PHP - Displaying Mysql Results Code Error

Hello everyone,

So what I'm trying to do is have a dropdown menu displaying a number of <options> for people to select and to update that selection to the database, easy enough right? But I want that option to be displayed as the "selected" option when the page is revisited or refreshed and I just can't figure it out!!! (Permission to bang head on desk?) It would seem like it sould be a really basic thing to do but it's got me completely and a lot of menus around the site are going to rely on this so I came to you guys for help.

A simple example would be like the facebook edit profile page, the user selects whether they are Male or Female, the database gets updated and when you return the option you selected before is the one that appears as if selected="selected" had been done. I've tried everything I can think of (all be it from a learners perspective) with no joy, ive managed to get the database connection sorted, the tables done, the login with unique id $_SESSION, logout etc... so then when I got to this I thought... easy LOL yeah right.

Some of this probably doesnt even make sense but I'll show you the kind of things I've tried...


            <select name="gender" size="1" id="gender">
              <option value="male" <?php if ($gender == "male") {echo 'selected="selected"';} ;?>>Male</option>
              <option value="female" <?php if ($gender == "female") {echo 'selected="selected"';} ;?>>Female</option>
            </select>



OR


     <select name="gender" id="gender">
              <option value="" selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option>
              <option value="male" selected="<?php if ($gender == "male") {echo "selected";} else {echo "";} ;?>">Male</option>
              <option value="female" selected="<?php if ($gender == "female") {echo "selected";} else {echo "";} ;?>">Female</option>
            </select>


OR



            <select name="gender" size="1" id="gender">
              <option selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option>
              <option value="<?php if ($gender == "Male") {echo "selected";} else {echo "male";} ;?>">Male</option>
              <option value="<?php if ($gender == "Female") {echo "selected";} else {echo "female";} ;?>">Female</option>
            </select>


OR

            <select name="gender" id="gender">
              <option value="male"><?php if ($gender == "male") {echo "Male";} ;?></option>
              <option value="female"><?php if ($gender == "female") {echo "Female";} ;?></option>
            </select>


Honestly man, I've got no idea.


The other thing is, I have more than 1 dropdown menu in the same form (5 in total) and if I use 2 or more selecting different options as I go I get a blank screen.  And one more, if I have selected Male and it updates the users row and I resubmit Male again it's blank screen time again, lol.

Any help would be tremendous and greatly appreciated.

Thanks very much, Learner

P.S  Man!

am total newbie to programming, apart from knowing SQL, the thing is i have been given a MYSQL database containing various information about kids diseases and a web interface written in php to create reports from the database that can be accessed via the interface. there are almost 25 different variables that need to be computed in the report, i have written sql queries to compute all these values, but i don't know anything about PHP, if i have all these queries isn't there a way for me to combine all these sql queries to be display results on a webpage and come up with this report without writing PHP code?

Thanks for your help

very sorry if this is too basic

Hi, 

I am trying to array a mysql tables data into a php table.

Not having luck...

<?php

include('dbconnect.php')

?>


<?php
// Make a MySQL Connection
$query = "SELECT * FROM cars"; 
 
$result = mysql_query($query) or die(mysql_error());


while($row = mysql_fetch_array($result)){
echo $row['CarName']. " - ". $row['CarTitle'];
echo "<br />";

echo "<TABLE CELLPADDING=0 CELLSPACING=0 WIDTH=100%>";
echo "<TR />";
echo "<TD />";
echo $row['CarName'].;
echo "</TD>";
echo "<TD />";
echo $row['CarTitle'].;
echo "</TD>";

echo "</TR>";
echo "</TABLE>";

}
?>


Error says: Parse error: syntax error, unexpected ';' in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/viewcars.php on line 69

echo </TD /> is line 69.

