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up vote -2 down vote favorite

I am trying to retrieve data from the database (list of products) but I have a blank page when I execute the code.

<?php


$DB_HOST = "@@@";
$DB_NAME = "@@@";
$DB_PASS = "@@@";
$DB_USER = "@@@";


$db_obj = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($db_obj->connect_errno > 0) {
  die('Connection failed [' . $db_obj->connect_error . ']');
}

if($db_obj->connect_errno > 0){
    die('Unable to connect [' . $db_obj->connect_errno . ']');
}


?>



<html>
<head>
<title>List of products</title>
</head>
<body>
<?php


$query="SELECT * FROM product"; 
$result_obj = $db_obj->query($query);
$row=mysql_fetch_row($result);

echo $row;




?>

</body>
</html>

I am doing it right? where is the problem exactly? is the problem can be from the server side?

new try:

<?php
// session_start();
// require("mysqli.php"); 

$DB_HOST = "webdev.cs.kent.edu";
$DB_NAME = "ralsuhai";
$DB_PASS = "810646396";
$DB_USER = "ralsuhai";


$db_obj = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($db_obj->connect_errno > 0) {
  die('Connection failed [' . $db_obj->connect_error . ']');
}

if($db_obj->connect_errno > 0){
    die('Unable to connect [' . $db_obj->connect_errno . ']');
}


?>



<html>
<head>
<title>List of products</title>
</head>
<body>
<?php


$query="SELECT * FROM product"; 
$result_obj = $db_obj->query($query);
//first try
//while($row=$result_obj->fetch_row() )
//{
  //   var_dump($row);
 //}

 //second try
//$res = array();
//while($row = $result->fetch_array(MYSQLI_ASSOC))
//{
  //$res[] = $row;
//}
//print_r($res);




?>
<table><tr><th>Product Id</th><th>Name</th><th>Price</th></tr>
<?php
$result_obj = $db_obj->query($query);
while($row=$result_obj->fetch_row() )
{ ?>
    <tr>
     <td><?=implode('</td><td>',$row)?></td>
    </tr>
<?php
 }
?>
</table>
<?php










?>

</body>
</html>
share|improve this question
5  
You are mixing up mysql_* and mysqli_* functions. –  Shankar Damodaran Apr 3 '14 at 4:17
    
what is the different between both of them? –  user3483167 Apr 3 '14 at 4:19
2  
Difference is that they are both different –  Shankar Damodaran Apr 3 '14 at 4:20
    
@user3483167 first off, they can't cooperate. And mysqli_ supports prepared statements and object-oriented approach, while mysql_ cannot. –  Jan Dvorak Apr 3 '14 at 4:20
    
What Shankar meant by that is, those two functions do not mix. @user3483167 –  Fred -ii- Apr 3 '14 at 4:38

2 Answers 2

up vote 0 down vote accepted

You've got a couple problems here. First is, the variable $db_obj is a mysqli object and your trying to use the mysql_fetch row function on it. That won't work, next even if it did work, $row would be an array.try this...

?>
<table><tr><th>Product Id</th><th>Name</th><th>Price</th></tr>
<?php
$result_obj = $db_obj->query($query);
while($row=$result_obj->fetch_row() )
{ ?>
    <tr>
     <td><?php echo implode('</td><td>',$row); ?></td>
    </tr>
<?php
 }
?>
</table>
<?php

this should dump each row returned by the query and you can see what's going on. Note that the result from $db_obj->query($query) is an object. This should print a table with your data.

share|improve this answer
    
when I applied this code I got the following: "array(3) { [0]=> string(10) "prod_99004" [1]=> string(14) "Tennis Racquet" [2]=> string(5) "95.85" } array(3) { [0]=> string(10) "prod_99007" [1]=> string(12) "Tennis Shoes" [2]=> string(5) "75.95" } array(3) { [0]=> string(10) "prod_99008" [1]=> string(12) "Tennis Balls" [2]=> string(4) "3.85" } array(3) { [0]=> string(10) "prod_99009" [1]=> string(14) "Tennis T-shirt" [2]=> string(5) "15.85" }" –  user3483167 Apr 3 '14 at 4:33
    
perfect. That's what should be expected, now do as you wish with the data... What are you trying to do anyways? –  Bryan Apr 3 '14 at 4:36
    
I want to show them as a list. –  user3483167 Apr 3 '14 at 4:37
    
all of the information or just the name or name and price? –  Bryan Apr 3 '14 at 4:38
    
all the information in the product table is the data base. –  user3483167 Apr 3 '14 at 4:41
up vote 0 down vote 
<?php
   $DB_HOST = "@@@";
   $DB_NAME = "@@@";
   $DB_PASS = "@@@";
   $DB_USER = "@@@";

   $db_obj = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
   if($db_obj->connect_errno) {
   die('Connection failed [' . $db_obj->connect_error . ']');
   }
?>
<html>
  <head>
    <title>List of products</title>
      </head>
      <body>
      <?php
         $query="SELECT * FROM product";
         $result_obj = $db_obj->query($query);
         $row=$result_obj->fetch_assoc();
         print_r($row);
      ?>
  </body>
 </html>
share|improve this answer
    
showed a blank page too >< –  user3483167 Apr 3 '14 at 5:22
    
if I want to update a table in the database, I can use the following: <?php $query="UPDATE admin SET password=npassword"; { ?> <?php echo "your password has been updated!" ?> <?php } ?> –  user3483167 Apr 5 '14 at 2:45
    
<?php $query="UPDATE admin SET password=npassword WHERE user_id=2"; $query = mysql_query($query) or die (mysql_error()); ?> –  Maulik patel Apr 5 '14 at 10:39
    
Thanks you, I will try it. –  user3483167 Apr 8 '14 at 4:18

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