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PHP - Mysqli_num_rows() Problem

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Mysqli_num_rows() Problem	View Content

Hey guys,

here's the script:

<?php

require("header.php");

$sql = "SELECT entries.*, categories.cat FROM entries, categories

    WHERE entries.cat_id = categories.id

    ORDER by dateposted DESC

    LIMIT 1;";

$result = mysqli_query($db, $sql);

$row = mysqli_fetch_assoc($result);

echo "<p><h2><a href='viewentry.php?id=".$row['id']."'>'".$row['subject']."</a></h2><br /></p>";

echo "<p>"."<i>In <a href='viewcat.php?id=" . $row['cat_id']."'>'" . $row['cat']."</a> - Posted on " . date("D jS F Y g.iA", strtotime($row['dateposted']))."</i></p>";

echo "<p>";

echo nl2br($row['body']);

echo "</p>";

echo "<p>";

$commsql = "SELECT name FROM comments WEHRE blog_id = " .$row['id']."ORDER BY dateposted;";

$commresult = mysqli_query($db, $commsql);

$numrows_comm = mysqli_num_rows($commresult);

if($numrows_comm == 0){

    echo "<p>No comments.</p>";

} else {

    echo "(<b>".$numrows_com."</b>)comments: ";

    $i = 1;

    while($commrow = mysqli_fetch_assoc($commresult)){



        echo "<a href='viewentry.php?id=".$row['id']."#comment".$i."'>'".$commrow['name']."</a>";

        $i++;



    }

}

echo "</p>";

require("footer.php");

?>

The error turns out to be Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\Book\index.php on line 22

why is that?

Full Excel VBA Course - Beginner to Expert

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Mysqli_num_rows() Expects Parameter 1 To Be Mysqli_result, Bool Given

Fatal Error: Uncaught Typeerror: Mysqli_num_rows(): Argument #1 ($result) Must Be Of Type Mysqli_result...on Line 28

Problem Fixed But Caused New Problem

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Learn VBA & Macros for Excel in 1 Week!

Mysqli_num_rows

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I keep getting this error because of my coding and I'm not sure why all I know is it has to do wiht the query itself.

<b>Warning</b>: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in <b>C:\wamp\www\E-Fed Manager (localhost)\processes\template.php</b> on line <b>37</b><br />

$query = "SELECT * FROM `templates` WHERE `templatename` = $templatename";
$result = mysqli_query ( $dbc, $query ); // Run The Query
$rows = mysqli_num_rows($result);

Mysqli_num_rows() Expects

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trying to figure this problem out. sucks being a newbie maybe i will gain enough knowledge to be able to help out others in the near future.
Code: [Select]
<?php
session_start();

$DBConnect = @mysqli_connect("localhost", "**********", "**********")
Or die("<p>Unable to connect to the database server.</p>"
. "<p>Error code " . mysqli_connect_errno()
. ": " . mysqli_connect_error()) . "</p>";
$DBName = "skyward_aviation";
@mysqli_select_db($DBConnect, $DBName)
Or die("<p>Unable to select the database.</p>"
. "<p>Error code " . mysqli_errno($DBConnect)
. ": " . mysqli_error($DBConnect)) . "</p>";

$CustomerName = "";
if (isset($_COOKIE['customerName']))
$CustomerName = $_COOKIE['customerName'];

$TableName = "mileage";
$Mileage = 0;
$SQLstring = "SELECT SUM(mileage) FROM $TableName WHERE
flyerID='{$_SESSION['flyerID']}";
$QueryResult = @mysqli_query($DBConnect, $SQLstring);
if (mysqli_num_rows($QueryResult) > 0) {
$Row = mysqli_fetch_row($QueryResult);
$Mileage = number_format($Row[0], 0) ;
mysqli_free_result($QueryResult);
}

mysqli_close($DBConnect);

?>

i know its a simple stupid error but cant remember nor figure it out. HERES THE ERROR

