How to fetch data form database using array - php

Question

I am trying to fetch data from the database, but not retrieve data particular id.
this is my one page:
example1.php
<a style="color: #3498DB;" class="btn btn-default" href="http://www.example.com/getafreequote?id=<?php echo $row['product_id']; ?>">Get Quote</a>
example2.php
<?php
$id = isset($_GET['id'])?$_GET['id']:'';
$query = "SELECT * FROM oc_product_description WHERE product_id=$id";
$run1 = mysql_query($query);
while ($fetch1 = mysql_fetch_object($run1)){
?>
<div class="col-xs-12 col-sm-6">
<label for="GetListed_product"></label>
<input class="le-input" name="product" id="GetListed_product" type="text" value="<?php
$b = $fetch1->product_id;
$q2 ="SELECT product_id,name FROM oc_product_description WHERE product_id = $b";
$q3 = mysql_fetch_assoc(mysql_query($q2));
echo $q3['name'];
?>" >
<span id="productmsg" class="msg"></span>
</div>
<?php
}
?>
</div>
but didnot get data form particular product id. I have got error show like this
Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in example.com/example2.php on line 71

Answer 1

Please don't use mysql functions they are deprecated. Use mysqli or PDO for database operations. Also the way you write the query string makes it easy for an SQL injection, use prepared statements instead. Here is an example:
$db = new PDO("...");
$statement = $db->prepare("select id from some_table where name = :name");
$statement->execute(array(':name' => "Jimbo"));
$row = $statement->fetch();
You can also use prepared statements for inserting or updating data. More examples here

Answer 2

As said by FilipNikolovski, don't use mysql functions they are deprecated. Use mysqli or PDO for database operations.
For your problem, the function mysql_query is returning false. The query is not returning any result and thus mysql_query is returning false.
Make a check like this:
$query = "SELECT * FROM oc_product_description WHERE product_id=$id";
$run1 = mysql_query($query);
if($run1)
{
if(mysql_num_rows($run1) > 0)
{
while ($fetch1 = mysql_fetch_object($run1))
{
// your stuff here
}
}
else
{
echo "No records found.";
}
}
else
{
echo "Error in query : ".mysql_error();
}
This will help you to detect the problem and to solve as well.