PHP - Give Image To Html Through Php-doesn't Work

I've written a script where it counts the page impressions on my website.

When a visitor lands on the page, it counts their impression and then sets a $_SESSION so that if they reload or return to the page it doesn't count it again. But the problem is it sometimes does.

Does this mean that the $_SESSION isn't being set probably? That's what it seems to be. And if so, what would be the reason for that?

Here's my basic code:

Code: [Select]
if(isset($_SESSION['userid'])){
     $userid = $_SESSION['userid'];
}
else{
     $userid = rand(1000, 10000000);
     $_SESSION['userid'] = $userid;
}

$date = date('Y-m-d');

$counters = mysql_query("SELECT * FROM counters
WHERE userid='$userid'
AND impressions='1'
AND date='$date'")
or die(mysql_error());

if(!mysql_num_rows($counters)){
     mysql_query("INSERT INTO counters
     (userid, impressions, date)
     VALUES('$userid', '1', current_date)")
     or die(mysql_error());
}

Hi everybody!

I'm creating a site and none of my php-code works.

I tried to do a simple php-code to test it but it doesn't work either.

The simple php-code:
<?php

echo "hello";

?>

I'm trying to implement an email activation method for my registration script. For that I need the last inserted id straight off the query.

This is how the part of my script looks like:


// write into database
$query = sprintf("INSERT INTO user (user_id, firstname, lastname, nickname, password, email, dob, doj, random, activated) VALUES (' ', '%s', '%s', '%s', '%s', '%s', '%s', now(), '$random', '0')",
mysqli_real_escape_string($dbc, $firstname),
mysqli_real_escape_string($dbc, $lastname),
mysqli_real_escape_string($dbc, $nickname),
mysqli_real_escape_string($dbc, $password),
mysqli_real_escape_string($dbc, $user_email),
$dob);

echo $lastid = mysqli_insert_id($dbc) or die (mysqli_error($dbc));


The query itself WORKS, the data gets inserted correctly, the scripts dies right at the die of mysqli_insert_id.

When I take away the "or die", then I always get printed out a 0 when trying to register, THOUGH the data entered into the registration page gets inserted correctly into the database as said.

I've read on w3schools.com, that the connection ($dbc) is OPTIONAL in mysqli_insert_id($dbc), because it automatically takes the last active connection, but when I leave it empty I get an error saying that it needs at least one parameter.

I must be doing something wrong, if you need more of the script let me know.

Since the scripts dies at the die, I didn't post anything below, and everything above is just if statements to check the entered data.

Hello,
I've been working on some stuff, and get the information parsed like this:
Code: [Select]
Value1,Value2,Value3 0.68,4377.54,11069.02So I'm trying to explode it and insert the latest numbers into the database. I'm using:

$value = explode(",", $variable);
$secondsplit = explode(" ", $value[2]);
mysql_query("INSERT INTO table(row) VALUES('".mysql_real_escape_string($secondsplit[1])."')");

The problem is that it has something like an enter in the database, or a space. But I can't figure out how to get rid of that problem. Does anyone know what's wrong?

Regards,
NLCJ

Hello all, I'm trying to clean up some user input for my database but the mysql_real_escape_string function doesn't seem to work, and neither does addslashes(). I've browsed the net and made sure I'm connected to the database before the function is called. I can't figure out what I'm doing wrong. Here's the relevant part of the code:

Code: [Select]
<?php

mysql_connect("localhost","adminname","adminpassword") or die (mysql_error());
mysql_select_db("databasename") or die (mysql_error());

$comment=mysql_real_escape_string($_POST['commentbox']);

?>
...and then the result is put into the database along with some other data.

The database entry is correct, except, when I check the database or echo the result I get the same thing that was entered into 'commentbox'. For instance, if I enter "I am very 'confused' by this" into the box, the same string will be found in the database table. No slashes, just quotes. Anyone have an idea what's wrong with this? Because it all seems simple and correct but doesn't work.

Hey,

$todo_upgrades_sql = "SELECT * FROM todo_upgrades WHERE time_completion >= $time";
$todo_upgrades_res = mysql_query($todo_upgrades_sql) or die (mysql_error()); results:

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 '' at line 1

Hi guys, me again.