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\PHP_Projects\Chapter.10\FrequentFlyerClub.php on line 23

Mysqli_num_rows Error

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This is the form:

Code: [Select]
<form id="submit_form" method="post" action="">

<label for="knuffix_category"><h4>Choose a category</h4></label><br />

<?php
        require_once ($sort_category_func);
        $switch = 0;
        sort_category ($switch);
        ?>

<br /><br />

<label for="contribution_description"><h4>Enter your description here</h4></label><br />
<textarea maxlength="150" type="text" name="contribution_description" value=""></textarea><br />

<label for="contribution"><h4>Enter your contribution here</h4></label><br />
<textarea maxlength="300" type="text" name="contribution" value=""></textarea><br />

<input type="submit" name="submit" value="Contribute" />

</form>

<?php
    include($submit_script);
?>

And this is the script:

Code: [Select]
<?php
if (isset($_POST['submit'])){

if (isset($user_name)) {
// Connect to the database
  $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);



// Grab the score data from the POST

//$knuffix_name = strip_tags(trim($_POST['knuffix_name']));

                $contribution_name = uniqid();

                if (!empty($_POST['contribution_description']))
                    $contribution_description = $_POST['contribution_description'];
               // if (!empty($_POST['knuffix_contribution']))
                    $contribution = $_POST['contribution'];
                // if (!empty($_POST['cat'])) {
                    $contribution_category = strip_tags(trim($_POST['cat']));
               // }


                  // Check if the fields and variables are empty

  if (!empty($contribution_category) && !empty($contribution)) {




                      // Check if the submission exists twice in the whole database

                      $query_get = "SELECT contribution FROM con WHERE = '{$contribution}'";

                      $query_run = mysqli_query($dbc, $query_get);

                      $num_rows = mysqli_num_rows($query_run);




                      if ($num_rows == 0) {





                          // Write the data into the database

  $query = sprintf("INSERT INTO con (user_id, name, contribution, category, contribution_date, description) VALUES ('$user_id', '%s', '%s', '%s', now(), '%s')",
mysqli_real_escape_string($dbc, $contribution_name),
mysqli_real_escape_string($dbc, $contribution),
mysqli_real_escape_string($dbc, $contribution_category),
                                        mysqli_real_escape_string($dbc, $contribution_description));


                            mysqli_query($dbc, $query) or die (mysqli_error($dbc));

                            mysqli_close($dbc);

  /* Set a cookie (later) to prevent duplicate submissions */
  // redirect the page to prevent duplicate submissions
// header ('Location: redirect.php');
echo 'Contributing was successful.';





                      } else {


                          echo "This contribution already exists in the database.
                              To keep this place clean we do not allow duplicate submissions.";



                      }



                      } else {

                            echo "Please enter all of the information to add your contribution.";

                            }



                      } else {

                            echo "You have to be signed-in to do a contribution.";

                            }
    }
?>

And this is the error message:

Code: [Select]
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\php_projects\myproject\controller\contribution\submit_script.php on line 38

As the error message states it is not becoming a result, and I do not see where the problem lies, everything looks right to me. Does somebody else see a mistake in there?

Mysqli_num_rows Is Giving Me An Error

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My code looks as follows:

include('connectvars.php');

/* REGISTER FORM */
// check if submit button has been clicked
if (isset($_POST['submit_signup'])) {

// process and assign variables after post submit button has been clicked
$user_email = strip_tags(trim($_POST['email']));
$firstname = strip_tags(trim($_POST['firstname']));
$lastname = strip_tags(trim($_POST['lastname']));
$nickname = strip_tags(trim($_POST['nickname']));
$password = $_POST['password'];
$repassword = $_POST['repassword'];
$dob = $_POST['dob'];
$find_us_question = strip_tags(trim($_POST['find_us_question']));

// connect to database
$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

$check_query = "SELECT * FROM users WHERE nickname = '$nickname'";

$check_connect = mysqli_query($dbc, $check_query);

$check_count = mysqli_num_rows($check_connect);

echo $check_count;

die();

It's a register (sign up) page, and it's the beginning of the script, the rest of the script is just checking if all fields are a empty and if the input is in the allowed character length etc. I could it off at die(); because the rest doesn't matter.

