PHP - If Statement Syntax Is Wrong

I've made a blackmark system, and upon login I want to process all the blackmarks (limit of 10) for the user attempting to login. Like such:

$expires[1] - is basically the time() value whenever the information was inserted into the database
$expires[0] - the amount of seconds the blackmark shall last

//let's retrieve and run an update on their blackmarks
$query_blackmarks = mysql_query("SELECT expires FROM blackmarks WHERE username = '$username' AND status = 0 LIMIT 10");

$x = 0;

//process their blackmarks
while($row = mysql_fetch_assoc($query_blackmarks))
{
//split the expired row
($row['expires'] == 0) ? $expires = 0 : $expires = explode(':', $row['expires']);

if(time()-$expires[1] < $expires[0] || $expires == 0)
{
$x++;
}
}


For some reason, $x only equals one. But it shouldn't as there are 3 blackmarks for that user which haven't expired:



Why isn't $x == 3?

Hi all,

As always thanks in advance. My php skills are still at a beginner level but I'm trying to get there!

What I'm trying to do:
Send data to the Mailchimp api (specifically email, name, birthday)

What it currently does do:
Sends and stores email and name

What I cant get it to do:
Send and store the birthday from $birthday string.

First I load the form data:

//Load form data $day = $_POST['Day']; $month = $_POST['Month']; $year = $_POST['Year']; //Create Birthday in mm/dd format for mailchimp $birthday = $month."/".$day;()

Then I pass it later on to mailchimp:

$mergeVars = array('FNAME' => $parsedName['first'], 'LNAME' => $parsedName['last'], 'BDAY'=>"$birthday");()

EVERYTHING in my script works perfectly except for the birthday bit.

So I tried this instead which DOES work, but doesn't use the form data defeating the whole point!

$mergeVars = array('FNAME' => $parsedName['first'], 'LNAME' => $parsedName['last'], 'BDAY'=>"04/04");()

I thought it may be the $birthday string wasn't in the mm/dd setup. But after echo'ing it everything is fine.

I get the feeling I'm doing something stupid here. And I think it's syntax based around the 'BDAY'=>"$birthday"() part.

Any help will be greatly appreciated! This is driving me INSANE.

Thanks,

So, let me explain:
Code: [Select]
<?php
class MySqlDatabase {
private $connection;

function __construct() {
$this->database_connect();
}

public function database_connect() {
$this->connection = mysqli_connect(DB_SERVER, DB_USERNAME, DB_PASSWORD, DB_NAME); 

if (!$this->connection) {
die("Database connection failed: " . mysqli_error());
}

}

public function database_query($sql) {
$query = mysqli_real_escape_string($sql, $this->connection);
$result = mysqli_query($query, $this->connection);
if(!$result) {
die("Database query failed: " . mysqli_error());
}
return $result;
}



public function database_close() {
if (isset($this->connection)) {
mysqli_close($this->connection);
unset($this->connection);
}

}
}
?>
Take a look at the database_query() method.

When I insert this into it from my index.php to test if it works:
Code: [Select]
<?php
$database = new MySqlDatabase();

$sql = "INSERT INTO `website`.`users` (`id`, `username`, `first_name`, `last_name`, `password`, `email`, `secret_question`, `secret_answer`, `create_time`) VALUES (NULL, 'joe', 'joe', 'doe', 'password123', 'myemail@email.com', 'Who am I?', 'myself', '2010-10-16 13:37:59');";

$database->database_query($sql);
?>Obviously the date and the password needs some working on (hashing and entering current time), but I get this error:

Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, string given in ...includes\classes.php on line 32

Warning: mysqli_query() expects parameter 1 to be mysqli, null given in ..includes\classes.php on line 33

Warning: mysqli_error() expects exactly 1 parameter, 0 given in ..includes\classes.php on line 35
Database query failed:

I I'm giving it a mysql syntax, but it doesn't work, also I get no error message.

