This question already has answers here:
Object of class mysqli_result could not be converted to string
(5 answers)
Closed 1 year ago.
i have to find max value from database. for this purpose i have used max() with where clause but when i echo the result then i get this error.
Catchable fatal error: Object of class mysqli_result could not be converted to string in
i have searched alot and tried this,, this and this and some of others but found nothing helpfull...
my code is :
include('connection.php');
$qry = "SELECT MAX(week) FROM reservation WHERE status= 1" ;
$result = mysqli_query($connection,$qry2);
echo $result ;
on the same page other query is working fine but this one is not..
what i want :
basically i want to get the maximum week number where status is = 1
Hope this helps you
$result = mysqli_query("SELECT MAX(week) AS max_week reservation WHERE status= 1");
$row = mysqli_fetch_array($result);
echo $row["max_week"];
Here is corrected code:
include('connection.php');
$qry = "SELECT MAX(week) as max_week FROM reservation WHERE status= 1" ;
$result = mysqli_query($connection,$qry);
while($row = mysqli_fetch_array($result)) {
echo $row['max_week'];
}
Related
This question already has answers here:
Object of class mysqli_result could not be converted to string
(5 answers)
Closed 1 year ago.
I'm having some issues with my SQL Query. I am trying to run an sql query that saves the output into a variable and then print the variable in a different section of code:
Query:
$sql = "select Count(distinct `Customer Name`) as columnNameCount from allservers";
$result = mysqli_query($DBcon, $sql);
Display variable:
<h3 align="center"><?php echo $resultarr;?></h3>
Error message:
Catchable fatal error: Object of class mysqli_result could not be converted to string
<?php
$sql = "select Count(distinct `Customer Name`) as columnNameCount from allservers";
$result = mysqli_query($DBcon, $sql);
$row = mysqli_fetch_row($result);
?>
<h3 align="center"><?php echo $row[0];?></h3>
Remember: Mysqli_query() return an Object Resourse to your variable $result! Not a String!
You can't directly use it as $result variable!
If you have multiple results, you can loop:
while ($row = $result->fetch_assoc())
echo $row['some_row'];
else you have to fetch as row and display according to index:
$row = $result->fetch_row();
echo $row[0]; // $row[index]
Read The Documentaion
This question already has answers here:
SELECT COUNT(*) AS count - How to use this count
(5 answers)
Closed 1 year ago.
I want to compare the result for count(*) with INT. This is my code;
$sql_check = "SELECT COUNT(*) FROM hd";
$result_check = mysqli_query($link, $sql_check) or die ("ERR");
$row = mysql_fetch_assoc($result_check);
if($row == 1)
{...}
It seem like there is no error. I tried to run it but it skip my if statement. If I tried to
echo $row
It show error
NOTICE: Array to string conversion
If I used code
$sql_check = "SELECT COUNT(*) FROM hd";
$result_check = mysqli_query($link, $sql_check) or die ("ERR");
$row = int($result_check);
if($row == 1)
{...}
It show error
NOTICE: Object of class msqli_result could not converted to int
I have tried to used the answers from previous questions that have been asked before this (same question) but it doesn't work. Can you help me? Please I really need to know where my false.
Thankyou for helping.
Some other questions (that have been asked) and I have tried it but it seem does not work for me:
1. Object of class mysqli_result could not be converted to int and all entries return true
2. mysqli_result could not be converted to int in
3. Object of class mysqli_result could not be converted to int - Can't find my Error
4. Convert SQL query object into integer
you can try this, not sure if it solves the problem!
$sql_check = "SELECT COUNT(*) FROM hd as 'res'";
$result_check = mysqli_query($link, $sql_check) or die ("ERR");
$row = mysql_fetch_assoc($result_check);
if($row['res'] == 1)
{...}
OR YOU CAN TRY THIS
$sql_check = "SELECT COUNT(*) FROM hd as 'res'";
$result_check = mysqli_query($link, $sql_check) or die ("ERR");
$row = mysql_fetch_assoc($result_check);
if((int)$row['res'] == 1)
{...}
This question already has answers here:
SELECT COUNT(*) AS count - How to use this count
(5 answers)
Closed 1 year ago.
I'm currently receiving this error message:
"Catchable fatal error: Object of class mysqli_result could not be
converted to string in ".
I'm trying to count the number of rows in a table. I know you can also use mysqli_num_rows but I thought I could just use SELECT COUNT ( * ) FROM table but it doesn't seem to be working. How can I get the Count(*) to work?
My code:
$query = "SELECT COUNT(*) FROM category";
$select_all_categories = mysqli_query($connection, $query);
echo "<div class='huge'> $select_all_categories </div>"
This won't work because mysqli_query returns an object filled with data, data that needs to be fetched by either mysqli_fetch_array or mysqli_fetch_assoc.
If you desire to get the count, you would need to do this instead:
$query = "SELECT COUNT(*) as count FROM category";
$select_all_categories = mysqli_query($connection, $query);
$data = mysqli_fetch_assoc($select_all_categories);
echo "<div class='huge'> " . $data['count']. " </div>"
$select_all_categories is an object, it is not a simple string. Therefor, you cannot use echo with it.
You will need to get the actual results before you can even get to the point of being a viable string.
