PHP - Getting An Empty Output

Hi everyone!  I'm new to PHP and needed from you.  I have the following code that I can't figure out why it doesn't work on the last elseif.   If I comment out the first elseif (line 19-39), it does go through the last elseif. 

Can someone here help?  Thank you!!!

The content of Instructors.txt as follow:
ID#|fName lName|email_address|course1:course2:course3
ID#|fName lName|email_address|course1:course3
ID#|fName lName|email_address|course1:course2:course4
ID#|fName lName|email_address|course2:course3:course5:course6

<?php
  //If we submitted the form
  if(empty($_POST['Location'])) {
    $FH = fopen("Locations.txt", "rb");
    print "<FORM name=\"myForm\" action=\"test4.php\" method=\"POST\">\n";
    print "<P>    Location:\n";
    print "<select name=\"Location\">\n";
    print "<option value=\"\">-- Select --\n";
    while (!feof($FH)){
      $line_of_text = fgets($FH);
      $line = explode('|', $line_of_text);
      print "<option value=$line[1]>$line[1]";
    }
    fclose($FH);

    print "<input type=\"submit\" value=\" Next 1 \">\n";
    exit();
  }
  elseif(empty($_POST['Instructor'])) {

    $location = ($_POST['Location']);
    $FH = fopen("Instructors.txt", "rb");
    print "<FORM name=\"myForm\" action=\"test4.php\" method=\"POST\">\n";
    print "<P>    Location: <input type=text value=$location>\n";
    print "<P>  Instructor: \n";
    print "<select name=\"Instructor\">\n";
    print "<option value=\"\">-- Select --\n";
    while (!feof($FH)){
      $line_of_text = fgets($FH);
      $line = explode('|', $line_of_text);
      if ($line[0] == "$location") {
        print "<option value=\"$line[1]\">$line[1]\n";
      }
    }
    fclose($FH);
    print "<input type=\"submit\" value=\" Next 2 \">\n";
    exit();

  }
  elseif(empty($_POST['Course'])) {
    $location = ($_POST['Location']);
    $instructor = ($_POST['Instructor']);

    $FH = fopen("Instructors.txt", "rb");
    print "<FORM name=\"myForm\" action=\"test4.php\" method=\"POST\">\n";
    print "<P>    Location: <input type=text value=$location>\n";
    print "<P>  Instructor: <input type=text value=\"$instructor\">\n";
    print "<P>      Course: \n";
    print "<select name=\"Course\">\n";
    print "<option value=\"\">-- Select --\n";
    while (!feof($FH)){
      $line_of_text = fgets($FH);
      $line = explode('|', $line_of_text);
      if ($line[1] == "Bert Vandenber") {
        $course = explode(':', $line[3]);
        for ($i=0; $i<count($course); $i++) {
          print "<option value=\"$course[$i]\">$course[$i]\n";
        }
      }
    }
    fclose($FH);
    print "<input type=\"submit\" value=\" Next 3 \">\n";
    exit();
  }
?>

I don't get the logic behind the construct.  I can't ever see myself using this because it doesn't make sense when I code it. 

I understand isset($var) and why I should use it sometimes to avoid undefined messages.

But with empty() it makes more sense to always test the variable directly because if the value is NULL, 0, '', or false then it evaluates as TRUE.  Is there any situation where empty should be used always?

i have made query and i want replace with no data N/A i have tryed with swich but no luck

Code: [Select]
if (($row['name']) = empty)
  {
    echo('n/a');
}

Hi all

I have a pice of code that I want to only show the image if it's empty:

Code: [Select]
<?php if(!empty($node->field_img['und'][0]['filename'])); { ?>

<li><img src="http://mysite.co.uk/sites/default/files/brand-images/<?php print $node->field_img['und'][0]['filename']; ?>" style="max-width:270px; text-align:center;" /></li>

<?php } ?>

However, it shows an image placeholder (Red cross) even when it's empty?

