PHP - Can't Get This Query To Execute. Always Returns False.

I'm working on a WordPress theme and I'm trying to build in a simple if statement which will check if the user has add his own logo into the images folder if he doesn't then the name of the blog will appear as normal text in place of the graphic logo.

This is how it looks like:


<?php

$logo_dir = get_template_directory_uri() . "/images/logo.png";

if (file_exists($logo_dir)) { ?>

<li><img src="<?php bloginfo ('template_directory'); ?>/images/logo.png" alt="<?php bloginfo('description'); ?>" /></li>

<?php } else { ?>

<li id='blog_name'><a href="<?php bloginfo('url'); ?>"><?php bloginfo('name'); ?><font>*</font></a></li>

<?php } ?>




I've echo'd out $logo_dir, the URL is correct, but still for some reason it's seeing it as if there would no file exist. If I turn it around into !file_exists then the graphic logo WILL show up. So it's always seeing it as non-existent.

Any ideas why this could be or am I using this function in a wrong fashion? As far as I've understood the PHP manual, this is the correct way of using it.

I tried parsing through first

https://www.jadaliyya.com/Details/28167/The-Empire-of-Sexuality-An-Interview-with-Joseph-Massad

then even tried

www.google.com

and on both I get the following returned

 

post_url equals:
www.google.compost_title equals:
testpost_Threadybody equals:
Enter your posts...

URL equals: www.google.com

www.google.com is not a valid URL

Web page redirects after 2 seconds.

And here is the if statement that handles my filter for URLs

if(isset($_POST["submit"]) && !empty($_POST["url"]) && !empty($_POST["Title"])){
$URL = $conn -> real_escape_string($_POST["url"]) ;
$BodyText = $conn -> real_escape_string(nl2br($_POST["ThreadBody"])) ;
echo "<P>URL equals: " ; 
echo $URL ; 
echo "<P>" ; 
//Change to embed for youtube.


if (filter_var($URL, FILTER_VALIDATE_URL)) {
    echo("$URL is a valid URL");
	//Check if url is a youtube url 
	if (strpos($URL,'youtube') !== false) {
	    echo 'Youtube exists.';
	  $URL = preg_replace("/\s*[a-zA-Z\/\/:\.]*youtube.com\/watch\?v=([a-zA-Z0-9\-_]+)([a-zA-Z0-9\/\*\-\_\?\&\;\%\=\.]*)/i","<iframe width=\"420\" height=\"315\" src=\"//www.youtube.com/embed/$1\" frameborder=\"0\" allowfullscreen></iframe>",$URL);

	} else {
	    echo 'Youtube is not included .';
	   $URL = "<ahref ='" . $URL . "</a>" ;  
	}


	$Title = $conn -> real_escape_string($_POST["Title"]) ;
	$User = $_SESSION['username'];
	 
	 
	 $sql = "INSERT INTO Threads (Title, Users, ThreadBody, url)
     VALUES ('$Title', '$User','$BodyText','$URL')";
     if (mysqli_query($conn, $sql)) {
        echo "New record has been added successfully !";
     } else {
        echo "Error: " . $sql . ":-" . mysqli_error($conn);
     }

		} 
else {
    echo("$URL is not a valid URL");
		}

 

 

Code: [Select]
<?php
# This script does get included into the document create_new_tag_from.php




    if(isset($_POST['tag_name_submit'])){
    $tag_name_submit = $_POST['tag_name_submit'];
    }
    
    if(!empty($_POST['tag_name'])){
    $tag_name = strip_tags($_POST['tag_name']);
    }
    
    if(!empty($_POST['tag_description'])){
    $tag_description = strip_tags($_POST['tag_description']);
    }


if(isset($tag_name_submit)){

    
    
    # The Validation of User Entered Data
    # Do validate for solely alphabetic characters with ctype_alpha

    # ctype_alpha($tag_name) && 
    if(ctype_alpha($tag_description)){
        $tag_name = strtolower($tag_name);
        $tag_description = strtolower($tag_description);

