=0;$i<$count;$i++){
$del_id = $checkbox[$i];
$del = mysql_query("DELETE FROM img WHERE id = '$del_id'");

}
}

?>