I want the script to check if the username already exists in the database, so I want mysqli_num_rows to tell me how many rows are already there with the same username, and then I want to continue doing an if statement saying

if ($check_count != 0) {
echo "Username already exists!"
}

But the mysqli_num_rows doesn't even print out how many rows there are availible, it gives me an error saying:

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in...

The num_rows function worked in the login script the same way, but for some reason it's not working in the register script.

Any ideas, what I'm doing wrong?

For testing purposes I just want it to print me "1" when I'm entering a username that's already in the database.

Warning: Mysqli_num_rows() Expects Parameter 1 To Be Mysqli_result, Boo

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Hi all,

I have received a warning message, which it still puzzles me. I suspect it might be my inner join command, which I have coded it wrongly?

Line 37 refers to this line - if (mysqli_num_rows($data) == 1) {

Do you guys have any idea? Thanks

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in D:\inetpub\vhosts\123.com\http\viewprofile.php on line 37

Code: [Select]
<?php
  // Connect to the database
  $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

  $query = "SELECT tp.name, tp.nric, tp.gender, tp.race_id, r.race_name AS race" .
        "FROM tutor_profile AS tp " .
        "INNER JOIN race AS r USING (race_id) " .
"WHERE tutor_id = '" . $_GET['tutor_id'] . "'";

  $data = mysqli_query($dbc, $query);

  if (mysqli_num_rows($data) == 1) {
    // The user row was found so display the user data
    $row = mysqli_fetch_array($data);
    echo '<table>';
    if (!empty($row['name'])) {
      echo '<tr><td class="label">Name:</td><td>' . $row['name'] . '</td></tr>';
    }
    if (!empty($row['nric'])) {
      echo '<tr><td class="label">NRIC:</td><td>' . $row['nric'] . '</td></tr>';
    }
    if (!empty($row['last_name'])) {
      echo '<tr><td class="label">Last name:</td><td>' . $row['last_name'] . '</td></tr>';
    }
    if (!empty($row['gender'])) {
      echo '<tr><td class="label">Gender:</td><td>';
      if ($row['gender'] == 'M') {
        echo 'Male';
      }
      if ($row['gender'] == 'F') {
        echo 'Female';
      }
      echo '</td></tr>';
    }
if (!empty($row['race'])) {
      echo '<tr><td class="label">Race:</td><td>' . $row['race'] . '</td></tr>';
    }

    echo '</table>'; //End of Table

    echo '<p>Would you like to <a href="editprofile.php?tutor_id=' . $_GET['tutor_id'] . '">edit your

  } // End of check for a single row of user results
  else {
    echo '<p class="error">There was a problem accessing your profile.</p>';
  }

  mysqli_close($dbc);
?>

Mysqli_num_rows() Expects Parameter 1 To Be Mysqli_result, Bool Given

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I have searched the internet about this and found hundreds have asked the question and not once was it answered in a meaningful way--or maybe I'm just dense.

Would somebody please tell me what is the problem he

Simple, simple form:

<html>
<body>

<form action="send_post.php" method="post">
<p>A: <input type="text" name="A">
<p>B: <input type="text" name="B">
<p><input type="submit">
</form>

</body>
</html>

===========

Simple, simple PHP script:

<?php

$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = 'xyz';
$dbname = 'ABC';
$conn = mysqli_connect($dbhost, $dbuser, $dbpass,$dbname);

if($conn)
echo 'Connected successfully. <br> Here are the results of your query:<br><br>';

$A=$_POST["A"];
$B=$_POST["B"];

$sql = 'SELECT * FROM numbers WHERE A = $A AND B = $B';
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_array($result)) {
echo $row["0"] . ",  ";
}
} else {
echo "0 results";
}
mysqli_close($conn);

?>

Which is 'parameter 1'? and which is the 'bool given'?