In your case, for a simple count:
$query = "SELECT COUNT(*) AS 'count' FROM category";
$select_all_categories = mysqli_query($connection, $query);
$rows = $select_all_categories->fetch_assoc();
$string = $rows[0]['count'];
echo "<div class='huge'> $string </div>"
Notice that I gave the column the alias of count in the MySQL Query.
Your query (including COUNT(*)) is fine, it is not generating the error here. The error you're getting is a PHP Error - the script itself has issues. PHP is telling you that you can't echo an object. You can only echo strings, though PHP will convert most data types to strings for you (like integers for example) if you try to echo them.
$select_all_categories variable it's a object, you can read about it in documentation. You can use with that methods fetch_all, fetch_assoc, fetch_row, etc.
In your case I recommend use fetch_row, example:
<?php
$connection = new mysqli('127.0.0.1', 'root', 'www', 'ccc');
$select_all_categories = $connection->query('SELECT COUNT(*) FROM `category`');
var_dump($select_all_categories->fetch_row()); // fetch_row() will return "1" for my database
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 months ago.
I don't know what the problem is with this line or how to fix it, before was okay and now I'm getting this error:
mysqli_fetch_object() expects parameter 1 to be mysqli_result
Here is my PHP code:
<?php
}
if($_GET['action']=="user_info")
{
$userid = $_GET['user_id'];
$query = "SELECT * FROM user WHERE user_id ='{$userid}'";
$result = mysqli_query($link, $query);
$user = mysqli_fetch_object($result);
$queryt = "SELECT * FROM user_title WHERE id='".$user->title."'";
$resultt = mysqli_query($link, $queryt);
$rowt = mysqli_fetch_object($resultt);
$title = $rowt->name;
$sorgu = "select * from pub_author where user_id='$userid'";
$publications = mysqli_query($link, $sorgu);
while($a = mysqli_fetch_object($publications))
{
$ids .= $a->pub_id . ',';
}
$ids = rtrim($ids,",");
$sorgu2 = "select count(id) as total , year from publication where id IN ($ids)
GROUP BY YEAR(`year`) order by `year` ";
$publications2 = mysqli_query($link, $sorgu2);
while($a2 = mysqli_fetch_object($publications2))
{
$mount = explode('-', $a2->year);
$accyaz[$mount[0]] = $a2->total;
}
}
?>
As far as your exact error is concerned one of your query is failing, the following steps might help. Ofcourse you question looks duplicate but here are some of the things that addresses your question
Your first query should be like this, with no curly braces, ofcourse untill you have explicitly ids wrapped in curly braces in your table.
SELECT * FROM user WHERE user_id ='$userid'
Secondly you are executing multiple queries so you might wanna consider error checking if your query executes properly or not(because of syntax error columns mismatch table name mismatch many more possibilities): do error checking like this as for while($a...) part
if ($result=mysqli_query($link, $sorgu);)
{
while($a=mysqli_fetch_object($result))
{
$ids .= $a->pub_id . ',';
}
$sorgu2 = "select count(id) as total , year from publication where id IN ($ids) GROUP BY YEAR(`year`) order by `year` ";
//... Your further code
}
else
{
echo "Something went wrong while executing query :: $sorgu";
}
Third i see your are getting pub_id make a comma seperated list of it so that you can give it as a parameter in your last query which is a long shot, why not use sub query for you IN clause like this:
SELECT
COUNT(id) as total, year
FROM publication
where id
IN (
SELECT pub_id FROM pub_author WHERE user_id='$userid'
)
GROUP BY `year`
order by `year`;
The error you are stating translates to this: The query fails somehow, instead of running the mysqli_query($link, $sorgu); line echo $sorgu, go to phpmyadmin and test your query, if it is bad, fix it in phpmyadmin until it works and set it up in the code correctly
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 2 years ago.
I try to google it but, i really dont understand what is the problem with this query. Here is code
include_once("includes/db_connection.php");
//Upit za prikaz pitanja!
$listaPitanja = "";
$sql = "SELECT id, username, question FROM useroptions ORDER BY DESC";
$user_query = mysqli_query($db_connection, $sql);
$pitanjaCount = mysqli_num_rows($user_query); //line 8
if ($pitanjaCount > 0) {
while ($row = mysqli_fetch_array($sql)) { //line 10
$id = $row['id'];
$question = $row['question'];
$username = $row['username'];
$listaPitanja .= '<div id="brojOdgovora">'.$id.'</div>
<div id="tekstPitanja"><h3>'.$question.'</h3></div>
<div id="userPitanja"><h6>'.$username.'</h6></div>';
}
} else {
$listaPitanja = "There is no inserted questions!";
}
This query gives me nothing. Just this error mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in something on line 8, and if i delete ORDER BY DESC there is some error on line 10?
Sorry if it repeated, but i have no idea to solve this!! Thank you!
Your SQL statement has no ORDER column:
$sql = "SELECT id, username, question FROM useroptions ORDER BY DESC";
Change it with the correct column name:
$sql = "SELECT id, username, question FROM useroptions ORDER BY column_name DESC";
Probably, mysqli_query is returning false instead of mysqli_result object.
To add segarci,
$row = mysqli_fetch_array($sql)
should be
$row = mysqli_fetch_array($user_query)