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 5 years ago.
I'm creating quite big import script. I want to see any problematic (unexecuted) SQL queries. I have problem with catching wrong SQL queries with try-catch PHP block.
I have a query:
SELECT id FROM tag WHERE name IN ()
Of course there is an error in it so I want to print queries like that with this code:
$sql = "SELECT id FROM tag WHERE name ".$tagsSql."";
try
{
$query = mysqli_query($this->mysqli, $sql);
$result = $query->fetch_assoc();
}
catch(Exception $e)
{
echo 'Problem with: '.$sql;
print_r($e); die;
}
When running the script PHP just throws me this:
Fatal error: Call to a member function fetch_assoc() on a non-object in C:\www\blackboard-import\index.php on line 227
Why this error isn't catched? Because it's fatal? How can I handle that kind of situations?
I use mysqli to contact with MySQL.
This is no regular exception but a fatal error (which causes PHP to shut down). mysqli_query() will return false on error, this is why your current script is failing, so a solution could be something like the following
try {
$query = $this->msqli->query($sql);
if ($query === FALSE) {
throw new Exception($this->mysqli->error);
}
$result = $query->fetch_assoc();
} catch(Exception $e) {
//...
}
Related
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 5 years ago.
I am working on the below code. Why am I not able to run the query properly? I already check the database connection and it is fine
<?php
$sql = "SELECT dt, events, eventtype FROM events";
$stmt = $mysqli->prepare($sql);
$stmt->execute();
$stmt->bind_result($dt,$events,$eventtype);
$stmt->store_result();
if($stmt->num_rows >0) {
$stmt->fetch();
}
else {
echo "Cant Find The data!";
}
$stmt->close();
$mysqli->close();
echo $dt;
echo $events;
echo $eventtype;
?>
getting this error
Fatal error : Call to a member function execute() on boolean in
/srv/disk1/2555378/www/domain.net/index.php on line 113
This means that the variable $mysqli contains a boolean value, probably false.
According to the php docs, http://php.net/manual/en/mysqli.prepare.php, the function mysqli::prepare will return false in case of an error.
You should use the error variable to get more information, like here: http://php.net/manual/en/mysqli.error.php
This question already has an answer here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Closed 3 years ago.
I'm trying to execute a few queries to get a page of information about some images. I've written a function
function get_recent_highs($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";
$sql .= "table1";
Though the example might look simple, when coding complex queries, the probability for this error is high. I was missing space before word "table1".
Please check if you have already close the database connection or not.
In my case i was getting the error because the connection was close in upper line.
This question already has an answer here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Closed 3 years ago.
I'm trying to execute a few queries to get a page of information about some images. I've written a function
function get_recent_highs($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";
$sql .= "table1";
Though the example might look simple, when coding complex queries, the probability for this error is high. I was missing space before word "table1".
Please check if you have already close the database connection or not.
In my case i was getting the error because the connection was close in upper line.
This question already has answers here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(2 answers)
Closed 3 years ago.
I'm trying to execute a few queries to get a page of information about some images. I've written a function
function get_recent_highs($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";
$sql .= "table1";
Though the example might look simple, when coding complex queries, the probability for this error is high. I was missing space before word "table1".
Please check if you have already close the database connection or not.
In my case i was getting the error because the connection was close in upper line.
This question already has answers here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(2 answers)
Closed 3 years ago.
I'm trying to execute a few queries to get a page of information about some images. I've written a function
function get_recent_highs($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.