PHP - Can Not Get Entry To Appear On Page (writes To Sql)

Hello, I've tried to get a dynamic table from an external page, and searching for entries in it, so i used a dynamic XLS file using php excel reader. I only exported the file, but i couldn't search for data. Can i get some help please ?

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?

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.

        $insertCount=0;
        foreach($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

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.

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!

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

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"; }
?>

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?

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>";
}

}

	$value0 = $_POST['date1'];
    $value1 = $_POST['date2'];
    $value2 = $_POST['date3'];

$sql = "INSERT INTO Datetable (startdate,enddate, total)
VALUES ('$value0','$value1', '$value2')" ;

$result = mysqli_query($sql);

Hi,

I was wondering is it possible to not insert values in MySQL if form entry is left blank? Right now if I dont enter any values in form for my dates than MySQL entry shows 0000-00-00 but I need it not to show anything.

 

Hi I am getting this error message and don't know why. please help this is the message " Error: Duplicate entry '' for key 1 "


Here is the code

Code: [Select]


<?php

include'../DB-connection.php';

$sql="INSERT INTO Company(CompanyName,Address,Logo,PhoneNumber,ContactPerson)

VALUES('".mysql_real_escape_string($_POST[CompanyName])."',

'".mysql_real_escape_string($_POST[Address])."',

'".mysql_real_escape_string($_POST[Logo])."',

'".mysql_real_escape_string($_POST[PhoneNumber])."',

'".mysql_real_escape_string($_POST[ContactPerson])."')";





if(!mysql_query($sql,$con))

{

die(' Error: '. mysql_error());

}

echo " <center>your compnay info added &nbsp &nbsp &nbsp <a href='User-Login.php'>Login to Post a job</a> </center>" ;

mysql_close($con)

?>

What exactly does the entry in the title mean? I cannot make sense out of it.

I would appreciate if somebody can shed some light in.

The error message occurs when I try to vote with the voting system I created.