Fatal error: Call to a member function prepare() on null












16














I am trying to access a list of categories and their contents. I have a class called Categories. I keep getting this error. The weird thing is that I've used this same exact code in two other places so far with no problems. All I did here was reuse the code and change all the variables.



Fatal error: Call to a member function prepare() on null


Here is the code to my class:



    <?php

class Category {
public function fetch_all() {
global $pdo;

$query = $pdo->prepare("SELECT * FROM dd_cat");
$query->execute();

return $query->fetchAll();
}

public function fetch_data($cat_id) {
global $pdo;

$query = $pdo->prepare("SELECT * FROM dd_cat WHERE cat_id = ?");
$query->bindValue(1, $cat_id);
$query->execute();

return $query->fetch();
}
}

?>


And here is the code I am trying to call:



<?php
session_start();
//Add session_start to top of each page//
require_once('includes/config.php');
require_once('includes/header.php');
include_once('includes/category.php');

?>
<link rel="stylesheet" href="css/dd.css">
<div id="menu">
<a class="item" href="drop_index.php">Home</a> -
<a class="item" href="create_topic.php">Create a topic</a> -
<a class="item" href="create_cat.php">Create a category</a>
<div id="userbar">
<?php
if( $user->is_logged_in() )
{
echo 'Hello ' . $_SESSION['user_name'] . '. Not you? <a href="logout.php">Sign out</a>';
}
else
{
echo '<a href="login.php">Sign in</a> or <a href="index.php">create an account</a>.';
}
?>
</div>
</div>

<?php

$category = new Category;
$categories = $category->fetch_all();

?>
<div id ="wrapper">
<h1>Categories</h1>
<section>
<ul>
<?php foreach ($categories as $category) { ?>
<li><a href="category.php?id=<?php echo $category['cat_id']; ?>">
<?php echo $category['cat_title']; ?></a>
</li>
<?php } ?>
</ul>
</section>
</div>
<?php
require_once('includes/footer.php');
?>









share|improve this question


















  • 2




    var_dump($pdo), $pdo is not a object(not initialized yet) when prepare is called.
    – Jigar
    Feb 18 '15 at 19:41








  • 1




    @JandenHale: did you initialize a (global) variable with the name $pdo before that?
    – Willem Van Onsem
    Feb 18 '15 at 20:09










  • It's global in the class, at least I thought it was.
    – ddonche
    Feb 18 '15 at 20:40
















16














I am trying to access a list of categories and their contents. I have a class called Categories. I keep getting this error. The weird thing is that I've used this same exact code in two other places so far with no problems. All I did here was reuse the code and change all the variables.



Fatal error: Call to a member function prepare() on null


Here is the code to my class:



    <?php

class Category {
public function fetch_all() {
global $pdo;

$query = $pdo->prepare("SELECT * FROM dd_cat");
$query->execute();

return $query->fetchAll();
}

public function fetch_data($cat_id) {
global $pdo;

$query = $pdo->prepare("SELECT * FROM dd_cat WHERE cat_id = ?");
$query->bindValue(1, $cat_id);
$query->execute();

return $query->fetch();
}
}

?>


And here is the code I am trying to call:



<?php
session_start();
//Add session_start to top of each page//
require_once('includes/config.php');
require_once('includes/header.php');
include_once('includes/category.php');

?>
<link rel="stylesheet" href="css/dd.css">
<div id="menu">
<a class="item" href="drop_index.php">Home</a> -
<a class="item" href="create_topic.php">Create a topic</a> -
<a class="item" href="create_cat.php">Create a category</a>
<div id="userbar">
<?php
if( $user->is_logged_in() )
{
echo 'Hello ' . $_SESSION['user_name'] . '. Not you? <a href="logout.php">Sign out</a>';
}
else
{
echo '<a href="login.php">Sign in</a> or <a href="index.php">create an account</a>.';
}
?>
</div>
</div>

<?php

$category = new Category;
$categories = $category->fetch_all();

?>
<div id ="wrapper">
<h1>Categories</h1>
<section>
<ul>
<?php foreach ($categories as $category) { ?>
<li><a href="category.php?id=<?php echo $category['cat_id']; ?>">
<?php echo $category['cat_title']; ?></a>
</li>
<?php } ?>
</ul>
</section>
</div>
<?php
require_once('includes/footer.php');
?>









share|improve this question


















  • 2




    var_dump($pdo), $pdo is not a object(not initialized yet) when prepare is called.
    – Jigar
    Feb 18 '15 at 19:41








  • 1




    @JandenHale: did you initialize a (global) variable with the name $pdo before that?
    – Willem Van Onsem
    Feb 18 '15 at 20:09










  • It's global in the class, at least I thought it was.
    – ddonche
    Feb 18 '15 at 20:40














16












16








16


1





I am trying to access a list of categories and their contents. I have a class called Categories. I keep getting this error. The weird thing is that I've used this same exact code in two other places so far with no problems. All I did here was reuse the code and change all the variables.