But most importantly, how should it be written so that it returns the desired results? I have checked the query from the CMD line, it returns multiple entries.

Really, I have reached FRUSTRATION OVERLOAD!

Edited April 12, 2020 by eljaydee

Fatal Error: Uncaught Typeerror: Mysqli_num_rows(): Argument #1 ($result) Must Be Of Type Mysqli_result...on Line 28

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<!DOCTYPE html>
<html>
<head>
   <title>Registered Users</title>
   <link rel="stylesheet" href="includes/style.css" type="text/css" media="screen"/>
   <link rel="stylesheet" type="text/css" href="https://cdn.datatables.net/1.10.25/css/jquery.dataTables.min.css">
</head>
<body>
   <?php
   include('includes/header.html');
   ?>
   <center>
       <table id="records">
           <thead>
               <tr>
                   <th>ID</th>
                   <th>First Name</th>
                   <th>Last Name</th>
                   <th>Email</th>
               </tr>>
           </thead>
           <tbody>
               <?php
                   require_once('mysqli_connect.php');
                   $query = "SELECT * FROM users";
                   $result = mysqli_query($dbc,$query);

                   if(mysqli_num_rows($result) > 0) -------------------------------------------------Line 28

                   {
                       while($row = mysqli_fetch_assoc($result))
                       {?>
                           <tr>
                               <td><?php echo $row["id"];?></td>
                               <td><?php echo $row["First Name"];?></td>
                               <td><?php echo $row["Last Name"];?></td>
                               <td><?php echo $row["Email"];?></td>
                           </tr>
                           <?php
                       }
                   }
                   mysqli_close($dbc);
               ?>
           </tbody>
       </table>
   </center>
</body>
<?php include('includes/footer.html');?>
</html>

Problem Fixed But Caused New Problem

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Guys thanks for helping me solve the problem i had but now i have another problem and i am lost. i have included the code below for you to have overview.

On the home page the url shows how it should be: http://localhost/board/italian.php?country=Italian
BUT
When you mouse over the NEXT link to go to next page you get this : http://localhost/board/?page=1&country=Italian and then an error 404 when you click on the link
page = 1 no matter if you are mousing over link for page 5 or 2

This is the code:

<table style="width:100%; margin-left:auto; margin-right:auto">
       <?php

           $items_per_page = 5;//how many records/rows to display per page

           $page_url = "http://".$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'];

           $cleaned_current_page_country = htmlspecialchars($_GET["country"]);

           $page_number = (isset($_GET['page']))? filter_var($_GET['page']) : 1;
           //$page_number = filter_var($_GET['page']);
           //$page_number = $_GET['page'];

           $selected_restaurant_type = $_GET['country'];

           // find number of rows in the db
               $query = "SELECT * FROM menu_update WHERE special_of_theday = 1 AND restaurant_type = '".$selected_restaurant_type."' ";
               $query_result = mysqli_query($connection, $query);
               $row_count = mysqli_fetch_array($query_result);
               $total_rows = mysqli_num_rows($query_result);

           $expected_number_of_pages = ceil($total_rows/$items_per_page); //expected number of pages

           $page_position = (($page_number - 1) * $items_per_page); //get starting position to fetch record

           //fetch a group of records using sql LIMIT clause
               $query = "SELECT * FROM menu_update WHERE special_of_theDay = 1 AND restaurant_type = '".$selected_restaurant_type."' LIMIT " .$page_position. ',' .$items_per_page;
               $query_result = mysqli_query($connection, $query);
       ?>