Simple one, I think. I just want to know if this is how I would do this code. on page one a javascript toggles between two divs depending on the option made in a select box. Div one is a radio button valued "dc resident" and the other is a text box where they enter some text. Then on page two, whichever one is entered becomes the variable to insert to my database.

Code: [Select]

if (isset($_POST['how1']){$how1 = mysql_escape_string($_POST['how1']);}
elseif  (isset($_POST['resident']){$how1 = mysql_escape_string($_POST['resident']); };


I think it's right, but something in my gut tells me its not, or that maybe there's a simpler way to do this?

I just had a curiosity that was bugging me and I'm unsure of how to google for the answer, so I thought I'd put it up to the test here.

I learned that when writing if statements in php the syntax is as follows:

if(condition){
    //Execute Code...
}else{
    //Execute Other Code...
}


However, I also see it written like so:

if(condition)
{
    //Execute Code...
}
else
{
    //Execute Other Code...
}


I thought that the later style was used for Javascript and the earlier for php. While they both work, I was wondering, which is more correct?

Thanks Everybody!

E

It's been awhile since I coded php so I'm trying to check what I'm doing wrong...



session_start(); 
$_SESSION['utilisateur']="Invite";

if($_SESSION['utilisateur'] == "Invite") { 

$dlogin = file_get_contents('display_login.html');

} else {

$dlogin = file_get_contents('display_usermenu.php');

}

echo($dlogin);

Notice: Undefined variable: _SESSION in C:\Program Files (x86)\EasyPHP-5.3.4.0\www\index.php on line 21

line 21 being if($_SESSION ...

hmmm, im confused i dont see any syntax errors, do you?

I have a Select statement that relies on two session variables to complete the statement. One variable, $egroup, supplies a column name. I cannot make the SELECT work. It works fine if I substitute the real name of one of the columns  such as "egroup1", or "egroup6", but that's what the variable $egroup is for. Can anyone tell me where I'm wrong?

I guess the answer is in the syntax, but I'm not very good at this and I've been on it for too long. Any help gratefully taken.

Code: [Select]
<?php                 
$query1 = mysql_query("SELECT * FROM topics WHERE managerId='".$managerId."' AND "$egroup"= 1 ORDER BY title ASC");
?>

I can't figure out how to write the a href part to make it blend in with the existing img syntax.  How do I alter this?

Code: [Select]
<?php

echo '<a href=\"{$r['feedusername']\"><img src="', getUserAvatar($r['feedusername']), "\" class=\"avatar mediumsmall newspadding f_left  mrl\" title=\"${r['associate_name']}\" alt=\"${r['associate_name']}\" /></a>";

?>

What's the correct syntax for having a variable in a select statement?

Here's an example of what I'm trying to do (after I'm already connected to the database).

Code: [Select]
$username = "thomas";
$query = mysql_query( "SELECT * from users WHERE username = $thomas" );
if( mysql_num_rows( $query ) > 0 )
{
bla bla...
It works if I don't put the WHERE part in, but I get an error if I use it, so I'm assuming I have the wrong syntax for using a variable in the select statement.

Hi, Im just having some trouble with this...maybe a fresh pair of eyes can help? 

Im getting a "Warning: mysqli_stmt::bind_param() [mysqli-stmt.bind-param]: Number of elements in type definition string doesn't match number of bind variables" error when I try run this:

Code: [Select]
$date = date("Y-m-d");
        $header = $_POST['header'];
        $summary = $_POST['summary'];
        $content = $_POST['content'];
 
        $query = "INSERT INTO articles (pubdate, title, summary, content) VALUES(?, ?, ?, ?)";
        $stmt = $mysqli->stmt_init();
       
        if ($stmt->prepare($query)){
           
            $stmt->bind_param('i,s,s,s', $date, $header, $summary, $content);
 
            $stmt->execute();
            $stmt->close();
        }
        else {
            echo "ERROR: SQL statement failure!";
            echo "<a href='addnews.php'> -> OK</a>";
        }
        $mysqli->close();

It looks fine to me, just can't see whats wrong lol!