Fatal error: Call to a member function prepare() on null


Here is the code to my class:



    <?php

class Category {
public function fetch_all() {
global $pdo;

$query = $pdo->prepare("SELECT * FROM dd_cat");
$query->execute();

return $query->fetchAll();
}

public function fetch_data($cat_id) {
global $pdo;

$query = $pdo->prepare("SELECT * FROM dd_cat WHERE cat_id = ?");
$query->bindValue(1, $cat_id);
$query->execute();

return $query->fetch();
}
}

?>


And here is the code I am trying to call:



<?php
session_start();
//Add session_start to top of each page//
require_once('includes/config.php');
require_once('includes/header.php');
include_once('includes/category.php');

?>
<link rel="stylesheet" href="css/dd.css">
<div id="menu">
<a class="item" href="drop_index.php">Home</a> -
<a class="item" href="create_topic.php">Create a topic</a> -
<a class="item" href="create_cat.php">Create a category</a>
<div id="userbar">
<?php
if( $user->is_logged_in() )
{
echo 'Hello ' . $_SESSION['user_name'] . '. Not you? <a href="logout.php">Sign out</a>';
}
else
{
echo '<a href="login.php">Sign in</a> or <a href="index.php">create an account</a>.';
}
?>
</div>
</div>

<?php

$category = new Category;
$categories = $category->fetch_all();

?>
<div id ="wrapper">
<h1>Categories</h1>
<section>
<ul>
<?php foreach ($categories as $category) { ?>
<li><a href="category.php?id=<?php echo $category['cat_id']; ?>">
<?php echo $category['cat_title']; ?></a>
</li>
<?php } ?>
</ul>
</section>
</div>
<?php
require_once('includes/footer.php');
?>









share|improve this question













I am trying to access a list of categories and their contents. I have a class called Categories. I keep getting this error. The weird thing is that I've used this same exact code in two other places so far with no problems. All I did here was reuse the code and change all the variables.



Fatal error: Call to a member function prepare() on null


Here is the code to my class:



    <?php

class Category {
public function fetch_all() {
global $pdo;

$query = $pdo->prepare("SELECT * FROM dd_cat");
$query->execute();

return $query->fetchAll();
}

public function fetch_data($cat_id) {
global $pdo;

$query = $pdo->prepare("SELECT * FROM dd_cat WHERE cat_id = ?");
$query->bindValue(1, $cat_id);
$query->execute();

return $query->fetch();
}
}

?>


And here is the code I am trying to call:



<?php
session_start();
//Add session_start to top of each page//
require_once('includes/config.php');
require_once('includes/header.php');
include_once('includes/category.php');

?>
<link rel="stylesheet" href="css/dd.css">
<div id="menu">
<a class="item" href="drop_index.php">Home</a> -
<a class="item" href="create_topic.php">Create a topic</a> -
<a class="item" href="create_cat.php">Create a category</a>
<div id="userbar">
<?php
if( $user->is_logged_in() )
{
echo 'Hello ' . $_SESSION['user_name'] . '. Not you? <a href="logout.php">Sign out</a>';
}
else
{
echo '<a href="login.php">Sign in</a> or <a href="index.php">create an account</a>.';
}
?>
</div>
</div>

<?php

$category = new Category;
$categories = $category->fetch_all();

?>
<div id ="wrapper">
<h1>Categories</h1>
<section>
<ul>
<?php foreach ($categories as $category) { ?>
<li><a href="category.php?id=<?php echo $category['cat_id']; ?>">
<?php echo $category['cat_title']; ?></a>
</li>
<?php } ?>
</ul>
</section>
</div>
<?php
require_once('includes/footer.php');
?>






php






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Feb 18 '15 at 19:39









ddonche

4491520




4491520








  • 2




    var_dump($pdo), $pdo is not a object(not initialized yet) when prepare is called.
    – Jigar
    Feb 18 '15 at 19:41








  • 1




    @JandenHale: did you initialize a (global) variable with the name $pdo before that?
    – Willem Van Onsem
    Feb 18 '15 at 20:09










  • It's global in the class, at least I thought it was.
    – ddonche
    Feb 18 '15 at 20:40














  • 2




    var_dump($pdo), $pdo is not a object(not initialized yet) when prepare is called.
    – Jigar
    Feb 18 '15 at 19:41








  • 1




    @JandenHale: did you initialize a (global) variable with the name $pdo before that?
    – Willem Van Onsem
    Feb 18 '15 at 20:09