           <table width="100%" border=0>
               <tr bgcolor = "#cccccc">
                   <td>Meal</td>
                   <td>Meal name</td>
                   <td>Meal Price</td>
                   <td>Order Now</td>
                   <td>Full Menu</td>
                   <td>Our Gallery</td>
                   <td>Find Us</td>
                   <td>Our Chef</td>
                   <td>Our Manager</td>
               </tr>
       <?php
               while($row_count = mysqli_fetch_array($query_result)){
                   echo "<tr>";
                       echo"
                           <td style ='width:200px; text-align:center; height:120px; background:#eaedf2'><a href='meal_fullPic.php'><img src= ".$row_count['img']." /></a></td>
                           <td style ='width:119px; text-align:center; height:120px; background:#eaedf2'> ".$row_count['meal_name']."</td>
                           <td style ='width:100px; text-align:center; height:120px; background:#eaedf2'>".$row_count['price']."</td>
                           <td style ='width:100px; text-align:center; height:120px; background:#eaedf2'><a href='order_processor.php'><img src= ".$row_count['order_now']." /></a></td>
                           <td style ='width:100px; text-align:center; height:120px; background:#eaedf2'><a href='full_menu.php'><img src= ".$row_count['full_menu_img']." /></a></td>

                           <td style ='width:170px; text-align:center; height:120px; background:#eaedf2'><a href='restaurant_gallery.php'><img src= ".$row_count['restaurant_pic']." /></a></td>
                           <td style ='width:100px; text-align:center; height:120px; background:#eaedf2'><a href='restaurant_location.php'><img src= ".$row_count['location_pic']." /></a></td>
                           <td style ='width:119px; height:120px; background:#eaedf2'> ".$row_count['chef_name']."</td>
                           <td style ='width:119px; height:120px; background:#eaedf2'> ".$row_count['admin']."</td>
                       ";
                   echo "</tr>";
               }
               mysqli_close($connection);
       ?>
           </table>
       <div class="board_pagination">
       <?php
           //display the links to the pages
           for($current_page = 1; $current_page <= $expected_number_of_pages; $current_page++){
    if ($page_number != $expected_number_of_pages){
                   $data = array(
                       'page' => $page_number,
                       'country' => $cleaned_current_page_country,
                   );
                   $the_query = http_build_query($data);
                   $page_url .= '?' . $the_query;
                   echo '<li><a href=" '.$the_query.' ">'.$current_page.'</a></li> ';
               }
           }
       ?>
       </div>
   </table>

Problem With If -

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Hi Guys, I am having an issue with an if statement, I cant get it to go true, even though it clearly is!

Please excuse the messy code, im pulling my hair out here!

$itemprice[0] = 8.99;
$itemprice[1] = 19.95;
$itemprice[2] = 8.99;
print_r($itemprice);
echo "<BR>";
$x=0;
foreach($itemprice AS $val) {
echo gettype($val) .$val. "<br>";
if($val == 8.99) {
$x++;
$freecount++;
}
}
echo $x;

outputs correctly:
Array ( => 8.99 [1] => 19.95 [2] => 8.99 )
double8.99
double19.95
double8.99
x = 2

replacing:
$newtotal = add_to_price($id, $non_disc);
$itemprice[] = $newtotal - $non_disc;
print_r($itemprice);
echo "<BR>";
$x=0;
foreach($itemprice AS $val) {
echo gettype($val) .$val. "<br>";
if($val == 8.99) {
$x++;
$freecount++;
}
}
echo $x;

The add to price function returns a running total, the take away gives me the current item price in a for loop to build the array.

Output:

Array ( => 8.99 [1] => 19.95 [2] => 8.99 )
double8.99
double19.95
double8.99
x = 1

The problem? Well when I build the array up automatically it still prints the same, has the same datatype, yet the if statement does not catch the second 8.99 value?

Please accept my apologies for this post, it is my first ever and i've been doing this for 10 years now, i have never been this stumped on something so simple, I can only think it is a datatype error but i have tried apostrophes in the if, make no difference.

All help massively gratefully received.

Please email me if you want to see it in action...

Problem With Sql In Php

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Hows it going guys.