(Please ignore HTML formatting)

Code: [Select]
if (is_null($RESULT['PBS_ACCOUNT_ID']))    {

echo "<font size=2 color=#000000 face=arial>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<</font>";
echo "<font size=2 color=#000000 face=arial>PBS_ACCOUNT_ID</font>";
echo "<font size=2 color=#000000 face=arial>></font>";
echo "<font size=2 color=#000000 face=arial>HELLO</font>";


echo "<font size=2 color=#000000 face=arial><</font>";
echo "<font size=2 color=#000000 face=arial>/PBS_ACCOUNT_ID</font>";
echo "<font size=2 color=#000000 face=arial>></font>";


}

else

{

echo "<font size=2 color=#000000 face=arial>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<</font>";
echo "<font size=2 color=#000000 face=arial>PBS_ACCOUNT_ID</font>";
echo "<font size=2 color=#000000 face=arial>></font>";

echo "<B>";
echo $RESULT['PBS_ACCOUNT_ID'];
echo "</B>";

echo "<font size=2 color=#000000 face=arial><</font>";
echo "<font size=2 color=#000000 face=arial>/PBS_ACCOUNT_ID</font>";
echo "<font size=2 color=#000000 face=arial>></font>";
echo "<BR>";

}
If the PBS_ACCOUNT_ID field is empty, it just shows blank. Why doesn't it echo "HELLO?" Or should I be using "IS_EMPTY?"

I have been looking at this code most of the morning and do not have a clue what is wrong with the code. I am hoping its not a stupid mistake, can someone please help me out? thank you   

 
<title>Inputing Travel Detials</title>
 
<header>
  <h1 align="center"> Adding Travel Detials </h1>
  
 
  <body>
    <p>  <center><img  src="cyberwarfareimage1.png" alt="Squadron logo" style="width:200px;height:200px" style="middle"></center>
        <table border="1">
          <tr>
            <td><a href="index.php"> Home Page </a></td>
            <td><a href="administratorhomepage.html">Administrator Home Page </a></td>
            <td><a href="viewhomepage.html">View Home Page </a></td>
            <td><a href="Inputhomepage.html">Input Home Page </a></td>
            <td><a href="traveldetials.html">Enter More Travel Detials </a></td>
        </table>
    </p>
 
 
<?php
 
include "connection.php";
 
$Applicant_ID = $_POST["Applicant_ID"];
$Method_Of_Travel = $_POST["Method_Of_Travel"];
$Cost = $_POST["Cost"];
$ETA = $_POST["ETA"];
$Main_Gate_Advised = $_POST["Main_Gate_Advised"];
 
$query = ("UPDATE `int_board_applicant` SET `Method_Of_Travel`=`$Method_Of_Travel', `Cost`=`$Cost', `ETA`='$ETA', `Main_Gate_Advised`='$Main_Gate_Advised' WHERE `Applicant_ID`='$Applicant_ID'");
   
 
$result = mysqli_query($dbhandle, $query) 
    or die(mysqli_error($dbhandle));
 
if($result){ echo "Success!"; }
 
  else{ echo "Error."; }
 
 
// successfully insert data into database, displays message "Successful".
 if($query){
  echo "Successful";
}
 
  else {
  echo "Data not Submitted";
}
 
 
//closing the connection
   mysqli_close($dbhandle)
   
 
 
 
 
?>

Ok this is puzzleing. I am geting "Could not delete data: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1". but its is deleting the entry that needs to be removed. The "1" is the entry. Just not sure what is causing the error. I do have another delete php but I have put that on the back burning for the time being.     <?php

         $con = mysqli_connect("localhost","user","password","part_inventory");
    // Check connection
if (mysqli_connect_errno()) {
  printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
} else {
    
    $result = mysqli_query($con, "SELECT * FROM amp20 ");
    