Many thanks for your help

Pete

I have this line in my code

$result2 = mysqli_query($link, 'Select * from categories where cat_loc ='.$nav['id']);

how do I say "if $result2 is empty or false"? (basically if it didn't find any cat_loc = to $nav['id'])

 to echo '</li>';

Hello everybody,   I am new to php, I have a question for you guys. why I've got an empty page.   <?php

Im trying to echo certain code if the array returned is empty.  I know for certain the array is empty and it works when its not empty. here is my code

<?php while ($club = mysql_fetch_assoc($qclubs)) {

if (empty($club)) {echo "<option value='None'>None</option>";}

else{

echo"
                                <option value='{$club['groupID']}'>{$club['group']}</option>";

}}?>


Can someone please point out the error of my ways?

Hi all,

I have this:
Code: [Select]
<?php
session_start();
include('login/functions.php');
$username = $_SESSION['username'];
$check = mysql_query("SELECT email FROM user_profile WHERE username = '$username'");
$a = mysql_fetch_field($check);
if(empty($a))
{
header('Location: complete.php');
}


?>
However if the field is empty, ie no email address for user then nothing happens?

Am going about this the right way?

Cheers

I am using a script to take postcodes and give distances between them and list all postcodes within a radius. Where it should list all the postcodes in the radius it provides and empty array - any clues??

the table is 'geodb'

relevant fields are                 
postcode 
place     
latitude     
longitude

and the code is
Code: [Select]
if(isset($_GET['Search']))
{
    $postcode = strtoupper($_GET['postcode']);
    $radius = $_GET['radius'];
echo $postcode . '<br>';
echo $radius . '<br>';

$conn = mysql_connect(/*credentials removed*/) or die('db connect error: ' . mysql_error());
    mysql_select_db('db379121', $conn) or die('could not select database');
  $sqlstring = "SELECT * FROM geodb WHERE postcode = '".$postcode."'";
    $result = mysql_query($sqlstring);

    $row = mysql_fetch_assoc($result);

++++++++++++++++++++++++++++++++++++++++++
var_dump($row); // seeing whqat is in the array - returns one result (array(9) { ["id"]=> string(1) "1" ["countrycode"]=> string(2) "GB" ["postcode"]=> string(4) "AB10" ["place"]=> string(8) "Aberdeen" ["postaltown"]=> string(8) "Aberdeen" ["region1"]=> string(8) "Scotland" ["region2"]=> string(13) "Aberdeen City" ["latitude"]=> string(12) "57.135813090" ["longitude"]=> string(12) "-2.121224030" })//
++++++++++++++++++++++++++++++++++++++++++
    $lng = $row["longitude"] / 180 * M_PI;
    $lat = $row["latitude"] / 180 * M_PI;
 echo $lat .
 '<br>';
 echo $lng . '<br>';
 ++++++++++++++++++++++++++++++++++++++++++
echo to check that there is a value here - i get lat  0.99720805922458        lng -0.037022343495923
++++++++++++++++++++++++++++++++++++++++++
    mysql_free_result($result);}



 $sqlstring2 = "SELECT DISTINCT geodb.postcode,geodb.place,(6367.41*SQRT(2*(1-cos(RADIANS(geodb.latitude))*cos(".$lat.")*(sin(RADIANS(geodb.longitude))*sin(".$lng.")+cos(RADIANS(geodb.longitude))*cos(".$lng."))-sin(RADIANS(geodb.latitude))* sin(".$lat.")))) AS Distance FROM geodb AS geodb WHERE (6367.41*SQRT(2*(1-cos(RADIANS(geodb.latitude))*cos(".$lat.")*(sin(RADIANS(geodb.longitude))*sin(".$lng.")+cos(RADIANS(geodb.longitude))*cos(".$lng."))-sin(RADIANS(geodb.latitude))*sin(".$lat."))) <= '".$distance."') ORDER BY Distance";

    $result = mysql_query($sqlstring2) or die('query failed: ' . mysql_error());

    $str = "<table width=\"300\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">";
    $str .= "<tr>";
    $str .= "<th>postcode</th>";
    $str .= "<th>city</th>";
    $str .= "<th>distance</th>";
    $str .= "</tr>";

    while ($row = mysql_fetch_array($result, MYSQL_ASSOC))
    {
        $str .= "<tr><td>".$row["postcode"]."</td><td>".$row["place"]."</td><td>".round($row['Distance'])."km</td></tr>";
    }

