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PHP - Checking Row Values Within While Using If Stmt
Hello ,
I am trying to update a record in my table if the particular value exists in the rows of a table. But my if-else not working properly. I am not getting where i am going wrong.
here is my table authorization, where i define authorization % with min and max.
authorisation.JPG 16.83KB
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after adding line items, before display i will check whether any one of the line item discount lies within this min and max and also checks whether authorization is required (required='Yes'/'No') for that discount in my authorization table. If authorization required then i will update that order number as authorized.
here is my line items table
line_items.JPG 20.37KB
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I am doing like this
while($row=mysql_fetch_array($query))
{
$dis1 = "SELECT auth_id, auth_min, auth_max, required FROM sales_authorisation";
$dis2 = mysql_query($dis1) or die (mysql_error());
while($d1 = mysql_fetch_array($dis2))
{
$min = $d1['auth_min'];
$max = $d1['auth_max'];
$req = $d1['required'];
//echo $req;
if( ($min <= ($row['discount'])) && ($max >= ($row['discount'])) && ($req='Yes'))
{
$auth = "UPDATE orders SET authorise='No' WHERE order_id=".$order_id."";
echo "hello";
}
else
{
$auth = "UPDATE orders SET authorise='Yes' WHERE order_id=".$order_id."";
}
$auth1 = mysql_query($auth) or die (mysql_error());
}
?>
<tr>
<td><?php echo $counter++; ?></td>
<td><?php echo $row['itemname']; ?> - <?php echo $row['uom']; ?></td>
<td><?php echo $row['description']; ?></td>
<td><?php echo $row['quantity']; ?></td>
<td><?php echo number_format($row['selling_price'],2); ?></td>
<td><?php echo $row['discount']; ?> %</td>
<td><?php echo $row['tname']; ?>-<?php echo $row['rate']; ?> %</td>
<td><?php echo $row['freight']; ?></td>
<td><?php echo number_format($row['total'],2); ?></td>
</tr>
While loop is for displaying line items for the order.
But my if condition doesn't works . Not getting how to do it. please suggest
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I am trying to check whether the values in rows are equal or same?
I have two tables orders and line_items. I will store more than 1 line items in line_items table with the same order_id getting from orders table. In line items if i have 3 line items for same order_id, when each line items gets dispatched i will set the status to 'Dispatched' in line_items table. When all the 3 becomes 'Dispatched', i will update the order status as 'Completed' in orders table.
Now i am not getting how to check whether all the 3 line items status id 'Dispatched'.
I tried doing like this
$q1 = "select orders.order_id, orders.item_status as istatus, orders.status as ostatus, orders.authorise, line_items.order_id, line_items.status as bstatus, COUNT(line_items.id) as bid from orders inner join line_items on orders.order_id=line_items.order_id where line_items.order_id=".$id." AND line_items.id<>'Dispatched'";
$q2 = mysql_query($q1);
while($q3 = mysql_fetch_array($q2))
{
echo $q3['bstatus'];
if($q3['bstatus']=='');
{
echo "HELLLOOOO";
}
}
But its not working. It goes into if loop even if a single value is 'Dispatched'
Please help me.
I read ages ago (and checked to see if it's true, it was and given how it works, it must still be) the end user can alter the value of any form field, using Firebug or similar, before submitting it. Two things I've figured out today: 1) a form input doesn't need a value - doesn't even need the attribute - if you're only checking whether the POST var isset and the actual value isn't important 2) Although it appears not to matter in the example I'm working on now, if the script doesn't check what the value is, and potentially sanitise it, the user could submit the form with any value, true, false, malicious, idk... So my question is: is this one of the ways malicious bad things can happen and do I *have to* specify a value, not because the script won't work without it, it does, but because in the real world it opens a security door if I don't check for malicious script by saying "if value not as expected, script has to die". Having formulated the question properly and thought about it I can't imagine simply making a form, without obvious connections to anything important, could be a problem in the way I'm asking about but I asked it now so Edited by appobs, 03 July 2014 - 12:08 PM. Php Whizzs! Here is my login.php partial relevant to the case code:
$query_1 = "SELECT id,recruits_number,sponsor_username,account_activation_status,id_video_verification_status,id_verification_video_file_url,username,password,primary_domain,primary_website_email,registering_country,registering_ip,registering_browser,registering_os,registering_isp,age_range FROM users WHERE $querying_column = ?";
$stmt_1 = mysqli_prepare($conn,$query_1);
mysqli_stmt_bind_param($stmt_1,'s',$login_username_or_email_or_domain);
mysqli_stmt_execute($stmt_1);
//Check if User's details was successfully extracted or not from 'users' tbl.