I have been using mail to send emails ( With your great help ) successfully in about three websites. For some reason, my email doesn't want to work with the site I am busy with now. This is my code :

Code: [Select]
<html>
      <head>
            <title>Practice</title>
      </head>
      <body>
<table border="1">
       <tr>
           <td>
<?php
      $emailto="hannes@ncc-cla.com";
      $emailsubject="confirmation";
      $emailfrom="footloose.sa@gmail.com;

                  if(!empty($_POST['uname'])){
                  $yourname=$_POST['uname'];
                  $yoursurname=$_POST['usurname'];
                  echo "hello
            </td>
            <td>"
                ." " . "$yourname" . " " . "$yoursurname
            </td>
         </tr>";

         $emailbody="hello
            </td>
            <td>"
                ." " . "$yourname" . " " . "$yoursurname
            </td>
         </tr>";
}
 else{
   echo"<tr>
            <td>
                enter your name
            </td>
        </tr>";

        $emailbody .="<tr>
            <td>
                enter your name
            </td>
        </tr>";
}
   if(isset($_POST['gender'])){      /*isset is what it is set to can be used */
   $gender=$_POST['gender'];
   echo "<tr>
             <td>
                 You are a
             </td>
             <td>
                 " . " " . "$gender
             </td>
   </tr>";
          $emailbody .="<tr>
             <td>
                 You are a
             </td>
             <td>
                 " . " " . "$gender
             </td>
   </tr>";
 }
       else{
             echo "<tr>
                       <td>
                           Select Gender
                       </td>
                    </tr>";
             $emailbody .="<tr>
                       <td>
                           Select Gender
                       </td>
                    </tr>";
       }
 if(isset($_POST['plang'])){
   $plang=$_POST['plang'];
   echo "<tr>
             <td>
                 You are interested in
             </td>
             <td>
                 " . " " . "$plang
             </td>
         </tr>";
         $emailbody .="<tr>
             <td>
                 You are interested in
             </td>
             <td>
                 " . " " . "$plang
             </td>
         </tr>";
 }
   else{
     echo "<tr>
               <td>
                   Select Programming language
               </td>
           </tr>";
           $emailbody .="<tr>
               <td>
                   Select Programming language
               </td>
           </tr>";
   }
 print "<tr>
            <td>
                Your qualification is
            </td>
            <td>";
            $emailbody .= "<tr>
            <td>
                Your qualification is
            </td>
            <td>";
 foreach ($_POST['qlevel'] as $level){
 print "$level<BR>";
 $emailbody .="$level<BR>";
 }
 $headers='From:'. $emailfrom;
                
                 mail($emailto,$emailsubject,$emailbody,$headers);
?>
             </td>
         </tr>
</table>
</body>
</html>

Im using func_get_args, but it gives an error?:

Too few arguments?, when its correct (as the number of %s = the number of values within $args)

<?php
function sprintify() {
global $urls;
$args = func_get_args();
$input = $args[0];
unset($args[0]);
return sprintf($urls[$input], implode(', ', $args));
}

echo sprintify('default', 46, 464, 46);

/*
For your information:
$urls['default'] = '%s/%s/%s';
*/
?>

I have some code which deletes a record from my database (which has been selected from the previous page). Only problem is, it comes up with error code, but still deletes from the database?

Error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /www/110mb.com/t/h/e/b/l/a/c/k/theblackwatch/htdocs/150749deleted.php on line 28

Code:
Code: [Select]
<?php
$id = $_GET[id];
// Start the connection to the database
$conn = mysql_pconnect("***", "***", "***");
// Select the database to use
mysql_select_db("***", $conn);
// Create the MySQL command to retrieve all records
$sql = "DELETE FROM **** WHERE id ='$id'";
$result = mysql_query($sql, $conn);
$array = mysql_fetch_array($result);
// Echo each entry as the value of the text boxes. This brings in the current info in the database
echo "<p>Entry has been deleted.</p>";
?>
Any way round it?

Hey there, I am trying out some tag cloud code, but something is amiss, as am getting no tags displaying. Using PHP/MYSQL and JQuery.

<?php

//connection information
  $host = "localhost";
  $user = "username";
  $password = "pw";
  $database = "DBName";

//make connection
  $server = mysql_connect($host, $user, $password);
  $connection = mysql_select_db($database, $server);

//query the database
  $query = mysql_query("SELECT * FROM tags");
 

//start json object
$json = "({ tags:["; 

//loop through and return results
  for ($x = 0; $x < mysql_num_rows($query); $x++) {
    $row = mysql_fetch_assoc($query);

//continue json object
    $json .= "{tag:'" . $row["tag"] . "',freq:'" . $row["frequency"] . "'}";

//add comma if not last row, closing brackets if is
if ($x < mysql_num_rows($query) -1)
$json .= ",";
else
$json .= "]})";
  }

//return JSON with GET for JSONP callback
$response = $_GET["callback"] . $json;
echo $response;

//close connection
mysql_close($server);

?>




and the JS

Code: [Select]
<script type="text/javascript">
$(function() {

//get tag feed
$.getJSON("/tagcloud.php?callback=?", function(data) {

//create list for tag links
$("<ul>").attr("id", "tagList").appendTo("#tagCloud");

