PHP - Csv Column Heading Php

Hi, I'm trying to get the last number from a column of numbers, add 1 to it and use this to insert a new value in the column.
The column is not the primary key, indexed, or auto insert, it's just a plain old int field.

$newid = mysql_query('SELECT IDCOL FROM prods ORDER BY IDCOL DESC LIMIT 1');

// the above just doesn't work--it returns 0, there are over 1200 products--I'm going nuts over this.

//$newid += 1; I've commented this out, it works though (in that it adds 1 to 0)

   $sql = mysql_query("INSERT INTO prods (IDCOL , Prod_Code, blah... blah..)
        VALUES('$newid','$prod_code','blah... blah...')") or die (mysql_error()); //this works (but inserts 0 as the newID)

You'd think this would be so easy.
Thanks
Steve

I fairly new to PHP. This seems very obvious but I cannot find an answer.

O programmer friend of mine said that I should never store a calculated value in a table because it was redundant. SO I have a very simple shopping cart. In the confirmation email, I ask the database for the line items of the order. So I'm pulling the quantity and price from the database. In the email I created a var called $ext which is $price * $disc which works.  I cannot figure out how to total $ext.

I created a var called $subtotal. Heres the code.

$result = mysql_query("SELECT * FROM order_detail, products
WHERE $ordid = orderid AND
order_detail.productid = products.serial");

while($row = mysql_fetch_array($result))
{
   $prodid = $row['productid'];
   $qty = $row['quantity'];
   $price = $row['price'];
   $prodname = $row['name'];
   $proddesc = $row['description'];
   $ext = $qty * $price;
   $subtotal = 0;
   $subtotal = ($subtotal + $ext);
}

Any help would be appreciated. 

hi guys,
i made this code list US states in six columns....the problem is that it lists it across rather than down

it lists it like thi

Alabama     Alaska     Arizona    blah1    blah2
blah3            blah4       blah5      blah6

when I want it to list like this

Alabama   blah2 
Alaska      blah3
Arizona    blah 4
blah1       blah5

but not just two columns....I want 4 columns

here is my code


$query = mysql_query("SELECT * FROM states ORDER BY name ASC");
echo '<center><table width="250" border="0" cellpadding="5" cellspacing="0">
<tr>';

while($show = mysql_fetch_array($query))
{

$column = 1+$column;

if($column == 6)
{
$column = '1';
$break = '</td></tr>';
}
else
{
$break = '';
}
echo '<td>
<table >
<tr class="style5">
  <td width="50%" align="left" height="23" nowrap>'.$show['name'].'</td>
  <td height="23" valign="middle" > 
    </td>
</tr>
</table>
'.$break.'';
}


echo '</table></center>';

I want to echo the contents of column 4 for my username only, and it doesn't show anything. I can do if ($memberQuery->fetchColumn(4) < 2)() though.

Code: [Select]
$host = 'localhost';
$database = '';
$db_username = '';
$db_password = '';

if (($db = new PDO("mysql:host=$host;dbname=$database", $db_username, $db_password)) == NULL) {
echo $db->getMessage();
}

$memberQuery = $db->prepare("SELECT * FROM members WHERE username=:username");
$memberQuery->execute(array(':username' => $_SESSION['username']));
$result = $memberQuery->fetchColumn(4);
echo "$result";

How can i get the last row data from a certain column?

Hey,

I've got a query in which a variable is interpreted as a column and I don't why this is caused.


$upgrade_time_sql = "SELECT * FROM todo_upgrades WHERE profile_id = ".$profile_id." AND level = ".$profile_data['level2_'.$show.'']." AND type = ".$show."";
$upgrade_time_res = mysql_query($upgrade_time_sql) or die (mysql_error());


$show is filled with the content "storage"

The mysql_error is

"Unknown column 'storage' in 'where clause'"

Thanks for helping.

Hi

Php Buddies,

Line 67:

$query = "SELECT page_url,link_anchor_text,page_description,keyphrases,keywords FROM links WHERE $_SESSION['search_column'] = ?";

I get error for the above:

Parse error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting '-' or identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) in C:\xampp\htdocs\test\select.php on line 67

 

How to fix this ?