if (!$stmt_1)
{
echo "ERROR 1: Sorry! Our system is currently experiencing a problem logging you in!";
exit();
}
else
{
$result_1 = mysqli_stmt_bind_result($stmt_1,$db_id,$db_recruits_number,$db_sponsor_username,$db_account_activation_status,$db_id_video_verification_status,$db_id_verification_video_file_url,$db_username,$db_password,$db_primary_domain,$db_website_email,$db_registering_country,$registering_ip,$registering_browser,$registering_os,$registering_isp,$db_age_range);
mysqli_stmt_fetch($stmt_1);
mysqli_stmt_close($stmt_1);
//Free Result_1 Set
mysqli_stmt_free_result($stmt_1);
I can login to user account with accurate password. Good! With wrong password, supposed to get error: "Incorrect log-in details". Instead get this:
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I read that you use "mysqli_stmt_free_result($stmt)" if you use mysqli_stmt_store_result($stmt). Is this causing the error ? Should I remove mysqli_stmt_free_result($stmt) ? Or, instead of this: mysqli_stmt_close($stmt_1); //Free Result_1 Set mysqli_stmt_free_result($stmt_1);
Should I do this: //Free Result_1 Set mysqli_stmt_free_result($stmt_1); mysqli_stmt_close($stmt_1);
Or, maybe I should add another line ? If so, then what and where ? Already checked the manual and stuck! How-about a sample code from your end ? Edited November 28, 2018 by phpsaneHi,
I have two scripts using almost identical code. Is this because PHP 7 needs to use only indexes in the WHERE part of SQL Query? Here are the two scripts - first one works
$sql = 'SELECT access,fname FROM clients WHERE email=?';
$stmt = $connect->prepare($sql);
$stmt->execute([$email]);
$data_exists = ($stmt->fetchColumn() > 0) ? true : false;
if ($data_exists) { // account found.
$row = $stmt->fetch();
$access = $row['access'];
$fname = $row['fname'];
}
else {
$err_msg = 'Invalid Email Address and/or Password.';
$email = '';
$pass = '';
require_once ("login_fm.php");
exit;
}
[/PHP]
$sql = 'SELECT email,fname,lname,confirm FROM clients WHERE user_key=?';
$stmt = $connect->prepare($sql);
$stmt->execute([$the_key]);
$data_exists = ($stmt->fetchColumn() > 0) ? true : false;
if ($data_exists) { // Account found.
$row = $stmt->fetch();
$email = $row['email'];
$name = $row['fname'].' '.$row['lname'];
$confirm = $row['confirm'];
$_SESSION['auth'] = "yes";
$_SESSION['email'] = $email;
$_SESSION['name'] = $name;
$sql = 'UPDATE clients SET confirm = ?,log_count = log_count+1,last_date=? WHERE email=?';
$stmt= $connect->prepare($sql);
$stmt->execute(['y',$today_time,$email]);
require_once("index.php"); // SUCCESSFUL LOGIN: THIS LOADS THE START PAGE
exit;
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else {
$err_page_message = "ERROR - Account not found $the_key : $name";
require_once("err.php");
exit;
} // end else
May be it's a different reason - like I'm too tied !
But it would be nice to know why it isn't working.
It's quite simple, I just want to check to see if a row exists with the condition given, that's all. I never had this problem the old mysql but the PDO seems a bit tricky or I'm just using the wrong code lol.
Thanks Edited June 15, 2019 by David-LondonCode: [Select] $params = array(); $type=array(); $fragments = array(); while (list($chiave,$val) = each($_POST)){ if($val!=""){ $fragments[] = $chiave." = ?"; //echo $chiave; if($chiave=="ritiro" or $chiave=="data") $params[] = normalToDbDate($val); else $params[] = $val; if($chiave=="legale") $type[]='i'; else $type[]='s'; } } $prova=implode("",$type); foreach($params as $param) echo "<p>".$param."</p>"; //echo "val(".count($params).")-->".implode(',',$params); //echo "type(".strlen($prova).")-->".implode("",$type); $sql = $db->prepare("..... AND ".implode(" AND ", $fragments)); array_unshift($params,$prova) call_user_func_array( array( $sql, 'bind_param' ),$params); //$sql->bind_param($prova,$params); $sql->execute(); hi all, with the tructure above i've tried to implement the cration of a dynamic query for a search form. i debugged with echoes and che numbers of bind_param and the those in the prepared statement match in sense that count($params)=strlen($prova). having said that i get this error: Code: [Select] Warning: call_user_func_array() expects parameter 2 to be array, integer given in and no results are shown to me even when the filter match. Hi all !
I have a piece of code he
$result = display_all(fcon, $var1, $var2); function display_all( // defined in another file $query = "SELECT one, two, three four, index1, index2 FROM numbers WHERE index1 = $var1 LIMIT 0, 1"; $result = mysqli_query($fcon, $query); return ($result); )and then I use the returned variable $result to display the value as follows:-
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
echo "<tr>";
echo "<td>".$one."</td>";
echo "<td>".$two."</td>";
echo "<td>".$three. "</td>";
echo "<td>".$four. "</td>";
echo "<td>".$index1. "</td>";
echo "</tr>";
}
and this displays the n rows of data returned.
Now I have started using prepared statements and the function is now
function display_all{
$query = "SELECT one, two, three four, index1 FROM numbers WHERE index1 = ?
LIMIT 0, 10";
$stmt = $conn->prepare($query);
$stmt->bind_param('i',$var)
if($stmt->execute())
{
$stmt->bind_result($one, $two, $three, $four, $index1);
$stmt->store_result();
}
return($stmt);
}
However the returned $stmt object is unable to display the n rows of data since it shows null values. I assume that this is not the right way to use the $stmt object to display data. I must be missing something. So I request you guys to help me with this.
Thanks loads.
Edited by ajoo, 23 December 2014 - 11:24 AM. Dear All Members here is my table data.. (4 Columns/1row in mysql table)
id order_no order_date miles How to split(miles) single column into (state, miles) two columns and output like following 5 columns /4rows in mysql using php code.
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