//create tags
$.each(data.tags, function(i, val) {

//create item
var li = $("<li>");

//create link
$("<a>").text(val.tag).attr({title:"See all pages tagged with " + val.tag, href:"http://localhost/tags/" + val.tag + ".html"}).appendTo(li);

//set tag size
li.children().css("fontSize", (val.freq / 10 < 1) ? val.freq / 10 + 1 + "em": (val.freq / 10 > 2) ? "2em" : val.freq / 10 + "em");

//add to list
li.appendTo("#tagList");

});
});
});
</script>

when i turn on error reporting I get a lot of undefined index notices for tag and frequency so not sure if that is causing the problem or not.

Hey, I have made a audio button in dreamweaver and copied the code into my php panel:

<script language="JavaScript" src="http://www.MYWEBSITE.com/myname/audio/audio-player.js"></script>
<object type="application/x-shockwave-flash" data="http://www.MYWEBSITE.com/media_folder/audio/player.swf" id="audioplayer1" height="24" width="290">
<param name="movie" value="http://www.MYWEBSITE.com/media_folder/audio/player.swf">
<param name="FlashVars" value="playerID=audioplayer1&soundFile=http://www.MYWEBSITE.com/media_folder/audio/Reconstructed Ancient Egyptian Melody.mp3">
<param name="quality" value="high">
<param name="menu" value="false">
<param name="wmode" value="transparent">
</object>

However once I press save, all that is saved is:
<script language="JavaScript" src="http://www.MYWEBSITE.com/myname/audio/audio-player.js"></script>
<object type="application/x-shockwave-flash" data="http://www.MYWEBSITE.com/media_folder/audio/player.swf" id="audioplayer1" height="24" width="290">
<param name="movie" value="http://www.MYWEBSITE.com/media_folder/audio/player.swf">

The audio button works however because the code which includes the mp3 file never saves, it can never find the audio file. Can anyone tell me why the audio file part of the code doesn't save?

Thanks.

I've been puzzling over this all morning. No matter what I do i keep getting these errors:

Quote

[01-Oct-2019 21:45:32 UTC] PHP Warning:  Use of undefined constant  PASSWORD_BCRYPT - assumed ' PASSWORD_BCRYPT' (this will throw an Error in a future version of PHP) in /home/ijwe4nfwf9ei/public_html/blog-approval.php on line 7
[01-Oct-2019 21:45:32 UTC] PHP Warning:  password_hash() expects parameter 2 to be integer, string given in /home/ijwe4nfwf9ei/public_html/blog-approval.php on line 7[01-Oct-2019 22:18:20 UTC] PHP Warning:  Use of undefined constant  PASSWORD_BCRYPT - assumed ' PASSWORD_BCRYPT' (this will throw an Error in a future version of PHP) in /home/ijwe4nfwf9ei/public_html/blog-approval.php on line 3
[01-Oct-2019 22:18:20 UTC] PHP Warning:  password_hash() expects parameter 2 to be integer, string given in /home/ijwe4nfwf9ei/public_html/blog-approval.php on j 3[01-Oct-2019 22:21:44 UTC] PHP Warning:  Use of undefined constant  PASSWORD_BCRYPT - assumed ' PASSWORD_BCRYPT' (this will throw an Error in a future version of PHP) in /home/ijwe4nfwf9ei/public_html/tester.php on line 3
[01-Oct-2019 22:21:44 UTC] PHP Warning:  password_hash() expects parameter 2 to be integer, string given in /home/ijwe4nfwf9ei/public_html/tester.php on line 3

Here's my code: 

<?php
echo phpinfo();
$pass=password_hash ("password", PASSWORD_BCRYPT);
?>

I checked and I'm using php 7, so why do I keep getting this error?

I had it print out my php info so I could check..

https://pyro-creations.com/tester.php

I even read the manual and tried the examples and I still get those errors. How can I fix this?

[please use code tags.]

Hi, when I output this, it's same order, I want it to be:

email/message/to/toLastName

please see code:
$testArray=array("toLastName","to","message","email");
$reversedTestArray=rsort($testArray);//rsort returns TRUE on success
//$strTestArrayReversed=implode(",",$testArray);
print_r($testArray);

Any idea what I am doingwrong, it looks fairly straightforward.

Any help much appreciated!

I have something like below

Code: [Select]
$j = explode("-", $ROW['options']);
if(array_search("src_str", $j) == false) echo '_no';      // This doesn't work
Also i tried
Code: [Select]
$j = explode("-", $ROW['options']);
if(!in_array("src_str", $j)) echo '_no';      // This also doesn't work
Is this bug or am i doing something wrong?