Please help

Ian

need a little help guys! I use the script below to display profile images, trouble is it shows 1 on top of the other, and i need it to double up 2 profile images on top of 2 profile images ect any ideas how i can do this. 



require("./include/mysqldb.php");

$con = mysql_connect("$dbhost","$dbuser","$dbpass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("$dbame", $con);
$result = mysql_query("SELECT * FROM Search_profiles_up WHERE upgrade_one ='1' ORDER BY RAND() LIMIT 40");


print "<table width=\"293\" height=\"111\" border=\"0\">
  <tr>\n";

while($row = mysql_fetch_array($result))
  {
print "<td width=\"142\"><img src=" . $row['search_small_image'] . " width=\"144\" height=\"169\" /></td>\n";
print "  </tr>\n";
print "  <tr>  \n";
print "<td>" . $row['star'] . "</td>\n";
print "  </tr>\n";
print "  <tr>\n";
print "<td>" . $row['username_search'] . "</td>\n";
print "  </tr>\n";
print "  <tr>  \n";
print "<td>" . $row['phone_search'] . "</td>\n";
print "  </tr>   \n";

  }
print "</table>";

mysql_close($con);
?>


 

Hello everyone,

I begin in everything web related but I have been programming for years. I tried to code something simple : small Mysql DB (works fine) and to begin a search bar to browse data. I adapted a code that I understood provided here : https://www.cloudways.com/blog/live-search-php-mysql-ajax/. Base principle is simple :  as you type in your query, it will pass the text to script.js  that will forward this request to ajax.php file. In the ajax.php, a javascript function named “fill()” will pass the fetched results. This function will also display the result(s) into “display” div in the  “search.php” file.

The problem is that when I type anything it displays, below the search bar, at the moment I type a character:

Quote

 

'; //Fetching result from database. while ($Result = MySQLi_fetch_array($ExecQuery)) { ?>

")'>

 

instead of the actual answer from my database (no error in the browser console).

I tested the SQL query + the user I provide and everything seems fine.

Any clue what could be the root cause ? I strongly suspect a mistake in the code as I already corrected one (script.js instead of scripts.js) but I really cannot figure out where.

Thanks in advance,

 

 problematic code (ajax.php):

 

<?php

//Including Database configuration file.

include "db.php";

//Getting value of "search" variable from "script.js".

if (isset($_POST['search'])) {

//Search box value assigning to $Name variable.

   $Name = $_POST['search'];

//Search query.

   $Query = "SELECT Name FROM search WHERE Name LIKE '%$Name%' LIMIT 5";

//Query execution

   $ExecQuery = MySQLi_query($con, $Query);

//Creating unordered list to display result.

   echo '

<ul>

   ';

   //Fetching result from database.

   while ($Result = MySQLi_fetch_array($ExecQuery)) {

       ?>

   <!-- Creating unordered list items.

        Calling javascript function named as "fill" found in "script.js" file.

        By passing fetched result as parameter. -->

   <li onclick='fill("<?php echo $Result['Name']; ?>")'>

   <a>

   <!-- Assigning searched result in "Search box" in "search.php" file. -->

       <?php echo $Result['Name']; ?>

   </li></a>

   <!-- Below php code is just for closing parenthesis. Don't be confused. -->

   <?php

}}


?>

</ul>

 

Hello everyone,

When I submit my login form to go to my login page it goes to a white page and does not display an error.

Code: [Select]
<?php 
include("../secure/database.php");

if(!empty($_POST['loginsubmit'])){
if(!empty($_POST['email'])){

$email = securevar($_POST['email']);

if(!empty($_POST['passconf'])){

$pass = securevar($_POST['passconf']);

$q = "SELECT * FROM `accountinfo_db` WHERE `email` = '$email' AND `password` = '$pass'";

$res = mysql_query($q) or die(mysql_error());

$login = mysql_fetch_array($res);
$id = $login['id'];
$active = $login['active'];
if($id>=1){
if($active==1){
$_SESSION['logged'] = $login['id'];