        
    # The Insertion Into the Database
    $db_connect = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
    
    # Do research if SQL query has to be modified into a PHP comform query
    
    $sql_query = sprintf(
            
            "set @newid = convert(
                
              ( select 
               max(convert( (substring(tag_id, 2)) , unsigned integer))+1
               from tags), char(10) );

                set @newid = if(length(@newid) = 1, concat('0', @newid), @newid);
                set @newid = concat('c', @newid);

                INSERT INTO tags (tag_id, tag_name, tag_description, added_by_user_id,
                                creation_date, last_edited)
                VALUES (@newid, '%s', '%s', 7, now(), '0')",
            
                mysqli_real_escape_string($db_connect, $tag_name),
                mysqli_real_escape_string($db_connect, $tag_description)
            
            
            );
    
    $sql_query_run = mysqli_multi_query($db_connect, $sql_query);
   
    # Print Test to See If It Works
    echo "works_ ";
    echo $tag_name . "_ ";
    echo $tag_description . "_ ";
                     
    
        } else { # End of the Validation
            echo "The entered data must be in alphabetic characters.";
        }
    
}



?>


Here is the form:

Code: [Select]
<?php
include_once('../../header.php');
include_once('../../model/contribution/create_new_tag_script.php');
?>

<form action="" method="post">
    <input type="text" name="tag_name" value="" />
    <textarea maxlength="60" type="text"  name="tag_description" value=""></textarea>
    <input type="submit" name="tag_name_submit" value="Submit the New Tag" />
</form>


<?php
include_once('../../footer.php');
?>
By the script as given above the conditional check will skip to the error message.

Though I have to say that ctype_alpha does return TRUE when I do try it with "tag_name" which is entered data from the input field, while the other is from the textarea.

I simply cannot see the reason for the returned FALSE.

Hi, I'm pretty new to PHP and I am working on a form that sends and email. I want to display a message if the mail can't be sent (i.e. if mail() returns false) and a different message if mail() returns true, except it is returning false even though the emails are sending.


Basically I have used:

$to = 'myemail@email.com';
$subject = "Subject Line Goes here";
$headers = "From: $email";
$message = 'My message';

$mailSent = mail($to, $subject, $message, $headers);

In testing it I have used the following to determine what is going on:

   if (isset($mailSent)) { echo '$mailSent ='.$mailSent; } - echo's nothing
   if ($mailSent == FALSE) { echo '$mailSent = FALSE'; } - echo's '$mailSent = FALSE'
   if ($mailSent == TRUE) { echo '$mailSent = TRUE'; } - echo's nothing
   if (is_null($mailSent)) { echo '$mailSent = NULL'; } - echo's '$mailSent = NULL'
   echo '$mailSent ='.$mailSent; - echo's '$mailSent ='

Any help would be greatly appreciated!
Thanks!

Hello , this is my first post in this forum , i would like to see the following code and tell me your opinion , not about the security issues not about how good or bad it is. Its just a simple demonstration in order to show you the problem.

<?php

include('config/functions.php');

$mysqli = connectionToDb();

$mysqli->autocommit(FALSE);

$stmt = $mysqli->prepare("INSERT INTO users (first_name,last_name) VALUES ('stelios','stelios2')");
$stmt->execute();
$stmt->close();

$stmt = $mysqli->prepare("INSERT INTO users (first_name,last_name) VALUES ('stelios3','stelios4')");
$stmt->execute();
$stmt->close();

var_dump($mysqli->commit());

$mysqli->close();

This is the code and the var_dump($mysqli->commit()); line prints false ! BUT , the inserts are in the database. If i comment out the commit line , then nothing is in the database.    Can you advise ? Thanks !

in my code below the **data is extracted correctly for the first foreach but doesn't return anything for 2nd foreach though selected one present inside the 2nd foreach** i still get the error Warning: Invalid argument supplied for foreach() for line `foreach ($authorlist as $post)`... however the 2nd foreach returns data correctly as soon as i remove the first foreach loop. 

this is supposed to display the distinct first two characters of a string(billnmbr) in a combobox
ive tried this on phpmyadmin and it returns 08..
which is correct
when i run this on my server, it returns nothing..
there must be something wrong with the way i called the billnmbr
does anybody have an idea on how i could call and display this correctly?
thanks and have a nice day.

