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 8 years ago.
I have a connection to my database:
$con=mysqli_connect("localhost","xxxx","xxxxx","xxxxxx");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
which returns no error.
I then try to perform a mysqli_query:
$result = mysqli_query($con,"INSERT INTO entries (name, genre, info , date , website , twebsite, video , tvideo, image, extra_genre)
VALUES ('".$name."', '".$type."', '".$info."', '".$date."', '".$url."', '".$href."', '".$_POST["video"]."', '".$videotype."', 'video images/".$photo."', '".$type2."')");
with the alert whether this insertion to the database has worked or not:
if($result)
{
echo "Success:";
}
else
{
echo "error inserting data into database: ";
}
When I perform my code I get the output error inserting data into database:
But I have no ides why $result is not successful?
How can I debug it?
You are looking for mysqli_error() function.
Problem here is that date is a reserved word, so you have to escape that: `date`
try this:
echo $result;
and then run the sql output.. that should give you an error message.
Related
This question already has answers here:
Syntax error due to using a reserved word as a table or column name in MySQL
(1 answer)
How can I prevent SQL injection in PHP?
(27 answers)
Closed 5 years ago.
I am trying to store form data into database using mysqli but it is generating query error my code is given below....
When ever I try to submit the database connection is generating.. the $_POST is working perfectly.. the error only generating by mysqli_query..
<?php
$name = $_POST["firstname"] . " " . $_POST["lastname"];
$email = $_POST["email"];
$happen = $_POST["whendidhappen"];
$howlong = $_POST["howlong"];
$howmany = $_POST["howmany"];
$describe = $_POST["describe"];
$whattheydid= $_POST["whattheydid"];
$seenmycat = $_POST["seenmycat"];
$anythingelse = $_POST["anythingelse"];
$dbc = mysqli_connect('localhost','root','','abductionreport')
or die('Database connection error');
$query = "INSERT INTO abductionform (firstname, lastname, email,whendidhappen, howlong, describe, whattheydid, seenmycat,anythingelse)VALUES('$name','$name','$email','$happen','$howlong', '$howmany','$describe','$whattheydid', '$seenmycat','$anythingelse')";
$result = mysqli_query($dbc,$query) or die ("Query Error");
mysqli_close($dbc);
?>
<h3>Aliens Abducted Me - Report an Abduction</h3>
<p>Thanks for Submiting the form.</p>
<?php
echo "$name it happend to you on $happen it take $howlong <br>";
echo "Number of aliens: $howmany<br>";
echo "Describe: $describe<br>";
echo "What they did to you: $whattheydid<br>";
echo "Have you seen my cat: $seenmycat<br>";
echo "Anything else : $anythingelse<br>";
echo "Your Email Address is : $email<br>";
?>
DESCRIBE is a mysql keyword. Wrap the column name in backticks. For that matter wrap your table and all columns in backticks. Always check that $_POST elements exist with isset () before trying to access them. Use mysqli prepared statements with placeholders for improved security. Always perform error checking BEFORE posting to SO.
You also have 9 columns and 10 values in your query - this will cause a failure every time.
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 have this code:
$mysqli = new mysqli('localhost','username','password','database');
$insert = "INSERT INTO submissions ( client_name, client_company ) VALUES ('$client_name', '$client_company' )";
if ( $mysqli->query($insert) === TRUE ) {
echo "success";
} else {
echo "Error: " . $mysqli . "<br>" . $mysqli->error;
}
In my development environment it runs perfectly, inserting the variables into the correct table columns.
However in production I get this error:
Catchable fatal error: Object of class mysqli could not be converted
to string in 'path' on line x
I know that:
There is no error with my mysqli connection as the rest of my application uses the credentials etc. fine to retrieve data from my db in both dev/live
That this code runs fine, inserts correctly, doesn't error in my dev environment.
I've looked for solutions to the Error message- and have tried a few, but I remain unconvinced that this is the actual cause for the issue.
PHP try to cast $mysqli to string here:
echo "Error: " . $mysqli . "<br>" . $mysqli->error;
You should do that like this instead
echo "Error: with mysqli <br>" . $mysqli->error;
Or maybe a var vump if you want more details but mysqli->error should be enough
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 6 years ago.
this is my php codes getting Post Successfully response when true but when it failed for some reason.
i got this error.
mysqli_error() expects parameter 1 to be mysqli, null given
$result = mysqli_query($connection, $query) ;
if($result){$response["msg"] = "Post Successfully";
echo json_encode($response);} else {$response["msg"] = "failed ".mysqli_error($con);
echo json_encode($response);
}
mysqli_close($connection);
}
$con is null
mysqli_error($con); should be mysqli_error($connection); as you have declared connection as $connection.
According to this line:
mysqli_query($connection, $query)
You are using variable $connection for link identifier, and in mysqli_error($con) you are using $con variable which is undefined or NULL, this should be:
mysqli_error($connection)
This question already has an answer here:
Syntax error due to using a reserved word as a table or column name in MySQL
(1 answer)
Closed 7 years ago.
I am currently trying to use PHP as a backend and MYSQL as my database to setup a simple PHP script that will send a friend request.
There are two parameters for a friend request in my MYSQL data base, From, Too. The Database name is send_friendreq and the table in that database is pending_req.
I have tried multiple ways of sending a post, including PostMan and a different addon but everytime I send the post, I get an error from my PHP code which is "Failed". From my understanding this means that it is connecting to the database fine, but it's not actually sending the data too the Database.
I'm not sure if I have the database set up wrong, or if my PHP is wrong but any help would be extrememly appreciated.
Here is my code for the PHP backend
//Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_errno();
}
if (isset($_POST['Username']) && isset($_POST['FriendReq']))
{
$username = $_POST['Username'];
$usernamebeingreq = $_POST['FriendReq'];
//$sqlCheck = "SELECT Username FROM Users WHERE Username = '" . $usernamebeingreq . "'";
//$resultCheck = mysqli_query($con, $sqlCheck);
//if(!$resultCheck)
//{
//echo "Invalid Username";
//}
//else
//{
$sql="INSERT INTO pending_req (To, From) VALUES ('$usernamebeingreq', '$username')";
$result = mysqli_query($con, $sql);
if(!$result)
{
echo 'Failed';
}
else
{
echo 'Friend added!';
}
//}
}
else
{
echo 'Missing Parameters';
}
?>
If you are in need of my database information, I can reveal that!
from and to are reserved words in SQL you have to add backticks arrond:
$sql="INSERT INTO pending_req (`To`, `From`) VALUES ('$usernamebeingreq', '$username')";
or better rename the column.
Hint: use prepared statement. it is much more safty.
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 have written a mysql_query and it is not working.
I want to know the problem via mysqli_error() however it gives the following error mysqli_error() expects parameter 1 to be mysqli
my mysql code is as follows:
$connect = mysqli_connect("localhost","root", "", "tomuman");
$query = mysqli_query($connect, "SELECT id, to FROM messages WHERE read='0'");
and mysqli_error as follows:
echo mysqli_error($query);
What could cause this problem?
You should not use $query as a parameter for mysqli_error()
use
echo mysqli_error($connect);
Just after trying to connect you can also check for specific connection errors with:
if (mysqli_connect_errno()) {
echo mysqli_connect_error();
exit();
}
Add following code in your query to get the mysql error message.
or die (mysqli_error($connect))
as like below
$query = mysqli_query($connect, "SELECT id, to FROM messages WHERE read='0'") or die (mysqli_error($connect));
if ($query) {
echo "success";
}
else {
echo("Error description: " . mysqli_error($connect));
}