PHP - For Loop Only Storing Some Entries In Database

Hello,

For starters I'm not sure if what I want to do is possible, but if it is I would like your input. I have a script that will show a number of fields to fill out in a second form based on the number the user puts into the first from. the problem is that only the last one saves into the database and not all of them.

Code: [Select]
<form auction="index.php" method="post">
System Name: <input type="text" name="systemname">
Number of E-sites: <input type="text" name="events">
Number of Sigs: <input type="text" name="sigs"><br>
<input type="reset" name="reset" value="Reset">
<input type="submit" name="start" value="Start">
</form>
<form auction="index.php" method="post">
<?php
$events = $_POST['events'];
$system = $POST['systemname'];

if (isset($_POST['start'])) {
$num = $_POST['sigs'];
$i = 0;
While ($i < $num)
{
echo "Sig ID: <input type=text name=sigid>";
echo "Type: <input type=text name=type>";
echo "Name: <input type=text name=name>";
echo "Notes: <input type=text name=notes>";
echo "<br>";
$i++;
}
}
?>
<input type="submit" name="enter" value="Enter">
</form>
<?php
$sigid = $_POST['sigid'];
$type = $_POST['type'];
$name = $_POST['name'];
$notes = $_POST['notes'];
mysql_connect('xt', 'x', 'x');
mysql_select_db('wormhole');
if (isset($_POST['enter'])) {

$query = "INSERT INTO sites VALUES ('$system','$events','$sigid','$type','$name','$notes')";
mysql_query($query);

}
?>
How do I get it so all the data saves, lets say that $num = 5, I want all 5 to save not just the last one.

Code: [Select]
$query = mysql_query("SELECT a.*, b.*
         FROM friendlist a
INNER JOIN friendlist b
ON (a.friendemail=b.friendemail)
INNER JOIN users c
ON (b.friendemail = c.EmailAddress)
WHERE a.email = 'asdf@gmail.com'
AND c.Username LIKE '%carol%'
GROUP BY a.id
ORDER BY count(*) DESC");
Code: [Select]
while ($showfriends = mysql_fetch_array($query)) {
     echo $showfriends['Username'];
}
and I would get nothing. It produces the correct number of <div> so i know it's getting through, but it's having trouble displaying the entries?

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.

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

Quote

i want to store a value from a database
and use it as variable in php code

can anyone help me out
in this code i want to store value of copies
in a php variable

and want that it should be more than 0 (zero)

for furthur calculations




$update_book="update book set copies=copies-1 where bookid='$bookid'";
        $result=mysql_query($update_book,$linkID1);
            if($result)
            {
            print "<html><body background=\"header.jpg\">
            <p>book successfully subtracted from database</p></body></html>";
            }
            else {
            print "<html><body background=\"header.jpg\">
            <p>problem occured</p></body></html>";
            }

Dear All,

I am developing a like/dislike voting system for a scientific experiment in which the user will cast vote on a number of images. I have already written the code for it but the code is unable to store the data in the database. Kindly help me out in this issue. Below is my code for it:

HTML Form for Like/Dislike Button:

Code: [Select]
<html>
<head>
<title>
Like/Dislike buttons
</title>
</head>
<body>
<h1>
Like/Dislike buttons
</h1>
<form name="form1" action="likedislike.php" method ="post">
<input type="submit" name="Like"    value="Like">
<input type="submit" name="Dislike" value="dislike">
</form>
</body>
</html>


PHP code for retrieving data and storing the vote in the database:

Code: [Select]
<?php
session_start();
ob_start();
$host="localhost";
$username="computat_abhi";
$password="1123581321";
$databasename="computat_button";
$tbl_name="record";

$db=mysql_connect ("localhost","computat_abhi","1123581321")or die("cannot connect");
mysql_select_db($databasename,$db);

if( isset($_POST['Like']) )
{

$sql = 'INSERT INTO record(Image Number,Like Counter)
VALUES ("b1.php",1)';
$result = mysql_query($sql);
}
elseif(isset($_POST['Dislike'])){
$sql = 'INSERT INTO record(Image Number,Dislike Counter)
VALUES ("b1.php",1)';
$result = mysql_query($sql);
}
ob_end_flush();
?>

Thank you for your help.

Hi

I am using Zen-Cart to which I have made significant changes.  Two years ago I managed to change the Products Sort Order from Integer to Decimal(5,2) to store books such as 1, 1.5, 1.75, 2 etc.  Regrettably we had a major server issue and I also lost the new coding.  Having restored everything else whenever we enter a book series number such as 1.5 it only stores as 1.

