PHP - How To Specify The Size Of Entry Fields ?

My query is not finding the last recieptnum entry, it is finding the number 9 everytime for some odd reason. Im trying to incrementally increase this each time a reciept is created.

$getreceiptnum = mysql_query("SELECT receiptnum FROM accounting WHERE agency = '$agency' ORDER BY receiptnum DESC LIMIT 1") or die(mysql_error());
        $recieptarray = mysql_fetch_array($getreceiptnum);
        $recieptnum = $recieptarray['receiptnum'];
        echo $recieptnum;

I have  a code where i can edit or delete certain details from the database. Right now, if the user clicks on the edit button it takes him edit page where he can edit the details. But, I am not able to Incorporate a Delete button such that, when the user clicks on a delete button, it should ask for a confirmation box. If the user clicks YES, then do the following:

Code: [Select]
DELETE from emp WHERE emp_id='$emp_id';
When there are multiple entries and I click on delete it deletes everything from the database. how can i make it to delete only the entry that is besides the delete button?

Code: [Select]
if(mysql_num_rows($emp_query) > 0){
echo "<table border='1'>";
echo "<th>Employee Id </th>";
echo "<th>Employee Name </th>";
while($get_emp = mysql_fetch_assoc($emp_query)){
$emp_id = $get_emp['emp_id'];
$emp_name = $get_emp['first_name']." ".$get_emp['last_name'];
echo "<tr>";
echo "<td width='100'>";
echo $emp_id;
echo "</td>";
echo "<td width='400'>";
echo $emp_name;
$edit_path = 'edit_employee.php?id='.$emp_id;

?>
<INPUT TYPE="button" style="display:inline;" value="VIEW / EDIT" onClick="location.href='<?php echo $edit_path; ?>'">
<form style='margin: 0; padding: 0; display:inline;' method="post" action="<?php echo $_SERVER['PHP_SELF'];?>" onSubmit="return confirm('Are you sure this is correct?');">
<input style='display:inline;' name="delbutton" type="submit" value="DELETE">
<?php if(isset($_POST['delbutton'])){
$del_emp = mysql_query("DELETE from employee WHERE emp_id = '$emp_id'") or die(mysql_error());
    //header('Location:view_employee.php');
  }
echo '</form>';

echo "</td>";
echo "</tr>";
}

}

        $insertCount=0;
        foreach($results[1] as $curName)
        {
            if($insert){$insertCount++;}
        echo <<< END
$curName<BR>
END;

        }


Right now the results would show up as...

Bill
Fred
Jessica
James
John

How do you make them show up like...

1 Bill
2 Fred
3 Jessica
4 James
5 John

Well this sounds weird, but it happens. I have an itemshop script, and I wanna code a system that automatically delete the database entry once the variable item amount falls to 0. However, this doesnt happen when I check phpmyadmin, and the item remains in the database even after the amount becomes 0. In fact, the amount can become negative at times, which annoys me.



The code of deleting sql entry looks like this:


if($continue == "yes"){
$query = "SELECT * FROM ".$prefix."user_inventory WHERE item_owner = '$loggedinname' AND item_name = '$item_name'";
                  $result = mysql_query($query);
                  $num = mysql_numrows($result);
                  $item_data = mysql_fetch_array($result);
                  
                  $item_amount = $item_data['item_amount'];
                  
                  $newquantity = $item_amount - $quantity;

$query = "UPDATE ".$prefix."user_inventory SET item_amount = '$newquantity' WHERE item_owner = '$loggedinname' AND item_name = '$item_name'";
                  $result = mysql_query($query);
                  
                  if($newquantity == "0"){
                    $query = "DELETE FROM ".$prefix."user_inventory WHERE item_owner = '$loggedinname' AND item_name = '$item_name'";
                  $result = mysql_query($query);
                  }
                  


What part of the codes should I edit to fix this issue? Thanks.

Hey folks,

Sorry for being a pain in the ass.

I am trying to submit data to my database via a form and when I click Submit, I get:

Duplicate entry '' for key 1

I understand that it means I have a duplicate entry with the ID of 1 or something like that. I can't find where the issue is.