I am currently having a problem. I had a program that worked. I migrated the website and now the program is broken.

Any time i try to run it, i get this error:

Quote

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/thegoo20/public_html/class/wordgame/wordgame.php on line 31

Here is the code at that point:

Code: [Select]
function changeword(){
$result = mysql_query("Select words from CurrentWords");
$totalwords = array();

while ($row = mysql_fetch_array($result)){
$totalwords[] = $row["words"];
}
$_SESSION['word'] = $totalwords[rand(0, count($totalwords)-1)];

$_SESSION['scrambled'] = str_shuffle($_SESSION['word']);

}

where line 31 is the while statement.

Any help would be appreciated.

Thanks

Sql Problem

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I have having a problem getting a mysql query to work.

If I just use mysql_query it looks like this and works fine:
INSERT INTO schedule (schedule_pk, schedule_month, schedule_day, schedule_year, schedule_hour, schedule_minute, schedule_type, employeenumber) VALUES ( NULL, 10, 10, 1999, 12, 6, 'DayIn', 3);

If I put it through mysql_real_escape_string it turns it to this and does not work when I put it into mysql_query:
INSERT INTO schedule (schedule_pk, schedule_month, schedule_day, schedule_year, schedule_hour, schedule_minute, schedule_type, employeenumber) VALUES ( NULL, 10, 10, 1999, 12, 6, \'DayIn\', 3);

aka:
$query = "INSERT INTO...."
mysql_query($query);

that works, but:
$query = "INSERT INTO...."
$sqlQuery = mysql_real_escape_string($query, $con);
mysql_query($sqlQuery);

results in the error.
mysql_query($query);

The error is:
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 '\'DayIn\', 3)' at line 1

Php Problem

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I do not receive email for my published php file which is:

<?php

///// easend.php /////

$youremail = "acdelco40108@yahoo.com"; /*put the email address here, between quotes,
the email address you want the message sent to*/

$to = $youremail;
$email = $_POST['EMail'];
$name2 = $_POST['Name'];
$street2 = $_POST['Street'];
$city2 = $_POST['City'];
$state2 = $_POST['State'];
$zip2 = $_POST['Zip'];
$Phone = $_POST['Home_Phone'];
$Cell = $_POST['Cell_Phone'];
$education = $_POST['email1'];
$comments = $_POST['email2'] ;

$headers = "From:" . $email;

$fields = array();
$fields{"Name"} = "Name";
$fields{"Street"} = "Street";
$fields{"City"} = "City";
$fields{"State"} = "State";
$fields{"Zip"} = "Zip";
$fields{"Home_Phone"} = "Home Phone";
$fields{"Cell_Phone"} = "Cell Phone";
$fields{"EMail"} = "Email";
$fields{"email1"} = "Education";
$fields{"email2"} = "Comments";

$subject = "We have received the following information from your employment application";
$body = "We have received the following information from your employment application:\n\n"; foreach($fields as $a => $b)
{ $body .= sprintf("%20s: %s\n",$b,$_POST[$a]); }

mail ($to, $subject, $body, $headers); //send mail to owner
#end create email vars

$headers = "From:" . $to;

mail ($email, $subject, $body, $headers); //send mail to user
#end create email vars

echo "<head><META HTTP-EQUIV=\"Refresh\" CONTENT=\"2;
URL=ThankYou.html\"></head>";

?>

Problem

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Okey so i made a table that you put your name,author,and message

when you submit it , it echoes a table with the name,author and message written (and it also echoes a delete buttom,so delete this post oif necessary)

im new to php and i have been with this problem for a couple of weeks so i guess its time to ask for some help

the problem is that i dunnot know how to make my delete buttom work! i tried if statement but it dosent work i also tried ternary operation and dint work :S

i read that there is something like $post[ID](and this is supposed to get the ID of the post submmited, and delete it) im not sure, im so confused! help! XD

this is the code

<?php

$tittle=$_POST['tittle'];
$author=$_POST['author'];
$message=$_POST['message'];

if ($_POST['submitnews']){

$currentdate= date("y-m-d");
$currenttime=date("H:i:s",strtotime("-6 hours"));