To date the zen-Cart forums have been unable to offer a solution so I am reaching out to see if anyone else can advise me of how to get this working again please.

 

Thanks in advance

Owen

 

Hi, wondering if somebody can tell me where I'm going wrong (I'm new to all of this). I have the following php code which uploads an image file into my database:


Code: [Select]
    //Connect to database
    include 'Resources/Include/db.inc.php';

    $tmp=$_FILES['image']['tmp_name'];
    //get users IP
    $ip=$_SERVER['REMOTE_ADDR'];

    //Don't do anything if file wasn't selected
    if (!empty($tmp)) {

    //Copy file to temporary folder
    copy($tmp, "./temporary/".$ip."");

    //open the copied image, ready to encode into text to go into the database
    $filename1 = "./temporary/".$ip;
    $fp1 = fopen($filename1, "rb");

    //record the image contents into a variable
    $contents1 = fread($fp1, filesize($filename1));

    $contents1 = addslashes($contents1);

    //close the file
    fclose($fp1);

    $ftype = $_FILES['image']['type'];

    //insert information into the database
    if(!mysql_query("INSERT INTO LetterImages (Data,Type,LetterID,Page)"." VALUES ( '$contents1', '$ftype',1,1)")){
    echo mysql_error();
    }

    //delete the temporary file we made
    unlink($filename1);

    }

This seems to work ok, as when I go to the LetterImages table there is now an additional row with a file in the blob field. I then have the following code which is supposed to display the image:


Code: [Select]
    $result=mysql_query("SELECT * FROM LetterImages WHERE LetterID=1 AND Page=1");

    //fetch data from database
    $sqldata=mysql_fetch_array($result);

    $encoded=stripslashes($sqldata['Data']);
    $ftype=$sqldata['Type'];

    //tell the browser what type of image to display
    header("Content-type: $ftype");

    //decode and echo the image data
    echo $encoded;

Instead of displaying an image, however, this just displays pages and pages of incomprehensible data. Can anybody tell me where I'm going horribly wrong?

I've got an array like this:

$firstquarter = array('January', 'February', 'March');

Then I'm adding it to a MySQL DB as follows:

$insert = "INSERT INTO sometable (months) VALUES ('$firstquarter') ";
$insertresult = mysql_query($insert) or die ('Error in Insert query: ' . mysql_error());

When I use PHPMyAdmin I see a this value stored in the DB:

'Array'

Can I retrieve the values, by index, from this database field??

Okay I'm a newbie at PHP but was trying to write a counter for my website to count the amount of times a specific button is clicked. My desired outcome is as follows:

If the form submit button "like" or "dislike" is clicked, get user's ip and check to see if it already exists in the database.
If it exists, don't increment the count (of either button), otherwise log the ip and increment the count (of whichever button is clicked) by 1.

Currently I've got the increment working (testing only the like button at the moment) and I've got the ip being stored into the database, but cannot get the comparison to work. Each time the button is clicked it adds the ip address into the database instead of disallowing it.

Note: there are two different tables--one storing the ip (counter) and one storing the like clicks (Designs) (dunno if there's a better way?)
Excuse my newbieness.

The problem area: (like button only):

if(isset($_POST['like']))
{
$ip_address = $_SERVER['REMOTE_ADDR'];
$query = mysql_query("SELECT ip FROM counter WHERE ip = '$ip_address'");
$num_rows = mysql_num_rows($query);
if ($num_rows > 0)
{
die ("You've already clicked the button!");
}
else
{
mysql_query("INSERT INTO counter VALUES ($ip_address)");
mysql_query("UPDATE Designs SET up = up + 1 WHERE ref = 'cataclysm'");
}
}

$select = mysql_query("SELECT up FROM Designs");

while($row = mysql_fetch_array($select))
{
echo $row['up'];
}

I need some advice, I'm not sure if this is the right section to post this or not but I'll give it a try. I am creating a website for a small group of people to be able to message back and forth. I ran into a problem when doing the whole reply to a message option. I dont want to have each individual message as a new entry in my database, and then look for each message in that conversation when loading the conversation, I thought about inserting something like +=+=+=+=+=+= in between messages in the conversation so that when i pull up that conversation i could just separate the messages into an array by looking for that +=+=+=+=+=+= in the string of text and separating all the messages out. I want to know if anyone knows a better way of doing this or if this is a good way to go. thanks for any suggestions.