PHP CodeIgniter exception handling - php

I am working on a system where every thing works fine as long as data is present in the database. But I wanted to test it against empty database. When I did that I got similar issues on almost every page.
An uncaught Exception was encounteredType: ErrorMessage: Call to a
member function row() on boolean
I know what is issue here, but I want to know how I should handle the exception. I don't want the error message to disturb my system in case of empty database. One solution is turning off error_reporting but I am looking for a better solution.

If you are sure there is one row for your query :
$sql = "SELECT * FROM table WHERE id = '1'";
$query = $this->db->query($sql);
if($query->num_rows() == 1)
{
$row = $query->row();
}
2.If the result may contain multiple rows :
$sql = "SELECT * FROM table";
$query = $this->db->query($sql)->result();
if(!empty($query))
{
foreach($query as $row)
{
//do something
}
}
isset will throw warning if the variable does not exist. empty() will not throw that warning

Related

In PHP, why doesn't mysqli_num_rows return an integer for a query with 0 returned rows?

I am having a strange issue with mysqli_num_rows. Searching for this issue, I've only found people having issues where NULL is returned no matter what. I also checked the official documentation for the function, and it says it returns an integer of the number of rows returned by the query. Whenever my query returns 1 row (it never should return more), it behaves as I expect. When the query returns 0 rows, I expect the function to return 0, but it returns NULL. Why doesn't it return 0?
I know that my database connection is good and my query works correctly, because when I look for a record that's in the database, I get an integer back. I just can't figure out why this is returning NULL rather than 0.
function getArtistID($name) {
global $conn;
$query = "SELECT artist_id FROM artist WHERE artist_name LIKE '${name}'";
$result = mysqli_query($conn, $query);
if ($result->num_rows) {
$row = mysqli_fetch_assoc($result);
return $row['artist_id'];
} else {
return 0;
}
}
Here's some code that I used to reproduce a case where num_rows seems to be NULL:
<?php
error_reporting(E_ALL);
$conn = new mysqli('127.0.0.1', 'root', null, 'test');
$sql = "SELECT * FROM dual";
$result = $conn->query($sql);
if ($result === false) {
print "Error: {$conn->error}\n";
}
$n = $result->num_rows;
echo "Dump the num_rows property: ";
var_dump($n);
Output:
Error: No tables used
Notice: Trying to get property of non-object in /Users/bkarwin/Documents/SO/my.php on line 14
Dump the num_rows property: NULL
The notice is because it's invalid to access an object-oriented property of a variable that is not an object. This is a frequent source of confusion for PHP developers, and it's a byproduct of the fact that PHP is a loosely typed language, and functions like query() can return either a result object, or a boolean scalar.
The query() function actually returned a false as $result because of some error. In my code, I checked for this error, and you didn't.
When you run mysqli::query() or mysqli::prepare() or mysqli_stmt::execute(), you must check for error conditions every time.
Something about your query caused an error. It's up to you to check for the error and report it.
Update: I edited some text above to make the explanation better, but it might make some comments below seem out of place.
I just can't figure out why this is returning NULL rather than 0.
We can only guess without seeing the log output; but, it is likely the return value is null because it raised an error instead.
You need to ensure that errors are handled when calling a function, before attempting to use the return value.

My function returns false after moving to a new server

I got php fatal error after transfer server with php v5.6.19, before that I had no problem at all with following script
Fetch data from db table:
function get_department_list($mysqli)
{
$sql = $mysqli->query("SELECT * FROM `dept` ORDER BY `dept_id` ASC");
if($sql->num_rows > 0){
return $sql;
}else{
return false;
}
}
Populate data in HTML:
<ul class="department overflow-scroll text-center">
<?php
$shop = new Shop;
$depts = $shop->get_department_list($mysqli);
while($dept = $depts->fetch_object()){
echo '<li>'.$dept->dept_name.'</li>';
}
?>
</ul>
In the end I got an error:
Fatal error: Call to a member function fetch_object() on boolean in C:\xampp\htdocs\project\include\header.php on line 206
First, you are returning a boolean from your function. So, no wonder PHP says you so.
Second, you should keep the matters separated. a function that works with mysqli should keep all mysqli stuff inside. An return just an array, that can be used anywhere without the need to call mysqli functions again.
function get_department_list($mysqli)
{
$sql = $mysqli->query("SELECT * FROM `dept` ORDER BY `dept_id` ASC");
return $sql->fetch_all();
}
And then use not while but foreach
foreach ($depts as $dept) ...
Besides (and more for the people who may chance to land on this question looking for an answer to their question) you should always set proper error reporting for mysqli, like it shown in this answer
Update your while loop for that case when you get false from $shop->get_department_list() call
updated while like this check for $depts if any data then get $dept:
while($depts && $dept = $depts->fetch_object()){

PHP mysql fatal error Call to a member function fetch_assoc() [duplicate]