 

Context:

	<?php
//include('error_reporting.php');
ini_set('error_reporting','E_ALL');//Same as: error_reporting(E_ALL);
ini_set('display_errors','1');
ini_set('display_startup_errors','1');
require('conn.php');
?>
	<form name = "search" method = "POST" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<label for="keywords">Keywords:*</label>
<input type="text" name="keywords" id="keywords" placeholder="Input Keywords" required>
<br>
<label for="search_column">Search in ... ?</label>
<select name="search_column" id="search_column">
<option value="page_url">Page Url</option>
<option value="link_anchor_text">Link Anchor Text</option>
<option value="page_description">Page Description</option>
<option value="keyphrase">Keyphrase</option>
<option value="keywords">Keywords</option>
</select>
<br>
<label for="tos_agreement">Agree to TOS or not ? *</label>
<select name="tos_agreement" id="tos_agreement" required>
<option value="Yes">Yes</option>
<option value="No">No</option>
</select>
<br>
<input type="button" name="search_links" id="search_links" value="Search Links!">
<br>
<input type="reset">
<br>
</form>
	<?php
	if($_SERVER['REQUEST_METHOD'] === 'POST')
{
    if(ISSET($_POST['search_links']))
    {
        if(ISSET($_POST['page_url']))
        {
            $_SESSION['search_column'] = $_POST['page_url'];
        }
        elseif(ISSET($_POST['link_anchor_text']))
        {
            $_SESSION['search_column'] = $_POST['link_anchor_text'];
        }
        elseif(ISSET($_POST['page_description']))
        {
            $_SESSION['search_column'] = $_POST['page_description'];
        }
        elseif(ISSET($_POST['keyphrase']))
        {
            $_SESSION['search_column'] = $_POST['keyphrase'];
        }
        elseif(ISSET($_POST['keywords']))
        {
            $_SESSION['search_column'] = $_POST['keywords'];
        }
        
        //Re-write the following 4 lines ...
        mysqli_report(MYSQLI_REPORT_ALL|MYSQLI_REPORT_STRICT);
        mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
       $conn->set_charset('utf8mb4'); //Always set Charset.
	        $query = "SELECT page_url,link_anchor_text,page_description,keyphrases,keywords FROM links WHERE $_SESSION['search_column'] = ?";
	        $stmt = mysqli_stmt_init($conn);
	        if(mysqli_stmt_prepare($stmt,$query))
        {
            mysqli_stmt_bind_param($stmt,'s',$_SESSION['search_column']);
            
            $result = mysqli_stmt_bind_result($stmt,$page_url,$link_anchor_text,$page_description,$keyphrase,$keywords);
            mysqli_stmt_execute($stmt);
            
            mysqli_stmt_fetch($stmt);
            
            while(mysqli_stmt_fetch($stmt))
            {
                echo "url"; echo "<br>";
                echo "anchor_text"; echo "<br>";
                echo "description"; echo "<br>";
                echo "keyphrases"; echo "<br>";
                echo "keywords"; echo "<br>";
                echo "|";
                echo "<br>";
            }
            
            mysqli_stmt_close($stmt);
            mysqli_close($conn);
        }
        else
        {
            echo "1. QUERY failed!";
        }
	        if(mysqli_stmt_prepare($stmt,$query))
        {
            mysqli_stmt_bind_param($stmt,'sssss',$_POST['page_url'],$_POST['link_anchor_text'],$_POST['page_description'],$_POST['keyphrases'],$_POST['keywords']);
            mysqli_stmt_execute($stmt);
            
            $result = mysqli_stmt_get_result($stmt);
            
            while($row = mysqli_fetch_array($result,mysqli_assoc))
            {
                $page_url = $row['page_url']; echo $page_url; echo "<br>";
                $link_anchor_text = $row['link_anchor_text']; echo $link_anchor_text; echo "<br>";
                $page_description = $row['page_description']; echo $page_description; echo "<br>";
                $keyphrases = $row['keyphrases']; echo $keyphrases; echo "<br>";
                $keywords = $row['keywords']; echo $keywords; echo "<br>";
                echo "|";
                echo "<br>";
            }
            
            mysqli_stmt_close($stmt);
            mysqli_close($conn);
        }
        else
        {
            die("2. QUERY failed!");
        }
    }
}
	