$post=mysql_query("INSERT INTO news VALUES('','$tittle','$author','$message','$currentdate','$currenttime')");
echo"Posted!";

}

$select=mysql_query("SELECT * FROM news ORDER BY id DESC");

while ($row= mysql_fetch_assoc($select))

{
$id=$row['id'];
$tittle=$row['tittle'];
$author=$row['author'];
$message=$row['message'];
$date=$row['date'];
$time=$row['time'];

if ($_SESSION['admin'])

{

echo "

<table width='488px' id='news_table'>
<tr>
<td>
</td>
<td>
<center><font size='5'>$tittle</font></center><br>
</td>
</tr>
<tr>
<td>
</td>
<td>
$message
</td>
</tr>
<tr>
<td>
</td>
<td>
<font size='1'>Posted By:<font color='green'>$author</font> on <font color='gray'>$date</font> at <font color='gray'>$time</font></font>
<input name='delete' type='submit' value='delete'>
<td>
</td>
</tr><br><br>
</table>";

Problem

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Php Sql Problem

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Hi there,

I have a map of America made in flash, where you click on a state and the page should display the SQL database information for that state in the HTML table - but instead all it shows is the first entry of the database regardless of which state you click and it doesn't display the 2 radio buttons.

My code is as follows

Flash Actionscript 3 (just showing one state)
Code: [Select]
function waClick(event:MouseEvent):void {
var waURL:URLRequest = new URLRequest("restaurants.php?state=Washington");
navigateToURL(waURL, "_self");
}
wa_btn.addEventListener(MouseEvent.CLICK, waClick);
PHP code

Code: [Select]
<?php
  include("mvfconnect.php");
  $theChoice = $_GET['state'];
  $query = "SELECT * FROM restaurants WHERE" .$theChoice;
  $result = @ mysql_query($query);
  if (!$result) {
   $message="Unfortunately we are having problems with this page, we promise to have it fixed as soon as possible";
    die($message);
  }
  $num = mysql_num_fields($result);
  $i=0;
  while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
   $show1=substr($row['state'],0,50)."...";
   $show2=substr($row['city'],0,50)."...";
   $show3=substr($row['rname'],0,50)."...";
   $show4=substr($row['address'],0,50)."...";
   $show5=substr($row['pnum'],0,50)."...";
   $show6=substr($row['web'],0,50)."...";
   $show7=substr($row['dishes'],0,50)."...";
   $show8=substr($row['dish_details'],0,50)."...";
   $show9=substr($row['challenges'],0,50)."...";
   $show10=substr($row['challenge_details'],0,50)."...";
   $show11=substr($row['youtube'],0,50)."...";
   $show12=substr($row['images'],0,50)."...";
   echo "<tr>".
          "<td>".$row['city']."</td>".
          "<td>".$row['rname']."</td>".
          "<td>".$row['address']."</td>".
          "<td>".$row['pnum']."</td>".
          "<td>".$row['web']."</td>".
          "<td>".$row['dishes']."</td>".
          "<td>".$row['challenges']."</td>".
              "<td id='state".$i."' style='display:none'>".$row['state']."</td>".
          "<td id='city".$i."' style='display:none'>".$row['city']."</td>".
          "<td id='rname".$i."' style='display:none'>".$row['rname']."</td>".
          "<td id='address".$i."' style='display:none'>".$row['address']."</td>".
          "<td id='pnum".$i."' style='display:none'>".$row['pnum']."</td>".
          "<td id='web".$i."' style='display:none'>".$row['web']."</td>".
          "<td id='dishes".$i."' style='display:none'>".$row['dishes']."</td>".
          "<td id='dish_details".$i."' style='display:none'>".$row['dish_details']."</td>".
          "<td id='challenges".$i."' style='display:none'>".$row['challenges']."</td>".
          "<td id='challenge_details".$i."' style='display:none'>".$row['challenge_details']."</td>".
          "<td id='youtube".$i."' style='display:none'>".$row['youtube']."</td>".
                  "<td id='images".$i."' style='display:none'>".$row['images']."</td>".
                  "<td><input type='radio' name='vid' id='vid".$i."' onclick='openVideo(".$i.")' /></td>".
          "<td><input type='radio' name='pic' id='pic".$i."' onclick='openImage(".$i.")' /></td>".
          "<td class='last'style='display:none'>".$show1." ".$show2." ".$show3." ".$show4." ".$show5." ".$show6."
                  ".$show7." ".$show8." ".$show9." ".$show10."</td>".
         "</tr>";
    $i++;
   }
?>