  • It's global in the class, at least I thought it was.
    – ddonche
    Feb 18 '15 at 20:40








2




2




var_dump($pdo), $pdo is not a object(not initialized yet) when prepare is called.
– Jigar
Feb 18 '15 at 19:41






var_dump($pdo), $pdo is not a object(not initialized yet) when prepare is called.
– Jigar
Feb 18 '15 at 19:41






1




1




@JandenHale: did you initialize a (global) variable with the name $pdo before that?
– Willem Van Onsem
Feb 18 '15 at 20:09




@JandenHale: did you initialize a (global) variable with the name $pdo before that?
– Willem Van Onsem
Feb 18 '15 at 20:09












It's global in the class, at least I thought it was.
– ddonche
Feb 18 '15 at 20:40




It's global in the class, at least I thought it was.
– ddonche
Feb 18 '15 at 20:40












2 Answers
2






active

oldest

votes


















29














It looks like your $pdo variable is not initialized.
I can't see in the code you've uploaded where you are initializing it.



Make sure you create a new PDO object in the global scope before calling the class methods. (You should declare it in the global scope because of how you implemented the methods inside the Category class).



$pdo = new PDO('mysql:host=localhost;dbname=test', $user, $pass);





share|improve this answer























  • That's it right there, I have it in my config file as $db instead of $pdo. Thanks!
    – ddonche
    Feb 18 '15 at 20:42






  • 1




    Dont use global ! Add in your function: fetch_data(PDO $pdo, $cat_id)
    – delato468
    Apr 23 '18 at 16:00










  • In this case, it's more of a constant.
    – ddonche
    Nov 1 '18 at 15:59



















0














You can try/catch PDOExceptions (your configs could differ but the important part is the try/catch):



try {
$dbh = new PDO(
DB_TYPE . ':host=' . DB_HOST . ';dbname=' . DB_NAME . ';charset=' . DB_CHARSET,
DB_USER,
DB_PASS,
[
PDO::ATTR_PERSISTENT => true,
PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES ' . DB_CHARSET . ' COLLATE ' . DB_COLLATE

]
);
} catch ( PDOException $e ) {
echo 'ERROR!';
print_r( $e );
}


The print_r( $e ); line will show you everything you need, for example I had a recent case where the error message was like unknown database 'my_db'.






share|improve this answer





















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    2 Answers
    2






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    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    29














    It looks like your $pdo variable is not initialized.
    I can't see in the code you've uploaded where you are initializing it.



    Make sure you create a new PDO object in the global scope before calling the class methods. (You should declare it in the global scope because of how you implemented the methods inside the Category class).



    $pdo = new PDO('mysql:host=localhost;dbname=test', $user, $pass);





    share|improve this answer























    • That's it right there, I have it in my config file as $db instead of $pdo. Thanks!
      – ddonche
      Feb 18 '15 at 20:42






    • 1




      Dont use global ! Add in your function: fetch_data(PDO $pdo, $cat_id)
      – delato468
      Apr 23 '18 at 16:00










    • In this case, it's more of a constant.
      – ddonche
      Nov 1 '18 at 15:59
















    29














    It looks like your $pdo variable is not initialized.
    I can't see in the code you've uploaded where you are initializing it.



    Make sure you create a new PDO object in the global scope before calling the class methods. (You should declare it in the global scope because of how you implemented the methods inside the Category class).



    $pdo = new PDO('mysql:host=localhost;dbname=test', $user, $pass);





    share|improve this answer























    • That's it right there, I have it in my config file as $db instead of $pdo. Thanks!
      – ddonche
      Feb 18 '15 at 20:42






    • 1




      Dont use global ! Add in your function: fetch_data(PDO $pdo, $cat_id)
      – delato468
      Apr 23 '18 at 16:00










    • In this case, it's more of a constant.
      – ddonche
      Nov 1 '18 at 15:59














    29












    29








    29






    It looks like your $pdo variable is not initialized.
    I can't see in the code you've uploaded where you are initializing it.



    Make sure you create a new PDO object in the global scope before calling the class methods. (You should declare it in the global scope because of how you implemented the methods inside the Category class).



    $pdo = new PDO('mysql:host=localhost;dbname=test', $user, $pass);





    share|improve this answer














    It looks like your $pdo variable is not initialized.
    I can't see in the code you've uploaded where you are initializing it.



    Make sure you create a new PDO object in the global scope before calling the class methods. (You should declare it in the global scope because of how you implemented the methods inside the Category class).