    $str .= "</table>";
++++++++++++++++++++++++++++++++++++++++++
var_dump($row["place"]); // this produces null
++++++++++++++++++++++++++++++++++++++++++

    mysql_free_result($result);
    mysql_close($conn);
    echo $str;

Hi, please help me for school i need to make an assignment and i need to make an script with regular expressions to get an image from a site named asaphshop.nl. but the only thing happens is that it's searching in the localhost what should be on the site. this is my code:

​ <?php
header('Content-Type: text/html; charset=utf-8');
$url = "http://www.asaphshop.nl/epages/asaphnl.sf/nl_NL/?ObjectPath=/Shops/asaphnl/Products/80203122";
$htmlcode = file_get_contents($url);
$pattern = '#class="noscript">.*(<img.*>).*</div>#isU';
preg_match_all($pattern, $htmlcode, $matches);
//print_r ($matches);
$image = ($matches[0]);
print_r ($image);
?>

 in the attached file is the thing i get Attached Files  Naamloos.png   21.8KB   0 downloads

how would I make it so that the option 'Remove (make blank)' would post an empty value?

Quote

<form action="/new/newrotastaffprocess.php" Method="POST">

<select name="username">
<?php
$result = mysql_query("SELECT * FROM members")
or die(mysql_error());
while($row = mysql_fetch_array( $result )) {

echo "<option value=\"" . htmlentities($row["username"]) . "\">" . htmlentities($row["username"]) . "</option>";
}
echo "<option value='empty'>Remove (make blank)</option>";
?>
</select>
</div>
<div style="position:absolute; top:397px; left:570px;">
<form action="/new/newrotastaffprocess.php" METHOD="POST">
<input type = "hidden" value="<?php echo $wkbeg;?>" name="wkbeg">
<input type = "hidden" value="<?php echo $staffno;?>" name="staffno">
<input type = "submit" value="New!">
</form>

I introduce if condition when a string is empty as
Code: [Select]
if (empty($string)) {
How I can add the condition for a certain value too. If empty or equal to "word1".

I have a number of fields to display on a page, from a mysql query,  some fields are empty, some not. depends on what they choose on the original form.

I am trying to display on the form results on a page formatted using the below:

If the mysql field is empty, display an unchecked check box next to the result. If there is data in the field display a check box ticked.

What can i do to achieve this, can you point me in the right direction?

Results page using echo:


</TD><TD>Options: <input type='checkbox' CHECKED>". $row[17]. " </TD> = is not empty field
</TD><TD>Options: <input type='checkbox' UNCHECKED>". $row[17]. " </TD> = emtpy ,mysql field
</TD><TD>Options: <input type='checkbox' CHECKED>". $row[18]. " </TD> = not empty mysql field from query
</TD><TD>Options: <input type='checkbox' UNCHECKED>". $row[18]. " </TD> = Empty field in mysql


I am trying to show Rows 1 to 16 to display normally, rows 17 to 30 use the above logic.

There are about 30 fields to display like this, the others are required fields and are displayed no matter what.

I'm not sure what i have done wrong here...
all it says is "Query was empty"


<?php
SESSION_START();

?>

//some html

<?php

if($_SESSION['email']){

$mysql = "SELECT id,firstname FROM users WHERE 'id'='".$_SESSION['email']."'";
$res = mysql_query($sql) or die(mysql_error());

if(mysql_num_rows($res) == 0){
session_destroy();
echo"Please <a href=\"/index.html\">Login</a> or <a href=\"/register.php\">Join Us.</a>";
}else{
$row = mysql_fetch_assoc($res);
echo "welcome back, <a href=\"./index.php?act=profile&id=".$row['id']."\">".$row['firstname']."</a>\n";

}
}else {
echo"Please <a href=\"/index.html\">Login</a> or <a href=\"/register.php\">Join Us.</a>";

}
?>

Hello:

I have this bit of code in the works (which does not work):
Code: [Select]
<?php if ( $mySubHeader; != "" ) { echo " 

<h2>
<?php echo $mySubHeader; ?>

</h2>

"; }

What I am trying to do is to show the H2 tag only if it has a value from the database.

Otherwise, I do not want to show it.

What is the correct way to do this?

Thanks.