    $amp20ptid = $_POST['amp20ptid'];                // escape variables for security
    $amp20ptid = mysqli_real_escape_string($con, $_POST['amp20ptid']);
    
    mysqli_query($con, "DELETE FROM amp20 WHERE amp20ptid = '$amp20ptid'");
      
    
if (!mysqli_query($con, $amp20ptid));

{
 die('Could not delete data: ' . mysqli_error($con));    
}
 echo "Part has been deleted to the database!!!\n";
    mysqli_close($con);
}
?>

Hi guys   I have this code below and all works fine when submitting this online application apart from when someone types either ' # & into one of the comment fields in which it throws up the error. Have tried various fixes from across the internet but no joy. Can anyone offer suggestions?   <?php $con = mysql_connect("localhost:3306","root","password"); if (!$con)   {   die('Could not connect: ' . mysql_error());   } mysql_select_db('sfapp', $con); $sql="INSERT INTO 'sfapp' ('surname_add','forename_add','dob_add','hometele_add','mobiletele_add','homeadd_add','siblings_add','schoolname_add','headname_add','schooladd_add','schooltele_add','schoolem_add','alevel_add','personstate_add','nameprovided_add','pe_add','se_add','PredGrade_Art','PredGrade_AScience','PredGrade_BusStudies','PredGrade_Electronics','PredGrade_EnglishLang','PredGrade_EnglishLit','PredGrade_French','PredGrade_German','PredGrade_Geog','PredGrade_Graphics','PredGrade_History','PredGrade_Maths','PredGrade_SepScience','PredGrade_ProductDesign','PredGrade_Spanish','PredGrade_Other','Gender_Male','Gender_Female','Sub_EnglishLit','Sub_Maths','Sub_FurtherMaths','Sub_Biology','Sub_Chemistry','Sub_Physics','Sub_French','Sub_German','Sub_Spanish','Sub_Geography','Sub_History','Sub_RE','Sub_FineArt','Sub_Business','Sub_Computing','Sub_GlobPersp','Sub_DramaAndTheatre','Sub_PE','Sub_Dance','Sub_Politics','Sub_Psychology','Sub_Sociology','readprospect_chk','Sib_Yes','Sib_No','Current_Student_Yes','Current_Student_No','I_Understand_chk','Current_Education_chk','Local_Care_chk','Staff_Cwhls_chk','Sub_Film')   VALUES ('$_POST[surname_add]','$_POST[forename_add]','$_POST[dob_add]','$_POST[hometele_add]','$_POST[mobiletele_add]','$_POST[homeadd_add]','$_POST[siblings_add]','$_POST[schoolname_add]','$_POST[headname_add]','$_POST[schooladd_add]','$_POST[schooltele_add]','$_POST[schoolem_add]','$_POST[alevel_add]','$_POST[personstate_add]','$_POST[nameprovided_add]','$_POST[pe_add]','$_POST[se_add]','$_POST[PredGrade_Art]','$_POST[PredGrade_AScience]','$_POST[PredGrade_BusStudies]','$_POST[PredGrade_Electronics]','$_POST[PredGrade_EnglishLang]','$_POST[PredGrade_EnglishLit]','$_POST[PredGrade_French]','$_POST[PredGrade_German]','$_POST[PredGrade_Geog]','$_POST[PredGrade_Graphics]','$_POST[PredGrade_History]','$_POST[PredGrade_Maths]','$_POST[PredGrade_SepScience]','$_POST[PredGrade_ProductDesign]','$_POST[PredGrade_Spanish]','$_POST[PredGrade_Other]','$_POST[Gender_Male]','$_POST[Gender_Female]','$_POST[Sub_EnglishLit]','$_POST[Sub_Maths]','$_POST[Sub_FurtherMaths]','$_POST[Sub_Biology]','$_POST[Sub_Chemistry]','$_POST[Sub_Physics]','$_POST[Sub_French]','$_POST[Sub_German]','$_POST[Sub_Spanish]','$_POST[Sub_Geography]','$_POST[Sub_History]','$_POST[Sub_RE]','$_POST[Sub_FineArt]','$_POST[Sub_Business]','$_POST[Sub_Computing]','$_POST[Sub_GlobPersp]','$_POST[Sub_DramaAndTheatre]','$_POST[Sub_PE]','$_POST[Sub_Dance]','$_POST[Sub_Politics]','$_POST[Sub_Psychology]','$_POST[Sub_Sociology]','$_POST[readprospect_chk]','$_POST[Sib_Yes]','$_POST[Sib_No]','$_POST[Current_Student_Yes]','$_POST[Current_Student_No]','$_POST[I_Understand_chk]','$_POST[Current_Education_chk]','$_POST[Local_Care_chk]','$_POST[Staff_Cwhls_chk]','$_POST[Sub_Film]')"; if (!mysql_query($sql,$con))   {   die('Error: ' . mysql_error());   }   ?> <?php //if "email" variable is filled out, send email   if (isset($_REQUEST['pe_add']))  {     //Email information     $admin_email = $_REQUEST['pe_add'];   $forename = $_REQUEST['forename_add'];   $email = "autoreply@testing.com";   $subject = "Application";   $desc =   "Dear $forename   Thank you for submitting your online application, we will be in touch shortly. " ;       //send email   mail($admin_email, "$subject", "$desc", "From:" . $email);     //Email response   echo "Thank you for contacting us!";   }     //if "email" variable is not filled out, display the form   else  { ?> If you are seeing this, you need to go back and fill out the Personal Email section! <?php   }   header("location:complete.php"); mysql_close($con) ?>     Thanks in advance.