<?php
  $mquery = "select distinct(substring(`billnmbr`, 1, 2)) from `tbltest`";
  $mres =mysql_query($mquery) or die(mysql_error());
   while($mrow=mysql_fetch_array($mres)){
$mm = $mrow['billnmbr'];
 echo "<option value=" . $mrow['billnmbr'] . ">";
 echo $mrow['billnmbr'] . "</option>"; 
} 
 echo " </select>";
  ?>

Hi,
Im starting to write a simple (so I thought! ) project at work, I have no problems with the HTML, CSS or even the php mysql setup, but I am struggling with a simple piece of code!

I have a list box, that populates from a table column using the following code:

Code: [Select]
<?php
function select(){
    $query="SELECT appname FROM kpe_apps";
    $result = mysql_query($query);
    echo '<select name="item" onchange="this.form.submit()">';
    while($nt=mysql_fetch_array($result)){
         echo '<option value="' . $nt['id'] . '">' . $nt['appname'] . '</option>';
    }
    echo '</select>';
}
?>

I then called the function on the page I needed. which works fine,
now all I need is whatever is selected in the listbox other columns in the same table relating to the selected item are seen.

I just cant seem to do it, loads of people are using jquery and other code, surely this can be done in php?

any help would really be appreciated..
Many Thanks

MOD EDIT: code tags fixed, linefeeds and indenting added.

Hi.
I have a query that determines if user is online/offline. Works perfectly when page loads. I want to change to auto run query every X seconds.
NOTES:
- The query is on a .php page with multiple other queries.
-IF it is possible to still run the query with-in this page? or make a separate file and include in an iframe? i would like to avoid the iframe and just $output the result of query every X seconds.

Here is my $query code:

      $query = "SELECT `manager_id` FROM #__profiles_xref WHERE profileid = '$profid'";
      $onofflineid=doSelectSql($query,1);
      foreach ($onofflineid as $isonoffline){
      $profmanagerid=$isonoffline->manager_id;
      }
      
      $query = "SELECT `session_id` FROM table_session WHERE userid = '$propmanagerid'";
      $onofflinestatus=doSelectSql($query,1);
      if (count($onofflinestatus)>0)
         {
         $output['PROPSTATUS']='Online';
      } else {
         $output['PROPSTATUS']='Offline';
         }
   

Unable to execute query (SELECT * FROM lb-players) in the database : 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 '-players' at line 1

I don't know why i am getting this error. All i can think of that is taking away lb from "lb-players" when you add a " -  "?

Code: [Select]
[m]<?php
$conn = mysql_connect("23.23.23.23", "unknown", "itsAsecertxD");//ignore this xD
if (!$conn) {
echo "Unable to connect to the database : " . mysql_error();
   exit;
}
if (!mysql_select_db("minecraft-rc")) {
   echo "Unable to select database mydbname : " . mysql_error();
   exit;
}
$sql = 'SELECT * FROM lb-players';
$result = mysql_query($sql);
if (!$result) {
   echo "Unable to execute query ($sql) in the database : " . mysql_error();
   exit;
}
if (mysql_num_rows($result) == 0) {
   echo "No rows found, nothing to display.";
   exit;
}
while ($row = mysql_fetch_assoc($result)) {
   echo $row["playername"];
}

mysql_free_result($result);

?>[/m]

tell me if i posted this in the wrong place its been forever since i last been on this site :$

Hello.  This one is driving me up the wall.  It used to work,

I have a sports league database (MsSql) and one of the tables contains team Captain's first name, last name, etc..  To create a unique but simple login, the query searches for the captain's first initial of their first name + lastname + league.  However since there is the remote possibility of two similar usernames (i.e. David Smith) if the last name is found, the query adds an integer to the end of the last name.

So the result would look like this:  DSmithM4A

At the moment, my database is completely blank.  However what is happening when I run the query below is I get: DSmith1M4A

Can someone figure out why this code is not working correctly?  If there is not more than one "Smith" no integer is supposed to be returned.  There is something in the "Captain Count" that I think is not working correctly.