This question already has an answer here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Closed 3 years ago.
I'm trying to execute a few queries to get a page of information about some images. I've written a function
function get_recent_highs($view_deleted_images=false)
{
$lower = $this->database->conn->real_escape_string($this->page_size * ($this->page_number - 1));
$query = "SELECT image_id, date_uploaded FROM `images` ORDER BY ((SELECT SUM( image_id=`images`.image_id ) FROM `image_votes` AS score) / (SELECT DATEDIFF( NOW( ) , date_uploaded ) AS diff)) DESC LIMIT " . $this->page_size . " OFFSET $lower"; //move to database class
$result = $this->database->query($query);
$page = array();
while($row = $result->fetch_assoc())
{
try
{
array_push($page, new Image($row['image_id'], $view_deleted_images));
}
catch(ImageNotFoundException $e)
{
throw $e;
}
}
return $page;
}
that selects a page of these images based on their popularity. I've written a Database class that handles interactions with the database and an Image class that holds information about an image. When I attempt to run this I get an error.
Fatal error: Call to a member function fetch_assoc() on a non-object
$result is a mysqli resultset, so I'm baffled as to why this isn't working.
That's because there was an error in your query. MySQli->query() will return false on error. Change it to something like::
$result = $this->database->query($query);
if (!$result) {
throw new Exception("Database Error [{$this->database->errno}] {$this->database->error}");
}
That should throw an exception if there's an error...
Most likely your query failed, and the query call returned a boolean FALSE (or an error object of some sort), which you then try to use as if was a resultset object, causing the error. Try something like var_dump($result) to see what you really got.
Check for errors after EVERY database query call. Even if the query itself is syntactically valid, there's far too many reasons for it to fail anyways - checking for errors every time will save you a lot of grief at some point.
I happen to miss spaces in my query and this error comes.
Ex: $sql= "SELECT * FROM";
$sql .= "table1";
Though the example might look simple, when coding complex queries, the probability for this error is high. I was missing space before word "table1".
Please check if you have already close the database connection or not.
In my case i was getting the error because the connection was close in upper line.

Call to a member function [duplicate]

This question already has an answer here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Closed 3 years ago.
I'm trying to execute a few queries to get a page of information about some images. I've written a function
function get_recent_highs($view_deleted_images=false)
{
$lower = $this->database->conn->real_escape_string($this->page_size * ($this->page_number - 1));
$query = "SELECT image_id, date_uploaded FROM `images` ORDER BY ((SELECT SUM( image_id=`images`.image_id ) FROM `image_votes` AS score) / (SELECT DATEDIFF( NOW( ) , date_uploaded ) AS diff)) DESC LIMIT " . $this->page_size . " OFFSET $lower"; //move to database class
$result = $this->database->query($query);
$page = array();
while($row = $result->fetch_assoc())
{
try
{
array_push($page, new Image($row['image_id'], $view_deleted_images));
}
catch(ImageNotFoundException $e)
{
throw $e;
}
}
return $page;
}
that selects a page of these images based on their popularity. I've written a Database class that handles interactions with the database and an Image class that holds information about an image. When I attempt to run this I get an error.
Fatal error: Call to a member function fetch_assoc() on a non-object
$result is a mysqli resultset, so I'm baffled as to why this isn't working.
That's because there was an error in your query. MySQli->query() will return false on error. Change it to something like::
$result = $this->database->query($query);
if (!$result) {
throw new Exception("Database Error [{$this->database->errno}] {$this->database->error}");
}
That should throw an exception if there's an error...
Most likely your query failed, and the query call returned a boolean FALSE (or an error object of some sort), which you then try to use as if was a resultset object, causing the error. Try something like var_dump($result) to see what you really got.
Check for errors after EVERY database query call. Even if the query itself is syntactically valid, there's far too many reasons for it to fail anyways - checking for errors every time will save you a lot of grief at some point.
I happen to miss spaces in my query and this error comes.
Ex: $sql= "SELECT * FROM";
$sql .= "table1";
Though the example might look simple, when coding complex queries, the probability for this error is high. I was missing space before word "table1".
Please check if you have already close the database connection or not.
In my case i was getting the error because the connection was close in upper line.

PHP Error - Login Script

I am creating a new login script/members directory.
I am creating it from scratch without any frameworks (advice on this matter would also be appreciated).
The situation:
// Look up the username and password in the database
$query = "SELECT admin_id, username FROM admin WHERE adminname = '$admin_user' AND password = SHA1('$admin_pass')";
$data = mysqli_query($dbc, $query);
if (mysqli_num_rows($data) == 1) {
This bit of code keeps giving me an error (the last line in particular):
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in /home8/craighoo/public_html/employees/security/dir_admin.php on line 20
When echoing the query I get:
SELECT admin_id, adminname FROM admin WHERE adminname = 'admin' AND password = SHA1('password')
EDIT:
Thanks to everyone. The problem was in my Database column names and the column names I was referencing.
Your query execution is failing. When that happens mysqli_query returns false (boolean value) and when is passed to mysqli_num_rows, you get this error.
Print the query just before executing and check for correctness.
Considering that mysqli_query returns false on failure, and that $data is a boolean, here, I suppose there is an error occuring during the execution of your SQL query.
You could try using mysqli_error to find out what this error is :
$data = mysqli_query($dbc, $query);
if ($data !== false) {
// Do whatever you want with $data
if (mysqli_num_rows($data) == 1) {
//
}
} else {
echo mysqli_error($dbc);
die;
}
Note : echoing the error message and dying, like I did here, is OK while developping your script ; but you should not do that in production.
Instead, in production, you should :
Log the error to a file
Display a nice message to the user
When you have a critical query, it's best to add a die to it like so:
mysqli_query($dbc, $query) or die('Critical error on line #'. __LINE__ .' when attempting to login ...<br>'. mysql_error());
Have you tried running that same query manually thru phpmyadmin or the console? What result do you get?

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