?>
	

Edited July 16, 2020 by 2020

I have become able to display due data in due column but the same value is showing for each customer - what is the wrong I am doing? Here is the code, please somebody help!

while ($row_custact = mysqli_fetch_assoc($query_custact)){
				$currentuser = $row_custact['cust_id']; 
				$sql_inccur ="SELECT i.inc_date, t.inctype_type, i.inc_amount, i.inc_text, c.cust_no, c.cust_name, i.inc_receipt
                From customer AS c
                LEFT JOIN incomes AS i ON c.cust_id = i.cust_id 
                LEFT JOIN inctype AS t ON i.inctype_id = t.inctype_id
                WHERE c.cust_id = '$currentuser'
                ORDER BY inc_date DESC";

                $query_inccur = mysqli_query($db_link, $sql_inccur);
	            checkSQL($db_link, $query_inccur);      
				
				
					
				while ($row_inccur = mysqli_fetch_assoc($query_inccur)){
					
				$inc_amount = $row_inccur['inc_amount'];
				$inc_text = $row_inccur['inc_text'];	
				
				
				//Iterate over income types and add matching incomes to $total
					 
						$total_row = $total_row + $row_inccur['inc_amount'];  
						$total_paid = $total_paid + $total_row;
				
				// part for total due finding.. and "inc_text" is due column
				        $total_row_due = $total_row_due + $row_inccur['inc_text'];  
				        $total_due = $total_due + $total_row_due;
				
				// this part gathers only total due paied for an account
				if($row_inccur['inctype_type']=='Duepay')
				{
					$total_duepay = $total_duepay + $row_inccur['inc_amount']; 
				}
				$remaining_due = $total_row_due - $total_duepay;	
				//echo $remaining_due; 	
					
				}
				
				
				//echo $currentuser; 
					
				echo '<tr>
								<td>
									<a href="customer.php?cust='.$row_custact['cust_id'].'">'.$row_custact['cust_no'].'</a>
								</td>
								<td>'.$row_custact['cust_name'].'</td>
								<td>'.$row_custact['custsex_name'].'</td>
								<td>
								    '.$remaining_due. '  // showing due left, here is the problem 
								</td>
								
								<td>'.$row_custact['cust_address'].'</td>
								<td>'.$row_custact['cust_phone'].'</td>
								<td>'.date("d.m.Y",$row_custact['cust_since']).'</td>
							</tr>';
				}

 

I have a table and one column has links that point to index.php?id= but I want it to display the info of that row of my database. On the index.php page i have this code Code: [Select]
<?php
include "dbaptsConfig.php";

// test id, you need to replace this with whatever id you want the result from
$id = "1";

// what you want to ask the db
$query = "SELECT * FROM `apartments` WHERE `id` = ".$id;

// actually asking the db
$res = mysql_query($query, $ms);

// recieving the answer from the db (you can only use this line if there is always only one result, otherwise will give error)
$result = mysql_fetch_assoc($res);

// if you uncomment the next line it prints out the whole result as an array (prints out the image as weird characters)
// print_r($result);

// print out specific information (not the whole array)
   echo "id: ".$result['id']."<br />";
echo "username: ".$result['username']."<br />";
   echo "type: ".$result['type']."<br />";
   echo "title: ".$result['title']."<br />";
   echo "description: ".$result['description']."<br />";  
echo "county: ".$result['county']."<br />";
   echo "town: ".$result['town']."<br />";
   echo "phone: ".$result['phone']."<br />";
   echo "rooms: ".$result['rooms']."<br />";
   echo "bath: ".$result['bath']."<br />";
   echo "squa  ".$result['square']."<br />";
echo "rent: ".$result['rent']."<br />";
echo "time: ".$result['time']."<br />";

?>I know there is something wrong with this cause I always get the same info no matter which apartment I click. Do I use the $_GET function and how do I implement this?

Hello, 
I think this question is a bit of both php and mysql related. I hope I can get some help here. Especially with the mysql part. 