Code: [Select]
    /*
     * Gather captian information
     */
    $strCaptainFirstInitial = substr($arrPost['cpt_first'],0,1);
    $strCaptainLastName = preg_replace("/[^a-zA-Z]/","",$arrPost['cpt_last']);
    $strCaptainName = strtolower($strCaptainFirstInitial.$strCaptainLastName);
    $sqlCaptainSearch = sprintf
    (
        $strCaptainSearch,
        $strCaptainName
    );
    $resCaptainSearch = mssql_query($sqlCaptainSearch,$conDB);
 
    /*
     * Create unique captain login and Password
     */
    $strCaptainCount = (mssql_num_rows($resCaptainSearch) > 0)
        ? mssql_num_rows($resCaptainSearch)
        : "" ;
    $strLeagueTypeInitial = substr($arrLeagueSearch['type'],0,1);
    $strCaptainLogin =
        $strCaptainName .
        $strCaptainCount .
        $strLeagueTypeInitial .
        $arrLeagueSearch['size'] .
        $arrLeagueSearch['division'];

I have this piece of code:
Code: [Select]
if (isset($_POST['type1'])) {
mysql_query("UPDATE tabke SET status = 1 WHERE id = $rid", $c2) or die(mysql_error());
}
if (isset($_POST['type2'])) {
mysql_query("UPDATE table SET status = 1 WHERE id = $rid", $c2) or die(mysql_error());
}
if (isset($_POST['type3'])) {
mysql_query("UPDATE table SET status = 1 WHERE id = $rid", $c2) or die(mysql_error());
}
?>
<br />
<form method="post" action="">
<input type="hidden" name="pageid" value="plyrmgmt">
<input type="hidden" name="action" value="changeBan">
<input type="hidden" name="uid" value="<?php echo $hr_uid; ?>">
<input type="hidden" name="repid" value="<?php echo $rid; ?>">
<input type="submit" name="type1" value="Choice1" onClick="this.form.submit()">
</form>
<form method="post" action="">
<input type="hidden" name="pageid" value="plyrmgmt">
<input type="hidden" name="action" value="changeMute">
<input type="hidden" name="uid" value="<?php echo $hr_uid; ?>">
<input type="hidden" name="repid" value="<?php echo $rid; ?>">
<input type="submit" name="type2" value="Choice2" onClick="this.form.submit()">
</form>
<form method="post" action="">
<input type="hidden" name="pageid" value="plyrmgmt">
<input type="hidden" name="action" value="changeLock">
<input type="hidden" name="uid" value="<?php echo $hr_uid; ?>">
<input type="hidden" name="repid" value="<?php echo $rid; ?>">
<input type="submit" name="type3" value="Choice3" onClick="this.form.submit()">
</form>

Now, for some reason when I click on Choice1, the query doesn't execute.
Also, when I click on Choice2 or Choice3, the URL in my browser doesn't change for some odd reason...it stays the same as it was prior to clicking the Submit Button.

Can anyone point out some errors I have?

Thanks,
Mark.

soo i get info from a form and i want to search for it in my db .. if i find user_name AND user_password the $result returns true .. otherwise false ..

<?php require_once("../includes/connection.php"); ?>
<?php
$user_name=$_POST["user_name"];
$user_password=$_POST["user_password"];
?>
<?php
$query = "SELECT *
          FROM users
          WHERE  '{$user_name}' = user_name
          AND '{$user_password}'= user_password
          ";
$result = mysql_query($query,$connection);
if($result==false){ 
redirect("../index.php");}
elseif($result==true){
redirect("../welcome.php");
}
?>
<?php //close connection!
mysql_close($connection);
?>


but the problem is that it always returns TRUE !!!!
whats the problem here exactly ???

I have a table korisnici in SQLite with INTEGER field aktivan that can have only 0 or 1 value (CHECK constraint).

Field aktivan has value 0 but PHP returns value 1, why? Is this a bug?

This is PHP code that I am running:

 $sql = "SELECT ime, aktivan FROM korisnici WHERE lower(ime) = '" . $ime . "'" . " AND sifra = '" . $_POST["sifra"] . "'";
                
$result = $db->query($sql);
$row = $result->fetchArray(SQLITE3_ASSOC);

$row['aktivan'] = 1 but in table the value is 0.

When I run same query in DB Browser for SQLite I get correct value 0.

Is this a bug?