    $pdo = new PDO('mysql:host=localhost;dbname=test', $user, $pass);






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Sep 28 '18 at 13:19









    Funk Forty Niner

    80.5k1247101




    80.5k1247101










    answered Feb 18 '15 at 20:18









    Dan Revah

    2,95023155




    2,95023155












    • That's it right there, I have it in my config file as $db instead of $pdo. Thanks!
      – ddonche
      Feb 18 '15 at 20:42






    • 1




      Dont use global ! Add in your function: fetch_data(PDO $pdo, $cat_id)
      – delato468
      Apr 23 '18 at 16:00










    • In this case, it's more of a constant.
      – ddonche
      Nov 1 '18 at 15:59


















    • That's it right there, I have it in my config file as $db instead of $pdo. Thanks!
      – ddonche
      Feb 18 '15 at 20:42






    • 1




      Dont use global ! Add in your function: fetch_data(PDO $pdo, $cat_id)
      – delato468
      Apr 23 '18 at 16:00










    • In this case, it's more of a constant.
      – ddonche
      Nov 1 '18 at 15:59
















    That's it right there, I have it in my config file as $db instead of $pdo. Thanks!
    – ddonche
    Feb 18 '15 at 20:42




    That's it right there, I have it in my config file as $db instead of $pdo. Thanks!
    – ddonche
    Feb 18 '15 at 20:42




    1




    1




    Dont use global ! Add in your function: fetch_data(PDO $pdo, $cat_id)
    – delato468
    Apr 23 '18 at 16:00




    Dont use global ! Add in your function: fetch_data(PDO $pdo, $cat_id)
    – delato468
    Apr 23 '18 at 16:00












    In this case, it's more of a constant.
    – ddonche
    Nov 1 '18 at 15:59




    In this case, it's more of a constant.
    – ddonche
    Nov 1 '18 at 15:59













    0














    You can try/catch PDOExceptions (your configs could differ but the important part is the try/catch):



    try {
    $dbh = new PDO(
    DB_TYPE . ':host=' . DB_HOST . ';dbname=' . DB_NAME . ';charset=' . DB_CHARSET,
    DB_USER,
    DB_PASS,
    [
    PDO::ATTR_PERSISTENT => true,
    PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
    PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES ' . DB_CHARSET . ' COLLATE ' . DB_COLLATE

    ]
    );
    } catch ( PDOException $e ) {
    echo 'ERROR!';
    print_r( $e );
    }


    The print_r( $e ); line will show you everything you need, for example I had a recent case where the error message was like unknown database 'my_db'.






    share|improve this answer


























      0














      You can try/catch PDOExceptions (your configs could differ but the important part is the try/catch):



      try {
      $dbh = new PDO(
      DB_TYPE . ':host=' . DB_HOST . ';dbname=' . DB_NAME . ';charset=' . DB_CHARSET,
      DB_USER,
      DB_PASS,
      [
      PDO::ATTR_PERSISTENT => true,
      PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
      PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES ' . DB_CHARSET . ' COLLATE ' . DB_COLLATE

      ]
      );
      } catch ( PDOException $e ) {
      echo 'ERROR!';
      print_r( $e );
      }


      The print_r( $e ); line will show you everything you need, for example I had a recent case where the error message was like unknown database 'my_db'.






      share|improve this answer
























        0












        0








        0






        You can try/catch PDOExceptions (your configs could differ but the important part is the try/catch):



        try {
        $dbh = new PDO(
        DB_TYPE . ':host=' . DB_HOST . ';dbname=' . DB_NAME . ';charset=' . DB_CHARSET,
        DB_USER,
        DB_PASS,
        [
        PDO::ATTR_PERSISTENT => true,
        PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
        PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES ' . DB_CHARSET . ' COLLATE ' . DB_COLLATE

        ]
        );
        } catch ( PDOException $e ) {
        echo 'ERROR!';
        print_r( $e );
        }


        The print_r( $e ); line will show you everything you need, for example I had a recent case where the error message was like unknown database 'my_db'.






        share|improve this answer












        You can try/catch PDOExceptions (your configs could differ but the important part is the try/catch):



        try {
        $dbh = new PDO(
        DB_TYPE . ':host=' . DB_HOST . ';dbname=' . DB_NAME . ';charset=' . DB_CHARSET,
        DB_USER,
        DB_PASS,
        [
        PDO::ATTR_PERSISTENT => true,
        PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
        PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES ' . DB_CHARSET . ' COLLATE ' . DB_COLLATE

        ]
        );
        } catch ( PDOException $e ) {
        echo 'ERROR!';
        print_r( $e );
        }


        The print_r( $e ); line will show you everything you need, for example I had a recent case where the error message was like unknown database 'my_db'.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 29 '16 at 14:49









        aesede

        3,91922629




        3,91922629






























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