Here is the form:


<form actin="" id="settings" name="settings">
                    <table class="listing form" cellpadding="0" cellspacing="0">
                        <tr>
                            <th class="full" colspan="2"><?php echo $lang_settings; ?></th>
                        </tr>
                        <tr>
                            <th colspan="2"><?php echo $lang_settings_description; ?></th>
                        </tr>
                        <tr>
                           <td><?php echo $lang_sitename; ?>: </td> <td><input type="text" name="sitename" value="<?php echo $site_name; ?>" width="172" /> <em>Site name for logo</em></td>
                        </tr>
                        <tr>
                           <td><?php echo $lang_email; ?>: </td> <td><input type="text" name="email" value="<?php echo $site_email; ?>" width="172" /> <em>Your email address</em></td>
                        </tr>
                        <tr>
                           <td><?php echo $lang_yourname; ?>: </td> <td><input type="text" name="name" value="<?php echo $your_name; ?>" width="172" /> <em>Your own name</em></td>
                        </tr>
                        <tr>
                           <td><?php echo $lang_meta_description; ?>: </td> <td><input type="text" name="meta-description" value="<?php echo $description; ?>" width="172" /> <em>SEO</em></td>
                        </tr>
                        <tr>
                           <td><?php echo $lang_keywords; ?>: </td> <td><input type="text" name="meta-keywords" value="<?php echo $keywords; ?>" width="172" /> <em>Separate with Commas</em></td>
                        </tr>
                       <tr>
                       <td><input type="submit" class="button" name="submit" value="<?php echo $lang_button_savesettings; ?>"></td>
                       </tr>
                    </table>
</form>


Here is the Insert code:

$insert = "INSERT INTO settings (site_name, description, keywords, email, name) 
VALUES ('$sitename', '$meta_description', '$meta_keywords', '$site_email', '$your_name')"; 
mysql_query($insert) or die(mysql_error());


Can anyone please tell me where I am going wrong here?

Much appreciated.

i'm trying to get the username of the last person to enter something into a database table,and display it an html table. the problem is, the query is failing, and i get the error
"Table 'forum.posts' doesn't exist". if in the error message "forum" means the name of the database, and "posts" the name of the table, there is something wrong, as they both exist.

the code i'm using is below

$list = "SELECT * FROM section_main ORDER BY section_title";
$result = mysql_query($list) or die ("Query failed");
$numofrows = mysql_num_rows($result);
for($j = 0; $j < $numofrows; $j++) { 
    $row = mysql_fetch_array($result);  
    echo "<tr><th>". $row['section_title']."</th></tr>";

$query2 = "SELECT * FROM section_sub WHERE section_id = '".$row['section_id']."' ORDER BY section_sub_title";
$result2 = mysql_query($query2) or die("Select Error :" . mysql_error()); 
$numofrows2 = mysql_num_rows($result2);
for($i = 0; $i<$numofrows2; $i++){
$row2 = mysql_fetch_array($result2);
echo "<tr><td><a href='display_forum.php?id={$row2[section_sub_id]}'>".stripslashes($row2[section_sub_title])."</a></td><td>"; 

$new = mysql_query("Select * FROM posts WHERE section_sub_id = '".$row2['section_sub_id'] . " ' ") or die ("Select Error :" . mysql_error()); //this is where the error message is from
$lastpost = mysql_fetch_assoc($new);
echo $lastpost['username']; 

echo"</td></tr>";
}
}


i've checked that the names of the tables and fields in the queries are correct, but i cant think of anything else that might be wrong.

any help would be great.

Hi..

So im currently working on a script..
My script generates a "oid" based on timestamp.

Ive made the "oid" field unique in my db, so if i do a quick refresh i get the message:

Duplicate entry '1283195988' for key 'oid'

Is there some way i can check if its a dublicate, and if it is + it with 1 or something?

Hello there,

I have some code here which sends a number of variables from flash to SQL...

I would simply like to add the functionality to overwrite records which have the same 'name' or 'pseudo'... can anyone help me please ? Thanks in advance Martin


<?php

$pseudo=$_POST['var1'];
$score=$_POST['var2'];
$table = $_POST['tab'];

$dategame = $_POST['tempjoueur'];
//$micro = microtime();
//$dategame = time()."".substr($micro, 2, 6);

$_COOKIE['User'] = $_SERVER['REMOTE_ADDR'];
$envoie = InsertDatas($table, "name, score, dategame", "'".$pseudo."','".$score."','".$dategame."'");
if ($envoie) {
print_r("OK, $pseudo, $score, $dategame,$ipclient");
}
else { echo "BAD, $pseudo, $score, $dategame,$ipclient"; }
?>

Hey guys!

I know, I know this problem is EVERYWHERE but i just dont understand! I have a solid knowlage of php but my SQL skills are low, so i dont know too much about Keys and stuff. But my error is: Duplicate entry '' for key 2.

