PHP - Select All Entry Beggining With Any Number?

Hello,

I had 8 select boxes(dropdowns) in a form. Its like a search kind of implementation in the website. I don't know what the user selects, he may choose different select boxes, any number of select boxes.

So, based upon the user selection, I need to generate the data. The data fields to be generated would be almost same.

So, How would I know what the user selection is and how many select boxes has been selected and how could I generate different data based upon selected boxes.

I had to make a small note abt this, that the data generation fields may change for some of the user select combinations, but most of the result fields would be same.

so, can you please help me out, how to do, how to make different combination data generations, because, i had 8 fields, i had to make 8! combinations, that would result in a big code.

I am using php to upload a file to my server, and at the same time inserting the files name and url into my mysql database.

$sql = "UPDATE uploads SET name = '$name', url='$target_path'";
    $statement = $dbh->prepare($sql); 
    $statement->execute();

This is working, however, when I upload a new file, rather than making a new entry in my database, it just overwrites the first one.

I'm quite new at mysql so was wondering how I would make it add new entrys instead of overwriting the current one?

I'm getting the dreaded " Invalid parameter number: number of bound variables does not match number of tokens" error and I've looked at this for days.  Here is what my table looks like:

|  id          | int(4)       | NO   | PRI | NULL    | auto_increment |
| user_id     | int(4)       | NO   |     | NULL    |                |
| recipient   | varchar(30)  | NO   |     | NULL    |                |
| subject     | varchar(25)  | YES  |     | NULL    |                |
| cc_email    | varchar(30)  | YES  |     | NULL    |                |
| reply       | varchar(20)  | YES  |     | NULL    |                |
| location    | varchar(50)  | YES  |     | NULL    |                |
| stationery  | varchar(40)  | YES  |     | NULL    |                |
| ink_color   | varchar(12)  | YES  |     | NULL    |                |
| fontchosen  | varchar(30)  | YES  |     | NULL    |                |
| message     | varchar(500) | NO   |     | NULL    |                |
| attachment  | varchar(40)  | YES  |     | NULL    |                |
| messageDate | datetime     | YES  |     | NULL    |        

Here are my params:

            $params = array(                 
               ':user_id' => $userid,                 
               ':recipient' => $this->message_vars['recipient'],                 
               ':subject' => $this->message_vars['subject'],                 
               ':cc_email' => $this->message_vars['cc_email'],                 
               ':reply' => $this->message_vars['reply'],                 
               ':location' => $this->message_vars['location'],                 
               ':stationery' => $this->message_vars['stationery'],                 
               ':ink_color' => $this->message_vars['ink_color'],                 
               ':fontchosen' => $this->message_vars['fontchosen'],                 
               ':message' => $messageInput,                 
               ':attachment' => $this->message_vars['attachment'],                 
               ':messageDate' => $date             );

Here is my sql:

$sql = "INSERT INTO messages (user_id,recipient, subject, cc_email, reply, location,stationery, ink_color, fontchosen, message,attachment) VALUES( $userid, :recipient, :subject, :cc_email, :reply, :location, :stationery, :ink_color, :fontchosen, $messageInput, :attachment, $date);";

And lastly, here is how I am calling it:
            $dbh = parent::$dbh;             
           $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_WARNING);              
           if (empty($dbh))    return false;             
                  $stmt = $dbh->prepare($sql);             
                 $stmt->execute($params) or die(print_r($stmt->errorInfo(), true));              
           if (!$stmt) {    print_r($dbh->errorInfo());              }

I know my userid is valid and and the date is set above (I've echo'd these out to make sure).  Since the id is auto_increment, I do not put that in my sql (though I've tried that too), nor in my params (tried that too).  What am I missing?  I feel certain it is something small, but I have spent days checking commas, semi-colons and spelling.  Can anyone see what I'm doing wrong?

Hello, I have problem durring binding update query. I can't find what is causing problem.

public function Update(Entry $e)
    {
        try
        {
            $query = "update entry set string = $e->string,delimiter=$e->delimiter where entryid= $e->id";
            $stmt = $this->db->mysqli->prepare($query);
            $stmt->bind_param('ssi',$e->string,$e->delimiter,$e->id);

            $stmt->close();
           

        }
        catch(Exception $ex)
        {
            print 'Error: ' .$ex->getMessage();
        }
    }

When I run function update I'm getting next error:Warning: mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement   Can you help me to solve this problem ?
Edited by danchi, 17 October 2014 - 10:25 AM.

I need to display a number(the number is retrieved from the db) in the form input field such that only the last 4 digits is visbile, the remaining can be masked as * or X or whatever is applicable.

I know the last 4 can be obtained as follows:

    Code: [Select]
$number=substr($number,-4,4);
 
But when i hit the submit button the form validates the input field and checks if the number is a valid number of a specific format.
Therefore when I click on the submit button then I should still be able to unmask the masked numbers or do something similar that would help me validate the whole number.

  Code: [Select]
  <input type="text" name="no" value="<?php if(!empty($number)){ echo $number;} ?>">

Hey

I have a string that looks like the following:
Quote

top-php-tutorials-2.html


I have a script that cycles through each page. The 2 in the quote above is the page number.

How can I extract the number between the - and the .html and replace it with another number?

I've tried
Code: [Select]
substr($engine->selectedcaturl, 0,-6).$v.".html"But then I realised this only works for numbers that are 1 digit long

Any input would be appreciated

hirealimo.com.au/code1.php

this works as i want it:
Quote

SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1



but apparelty selecting * is not a good thing????

but if I do this:
Quote

SELECT priceID, price FROM price INNER JOIN vehicle....etc


it works but i lose the info from the vehicle table.

but how do i make this work:
Quote

SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc


so that i am specifiying which colums from which tables to query??

thanks

hello,  i have a start time and an end time (no dates).  the following works great, except for when the start time is at night, and the end time is in the morning.  im thinking if i can do an if the number is negative, add 24, that it will resolve my issue.

any ideas?

Code: [Select]
$tbilled = $tfinish - $tstart;

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>";
}

}

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

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.

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;

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.

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!