$q = "UPDATE `accountinfo_db` SET `loggedtimes` = `loggedtimes`+'1' WHERE `id` = '$id'";
$res = mysql_query($q) or die(mysql_error());
$user = $login['username'];
header("Location: ../mainframe.php?strmsg=Welcome Back $user");
}else{
header("Location: activate.php");
}

}else{
   header("Location: ../index.php?strmsg=Login Information Incorrect!");    
}

       }else{
   header("Location: ../index.php?strmsg=You did not enter a valid password!");    
   }
   }else{
   header("Location: ../index.php?strmsg=You did not enter a valid email!");
   }
}else{
header("Location: ../index.php?strmsg=We are sorry but you are not allowed viewance of that directory!");
}
?>

any help would be great.
Brian

All I am trying to do is add a record on a page without the page refreshing.  For that ajax is used.  Here is the code.   It does not add the record to mysql table. Can anyone tell me what I am doing wrong?     record.php

<!DOCTYPE HTML>
<html lang="en"> 
	<head>
		<script type="text/javascript" src="js/jquery-1.11.0.min.js"></script>
		<script type="text/javascript" >
			$(function() {
				$(".submit_button").click(function() {
					var textcontent = $("#content").val();
					var name = $("#name").val();
					var dataString = 'content='+ textcontent + '&name='+name;
					if(textcontent=='')
					{
						alert("Enter some text..");
						$("#content").focus();
					}
					else
					{
						$("#flash").show();
						$("#flash").fadeIn(400).html('<span class="load">Loading..</span>');
						$.ajax({
							type: "POST",
							url: "action.php",
							data: dataString,
							cache: true,
							success: function(html){
								$("#show").after(html);
								document.getElementById('content').value='';
								$("#flash").hide();
								$("#content").focus();
							}  
						});
					}
					return false;
				});
			});
		</script>
	</head>
	<body>
	<?php
		$record_id = $_GET['id']; // getting ID of current page record 
	?>
	
		<form action="" method="post" enctype="multipart/form-data">
			<div class="field">
				<label for="title">Name *</label>
				<input type="text" name="name" id="name" value="" maxlength="20" placeholder="Your name">
			</div>
			<div class="field">
				<label for="content">content *</label>
				<textarea id="content" name="content" maxlength="500" placeholder="Details..."></textarea>
			</div>
			<input type="submit" name="submit" value="submit" class="submit_button">
		</form>
		<div id="flash"></div>
		<div id="show"></div>
		
	</body>
</html>	

action.php

if(isset($_POST['submit'])) {
	if(empty($_POST['name']) || empty($_POST['content'])) {
		
		$error = 'Please fill in the required fields!';
		
	} else {
		try {
			
			$name		= 	trim($_POST['name']);
			$content 	= 	trim($_POST['content']);
					
			$stmt = $db->prepare("INSERT INTO records(record_id, name, content) 
				VALUES(:recordid, :name, :content");
			$stmt->execute(array(
				'recordid'		=> $record_id,
				'name'			=> $name,
				'content' 		=> $content
			));
			
			if(!$stmt){
			
				$error = 'Please fill in the required fields.';
				
			} else {
			
				$success = 'Your post has been submitted.';
			
			}
			
		} catch(Exception $e) {
			die($e->getMessage());
		}
	}
}

I have a PHP while loop that pulls from an SQL database and displays the contents in a table with two columns.

// Check Connection
    if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

// Select Data Table
    $result = mysqli_query($con,"SELECT * FROM Recommendations") or die(mysqli_error);

// Split Data
    $mid = ceil(mysqli_num_rows($result)/2);

// Display Content
    while ($rows = mysqli_fetch_array($result)) {
        $Name = $rows['Name'];
        $Author = $rows['Author'];
        $Cover = $rows['Link to Cover'];
        $Link = $rows['Link to Profile'];
    echo "<table><tr><td>
    <a href='" . $Link . "' >$Name</a>
    <br />
    $Author
    <br />
    <a href='" . $Link . "' ><img src='" . $Cover . "' /></a>
    </td>
    <td>
    <a href='" . $Link . "' >$Name</a>
    <br />
    $Author
    <br />
    <a href='" . $Link . "' ><img src='" . $Cover . "' /></a>
    </td></tr></table>";
    }
?>

I want to be able to display the looped results side by side in columns of two. Example: 1 2 3 4
5 6   I've tried using pseudo classes to display only the even and odd results in the different table columns, but honestly have no idea how to do this. I'm new to PHP, so my apologies if the results are really obvious. Thanks in advance!