Hi..

I create mysql syntax for query testing before i input to my php code

here is my mysql code:

Code: [Select]
set @t = 0;
set @rqty=31968;
SELECT LOT_CODE as code,  DATE_ENTRY,   
CASE WHEN @t+OUTPUT_QTY > @rqty
THEN @rqty  -@t 
ELSE OUTPUT_QTY
END as qty,
@t := @t + d.OUTPUT_QTY as cumulative   
FROM dipping d  WHERE SUBSTR(LOT_CODE, 9,4) = 'P28' AND (@t < @rqty);

and i attach the sample output of the above query.

Now that query test is work i  will input that code to my php codes.


     $sql = "SELECT SKUCode, Materials, Comp, Qty 
     FROM bom  WHERE SKUCode = '$SKUCode'";
     $res = mysql_query($sql, $con);

    ($row = mysql_fetch_assoc($res));
     $Materials = $row['Materials'];
     $Qty = $row['Qty'];
     $Comp = $row['Comp']; //P28
//-----Compute Req Qty and Save to table---//     

     $ReqQty = $Qty * $POReq; // 31968


     
$sql = "UPDATE bom SET ReqQty = '$ReqQty' WHERE SKUCode = '$SKUCode' AND Materials = '$Materials'";
$resReqQty = mysql_query($sql, $con);

$t = 0;

$sql = "SELECT LOT_CODE as code,  DATE_ENTRY,   
CASE WHEN $t+OUTPUT_QTY > $ReqQty 
THEN $ReqQty  -$t  
ELSE OUTPUT_QTY 
END as qty,
$t := $t + d.OUTPUT_QTY as cumulative   
FROM dipping d  WHERE SUBSTR(LOT_CODE, 9,4) = '$Comp' AND ($t < $ReqQty)";



when I echo the query:

I got this:

 SELECT LOT_CODE as code, DATE_ENTRY, CASE WHEN 0+OUTPUT_QTY > 31968 THEN 31968 -0 ELSE OUTPUT_QTY END as qty, 0 := 0 + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = 'P28' AND (0 < 31968)

then I run it to the sql

and I got an error:

Error Code : 1064
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ':= 0 + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = '' at line 1
(0 ms taken)


Any help is highly appreciated

Thank you so much