 

I have a table like this:

CREATE TABLE days (
    id INT AUTO_INCREMENT,
      daya INT,
      dayb INT,
    PRIMARY KEY (id)
);


INSERT INTO days VALUES
    (70, 19, 18),
    (69, 20, 17),
    (68, 24, 25),
    (67, 19, 18),   
    (66, 19, 17),    
    (65, 17, 16);

and mysql query and php snippet is this:

 $sql = "SELECT
    d1.id AS id1,
    d2.id AS id2,
    IF((d1.daya-d2.daya)=1,1,0) +
    IF((d1.daya-d2.dayb)=1,1,0) +
    IF((d1.dayb-d2.daya)=1,1,0) +
    IF((d1.dayb-d2.dayb)=1,1,0)   AS tcount
FROM days AS d1 , days AS d2
WHERE d1.id IS NOT NULL AND d2.id IS NOT NULL AND d1.id > d2.id";

  $result = $conn->query($sql);
  if ($result->num_rows > 0) {

   while($row = $result->fetch_assoc()) 
   {
        echo "<tr><td>" . $row["id1"] . "</td><td>". $row["id2"] . "</td><td>" . $row["tcount"]. 
                        "</td></tr>";   
             
    }
echo "</table>";

which gives me a table like this:

Id1  Id2   Count
50    49    1
50    48    0
49    48    0
50    47    1
49    47    1
48    47    0
50    46    1
49    46    1
48    46    0
47    46    1
50    45    1
49    45    1
48    45    0
47    45    1
46    45    1

I need to convert the table into this form: id1 goes down the first column, and id2 is in the first row.  How can i do this, please? can you give me some help? thank you!

id2    49    48    47    46    45
id1                    
50     1      0     1     1     1
49            0     1     1     1
48                  0     0     0
47                        1     1
46                              1
45                    

 

I have an SQL query:
SELECT username, dl_sumtotal
FROM `vb_user`
ORDER BY dl_sumtotal DESC
LIMIT 0 , 30
Right now, its results display like this:

username3: 500
username1: 400
username2: 300

How can I give these results a numbered column so it displays like this:


1: username3: 500
2: username1: 400
3: username2: 300

I would like to output that number in my PHP script. thanks for any help.

Php Buddies,

Just how do you insert an uploaded img via html for ($_POST) to mysql tbl and what type of column should it be ?

Also, how to link the img from the mysql column to a webpage ?

Following html links the img from a folder/directory to a webpage:

<a href="../html-link.htm"><img src="flower.jpg" style="width:82px; height:86px" title="White flower" alt="Flower"></a>

I have a table with "votes" column. I basically want to retrieve the sum of all the votes columns associated with a user.

This is the full code. It gives me "NULL" output.

$count_votes = $db->prepare("SELECT SUM(votes) as total FROM entries WHERE user_id = :user_id");
$count_votes->bindParam(':user_id', $user_id);
$count_votes->execute();
$t_votes = $count_votes->fetchColumn();
$get_votes = $t_votes['total'];

echo $get_votes;

 

But if I remove the user_id bind paramter, it gives me the results. But I need that parameter to show which user has how many votes.

$count_votes = $db->prepare("SELECT SUM(votes) as total FROM entries");
$count_votes->execute();
$t_votes = $count_votes->fetchColumn();
$get_votes = $t_votes['total'];

echo $get_votes;

 

What am I doing wrong?

Here is where I'm starting:

Code: [Select]

SELECT * FROM wp_usermeta WHERE meta_key = 'wp_1_s2member_custom_fields' AND meta_value LIKE '%3%'


As the PHP produces the page, I'm going to have 15 different options coming from the Code: [Select]
meta_value LIKE  part.  I'm assuming I can set up my 15 different DIVs, each with their own IF  statement involving the meta_value. 

Right?  I'm not going to have to have 15 different queries, am I?


I assume it would start like this:
Code: [Select]
$query = 'SELECT * FROM wp_usermeta WHERE meta_key = "wp_1_s2member_custom_fields"';
$results = mysql_query($query);
while($line = mysql_fetch_assoc($results))

But from there, I don't know if it's possible to deal with partial data with IF statements.