The thing that im working on at this section is logging in with facebook. The code that presents my error is:

    $sql = "SELECT * FROM users WHERE uid=".$uid;
    $fbid = mysql_query($sql);
$num_rows = mysql_num_rows($fbid);

if(mysql_num_rows($fbid) < 1) {
echo "You are not logged in.    ";
mysql_query("INSERT INTO `users` (`uid`) VALUES ('".$uid."')")
or die(mysql_error());  

} else {
mysql_query("UPDATE users SET logged = '1' WHERE uid=".$uid);
//mysql_query("UPDATE users SET full_name = $me WHERE uid=".$uid);

echo "Your Logged in ";
echo $me['name'];
?> Continue to <a href="removed :)"> My Settings </a>. <?
}


Any help is welcome

For some reason this only allows one SQL to be added...

// SQL Connection
$username="monstert_admin"; $password="admin"; $database="monstert_admin";
$connection = mysql_connect("localhost", $username, $password) 
or die("Connection Failure to Database");
// Select Database
mysql_select_db($database, $connection) or die ($database . "No Database" . $username);

//Select everything from the the table
$MyQuery = "SELECT * FROM photos";
$retrieve = mysql_query($MyQuery) or die(mysql_error());
if(mysql_num_rows($retrieve) != 0):
$row = mysql_fetch_assoc($retrieve);
else:
echo '';
endif;

if(isset($_POST['Submit']) && !$errors)
{
$url = $newname;

   include('img.php');
   $image = new SimpleImage();
   $image->load($url);
   $image->resize(500,315);
   $image->save($newname);

mysql_query("INSERT INTO photos (url) VALUES ('$url')");

echo "File Uploaded Successfully as <i> ";
echo $newname;
echo "</i>";
}

What would the issue be? I only have two columns - ID and url

Thanks in advance!

Dear All,

I am developing a website www.computationalphotography.in on which I need to register new users. I have designed an HTML form for this purpose,
but will need help with writing the code for checking and validating the data in the respective fields. The code for my web form is below and the address
of the website is www.computationalphotography.in/registration.php:

Code: [Select]
<html>

<head>

<title>

Photography Experiment Registration
</title>

</head>

<body>

<h1>

Questionnaire
</h1>

<h3>
We would like to know about your background and interests in photography. For each multiple-choice question, please circle all answers that apply.</h3>
<form method="post" action="registration.php">


Name
<input name="name" type="text">
<br>
<br>
Email

<input name="email" type="text">
<br>
<br>
1.) Roughly How many pictures do you take each year?
<br>
<br>
<input name="radios" type="radio" value="a">
a) a few: <10 in a year
<br>
<input name="radios" type="radio" value="b">
b) many: probably 100's a photos a year
<br>
<input name="radios" type="radio" value="c">
c) a lot: probably 1000's of photos a year
<br>
<br>

2.) Do you share your pictures with anyone?
<br>
<input name="radios" type="radio" value="a">
a) No
<br>
<input name="radios" type="radio" value="b">
b) Yes, but only with friends or family
<br>
<input name="radios" type="radio" value="c">
c) Yes, I share with friends or family and also post on public sites such as Flickr, Picasaweb, or Snapfish for anyone to see.
<br>
<input name="radios" type="radio" value="d">
d) Yes, I share frequently and I have also published my photos in books or magazines, or exhibitde my photos in shows.
<br>
<br>

3.) How much training in photography have you had?
<br>
<input name="radios" type="radio" value="a">
a) None
<br>
<input name="radios" type="radio" value="b">
b) Some, but it's self-taught from books or websites
<br>
<input name="radios" type="radio" value="c">
c) Quite a bit: I have taken courses in photography
<br>
<input name="radios" type="radio" value="d">
d) A lot:I have completed a degreee or certificate in the subject
<br>
<br>

4.) How do you describe yourself as a photographer?
<br>
<input name="radios" type="radio" value="a">
a) A novice:someone who takes a few pictures occasionally
<br>
<input name="radios" type="radio" value="b">
b) An amateur:someone who takes many photos, but doesn't usually share them
<br>
<input name="radios" type="radio" value="c">
c) An enthusiast:someone who loves to take pictures, possibly shows them with oters on public sites, or who might own a DSLR, or frequently uses a camera in other than "automatic mode"
<br>
<input name="radios" type="radio" value="d">
d) An expert:skilled photographer who has published their work, or exhibited their work in shows, or has been paid for their work
<br>
<br>