Hello,

Im trying to display some results from mysql database, however none display.
Can anyone tell me where im going wrong please?

Code: [Select]
 
</head><body>
<div id="listhold">
<div class="list">
<a href="Restaurants.html">Restaurants</a><br />

<?php
mysql_connect("","",""); 
mysql_select_db("") or die("Unable to select database"); 
$result = mysql_query("SELECT name FROM business WHERE type ='restaurant' ORDER BY name");
$number_of_results = mysql_num_rows($result);
$results_counter = 0;
if ($number_of_results != 0) 
{while ($array = mysql_fetch_array($result))
$results_counter++;
if ($results_counter >= $number_of_results);}
?>
</div>

My error happens on line #81

Dear all,
I am new in this forum. This is my code

  $query = "
  SELECT webdb.id, webdb.writer, writer.picLoc, webdb.title FROM webdb, writer WHERE webdb.writer=writer.name and 
  category = 'Researchworks' and language = 'Farsi' ORDER BY writer DESC"; 
  $resultaat = mysql_query($query, $LinkID);
  $column_count = mysql_num_fields($resultaat) or die (mysql_error()."<br>Couldn't execute query: $SQLquery");
  
  $counter=1;
  echo "<table border=\"0\" width=\"700\" border color=white><tr>";
  while ($row = mysql_fetch_row($resultaat))
  {
  if ($author !== $row[1])
  {
  $author = $row[1];
  echo "<td align=right width=220 valign=top style=\"margin: 5px; float: right border-bottom-color:#000; border-left-color:#000;\">";
  echo "<img width=\"50\" height=\"80\" src=\"admin/writers/$row[2]\" border =\"0\"><br>".$row[1]."<br>";
  echo "<a href=\"poems.php?writer=$row[1]\">".$row[3]."</a><br>";
  echo "</td>";
  if($counter%3==0) 
  {
  echo"</tr><tr>";
  }
  $counter++;
}
  }
  echo"</tr></table>";


i have authors with different articles on a certain topic. What i want is, displaying the name of the author only once and all his titles under his name. I also want a dynamic table where i display three authors in each row and soon as there a fourth author a new row must start. My problem now is is the title is also being filtered and i can only display one title.

Thanks in advance

I'm realitivly new to PHP and was hoping somebody could help.

I have a mysql database that stores information about books. I am currently using the code below to query the database and extract the 3 most recent entries and showing them in a dynamic list:

Code: [Select]
$sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC LIMIT 3");
$productCount = mysql_num_rows($sql);
if ($productCount > 0) { // ensure a book exists
while($row = mysql_fetch_array($sql)){
             $id = $row["id"];
$title = $row["title"];
$author = $row["author"];
$price = $row["price"];
$date_added = strftime("%b %d, %Y", strtotime($row["date_added"]));
$dynamiclist .= //My table showing the products
}
} else {
$dynamicList = "There are currently no Books listed in this store";
}
This works well when showing the most recent books 1 below the other. However, I would like to show these products side by side, horizontally across the page.

Can somebody please point me in the right direction?

Many thanks

I'm trying to create a list that groups information by username. Only part of it is working. The first query ($get_item_sql) is grouping the information perfectly but the second query ($get_sold) is lumping the $item_price and $item_amount_due as one total for each one and outputting the same amounts into every username. I'm stuck on this and would appreciate your help.