I have used this code before and it has worked fine, can someone please help me out!

Php Problem

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I Have An error on line 42 not sure what it is can any one tell?

$foundnum = mysql_num_rows($run);

Rss Problem

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I am trying to save this an an xml document but am getting this error when I try to open the xml file "feed.xml" - "XML Parsing Error: no element found"

$xml = '<rss version="2.0">

<channel>

<title> RSS Feed</title>

<link></link>

<description>the best industry-lead opinions</description>

<language>en-us</language>
</channel>

</rss>";

$xml2 = new DOMDocument('1.0');
$xml2->Load($xml);
$xml2->save("feed.xml");

Oop Problem!

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I am trying to understand PHP OOP and I have some code that is not working. here is the error code.
Code: [Select]
Parse error: parse error, expecting `T_FUNCTION' in C:\wamp\www\testing\armor_lib.php on line 11
And here is the armor_lib.php file:
<?php
class armor {
// Head
public $head;
public $torso;
public $pants;
public $gloves;
public $boots;

// new stuff here
class head extends armor {
function __construct($head){
$this->set_head($head);
}
}
// Torso
class torso extends armor {
function __construct($torso){
$this->set_torso($torso);
}
}
// Pants
class pants extends armor {
function __construct($pants){
$this->set_pants($pants);
}
}
// Gloves
class gloves extends armor {
function __construct($gloves){
$this->set_gloves($gloves);
}
}
// Boots
class boots extends armor {
function __construct($boots){
$this->set_boots($boots);
}
}
}
?>

And here is the php in the armor.php file I have:
<?php
$head = new armor("Leather Helm");
$torso = new armor("Leather Shirt");
$pants = new armor("Leather Pants");
$gloves = new armor("Chain Mail Gloves");
$boots = new armor("Leather Boots");

echo "You are wearing: " . $head->get_head() . "on your Head";
echo "<br />";
echo "You are wearing: " . $torso->get_torso() . "on your Torso";
echo "<br />";
echo "You are wearing: " . $pants->get_pants() . "on your Legs";
echo "<br />";
echo "You are wearing: " . $gloves->get_gloves() . "on your Hands";
echo "<br />";
echo "You are wearing: " . $boots->get_boots() . "on your Feet";
?>

Any Help in understanding this will be much appreciated. Thanks.

Sql Like Problem

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Code: [Select]
$sql = "SELECT softwareID,softwareName,softwareType,softwareDesc,softwarePath,ITOnly FROM software WHERE softwareName LIKE '%($searchSoftware)%' ORBER BY softwareName";
$result = mysqli_query($cxn,$sql) or die(mysqli_error());

There is something wrong with my LIKE statement because it's not pulling it since I'm either formatting it wrong or something.
Can anyone catch it?

Problem With ' And \

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Hi all.
When I typed symbol ' in my textarea , after I submit it and view for what I typed , it will automatically add a slash in front of the ' .

Such as the message is "I'm fine" , then it will turn out as "I\'m fine". Can I know what is the problem ? and how can I solve it?

Thanks for every reply .