5.) Do you have artistic training or exposure?
<br>
<input name="radios" type="radio" value="a">
a) No
<br>
<input name="radios" type="radio" value="b">
b) I visit museums or art galleries
<br>
c) I have studied art: please describe
<br>
<textarea name="data5c" cols="50" rows="5">
</textarea>
<br>
d) I am a practicing artist: please describe
<br>
<textarea name="data5d" cols="50" rows="5">
</textarea>
<br>
<br>
6.) Some optional information, if you don't mind. We respect your privacy.
<br>
<br>
Age
<br>
<textarea name="6a" cols="1" rows="1">
</textarea>
<br>
<br>
Gender
<br>
<input name="radios" type="radio" value="a">
a) Male
<br>
<input name="radios" type="radio" value="b">
b) Female
<br>
<br>
Languages Spoken
<br>
<textarea name="6b" cols="25" rows="2">
</textarea>
<br>
<br>
Languages Read
<br>
<textarea name="6c" cols="25" rows="2">
</textarea>
<br>
<br>
<br>
<input type="submit" value="Send">

<input type="reset"  value"Reset">

</form>

</body>

</html>

 Kindly help me out with this issue. Thank you for your help in this regard.

I have a foreach loop as

foreach ($xml->channel->item as $value){
$title = $value->title;
$text = $value->text;
echo "$title <br /> $text";
}

I want to skip any entry in which the title contains a character like ":"

Need help in achieving a data entry grid.
I have two tables am_user (idam_user, name) and am_activity (idam_act, act) with the data
Data for am_user is
1, John
2, Mary
3, Kate
Data for am_activity is
1, Cooking
2, Cleaning
3, Painting
I have a third table that stores the role played by each user for the activities. am_role (idam_role, role)
1, Manage
2, Observe
3, Documents
I want my User table to be my header, Activity table to my 1st column and Role table to populate the data grid
John Mary Kate Cooking Manage Document Observe Cleaning Document null Manage Painting Observe Document Manage
By putting two while loops I am able to generate the header and 1st column but I need help in populating the grid in between.
Also I want my data entry area (rows and column besides header and 1st column) to have lookup where they can select values from the am_role table.
So, John can "Manage" "Painting" tomorrow if he wants to
The two while loops are
Code: [Select]
$query = "select idam_user, Name ";
$query .= " from am_user ";
        $result = mysql_query($query) or die(mysql_error()); 
               
        // display data in table
        echo "<center>";
        echo "<table>";
        echo "<tr>";
        echo "<th></th>";
       

        // loop through results of database query, displaying them in the table
        while($row = mysql_fetch_array( $result )) {
               
                // echo out the contents of each row into a table
                echo '<th>' . $row['name'] . ' ';
                echo "</th>";
        }
        echo "</tr>";

$query = "select idam_act, act ";
$query .= " from am_activity ";
        $result = mysql_query($query) or die(mysql_error());                 
        // display data in table
       

        // loop through results of database query, displaying them in the table
        while($row = mysql_fetch_array( $result )) {
               
                // echo out the contents of each row into a table
                echo '<tr>';
                echo '<td>' . $row['act'] . '</td>';
                echo '</tr>';
        }
       
       
        // close table>
        echo "</table>";
        echo "</center>";


Please advise if this thought is realistic and how can I achieve it.
Thanks in anticipation.

I'd like some help understanding how to use validation for form field characters please!


if(!preg_match("/^[a-zA-Z]+$/",$_POST['fname']))
{
echo "Invalid Entry";
}


Now, where do I specify what characters to check? Is it this part:


("/^[a-zA-Z]+$/",$_POST['fname'])


How would I add other things? I believe that allows upper and lower case letters only, how would I add things like '@' '-' '_' etc etc...would I just add them to the list inside the square brackets or is it a new set of brackets?... What about numbers? Would it be "/[0-9]/" ?

Also, can I swap $_POST['fname'] with just a plain variable ($fname)?.

(oh and does it need the '/' inside each quote?)

Thanks!

OH and also! if I was to check several fields, and if all is ok execute a script would I need to have elseif/else statements of could I have a new if statement for each and then just run the script below it?

Thanks again!

Hi

I'm very much a newbie to PHP and am struggling with a registration and login problem.

The registration part is fine - the password is emailed out fine, I check the database and the entry is there, the password is encrypted. but when the user tries to log in....

if (mysql_num_rows($result) == 0) .... it returns 0? and the access denied page is shown.

I'm thinking it is something to do with the sql database encryption? 

I'm completely lost here and would appreciate some guidance

Thanks in advance
Jan