For example:  Username   item fees   image fees item sales  item price  total due
                       Jim                   2               $0.40      $100.00        $3.00     $3.40
                       Kelly                 5               $1.00      $100.00        $3.00     $4.00

This example shows the columns in red as being the problem where Kelly didn't sell anything so her "item sales" and "item price" should be $0.00 but is carrying Jim's totals into hers.

Hope this helps!  Thank you!     

$get_item_sql = mysql_query("SELECT id, username, date, ROUND(price,2) AS price, SUM(item_fee) AS fee, item_fee, SUM(sold) AS sales, SUM(ROUND(price,2)) AS total FROM product WHERE MONTH(date) = MONTH(DATE_ADD(CURDATE(),INTERVAL -1 MONTH)) GROUP BY username" )
or die(mysql_error());

if (mysql_num_rows($get_item_sql) < 1) {
   //invalid item
   $display_block .= "<p><em>Invalid item selection.</em></p>";

  } else {
   //valid item, get info
   while ($item_info = mysql_fetch_array($get_item_sql)) {
   $item_username = $item_info['username'];
   $item_date = $item_info['date'];
   $item_price = $item_info['price'];
   
   $item_fee = $item_info['fee'];
   $image_fees = $item_fee * .20;
   $item_sold = $item_info['sales'];
   
  
 $get_sold = mysql_query("SELECT SUM(ROUND(price,2)) AS total, SUM(ROUND(sold,2)) AS sales, date, username FROM product WHERE sold = '1' AND MONTH(date) = MONTH(DATE_ADD(CURDATE(),INTERVAL -1 MONTH)) GROUP BY username")
or die(mysql_error());

if (mysql_num_rows($get_sold) < 1) {
   //invalid item
   $display_block .= "<p><em>Invalid item selection.</em></p>";

  } else {
   //valid item, get info
   while ($item_sold2 = mysql_fetch_array($get_sold)) {
   
   
   $item_sales = $item_sold2['total'];
   $item_price = ($item_sold2['total']) * .03;
  
$item_amount_due = $image_fees + $item_price; 

  
  

$content .= "<form action=\"add_artist.php\" method=\"post\"><table class=\"anotherfont\" width=\"670\" border=\"0\">

<tr><td width=\"201\">{$item_username}</td>
                    <td width=\"109\">{$item_fee}</td>
<td width=\"109\">{$image_fees}</td>
<td width=\"109\">&nbsp;&nbsp;{$item_sales}</td>
<td width=\"109\">&nbsp;&nbsp;{$item_price}</td>
<td width=\"109\"><input name=\"balance_due\" type=\"text\" value=\"{$item_amount_due}\" /></td>
</tr><br /></table></form>"; 

   }
  }
  }
  }

Hi, fairly new to PHP over the last couple weeks. Been having a problem with certain queries. I have a database with football results, games, teams etc. I can filter these using drop down and that's all well and good. The problem I'm having is displaying the data via gameweek. I've been asked to display the table like so - Gameweek1 will display week1 teams, results etc. Gameweek2 will display week2... and so on. I can manage to do this in a drop down. But I've been asked to display this using links like "Previous, 1, 2, 3 Next". I've tried pagination but I couldn't figure it out. Can anyone point me in the right direction? If I need a GET() method, how would I go about coding that so it will be used in a link(s)? Been searching and searching to find an answer but to no avail...

//Database connection etc...

$gameweek = "SELECT * FROM games WHERE gameweek= 1";
    
    //if(isset($_GET['gameweek']))
    //{
    //    $gameweek = $_GET['gameweek'];
    //    
    //}
    //....
    $result=mysqli_query($connection, "select * from games WHERE gameweek= 1");
 //Print table and table headings...
    
    mysqli_close($connection);
?>
<a href="http://weeks.php?gameweek=2">Week 2</a>
<a href="http://weeks.phpgameweek=3">Week 3</a>

</body>
</html>