This question already has answers here:
Object of class mysqli_result could not be converted to string
(5 answers)
Closed 2 years ago.
if(isset($_POST["id"])){
$stmt = $mysqli->prepare("SELECT * FROM account WHERE id = ?");
$stmt->bind_param("s", $_POST["id"]);
$stmt->execute();
$row = $stmt->get_result();
echo json_encode($row);
}
For some reason, the result is empty and produces this error.
Trying to access array offset on value of type null
What am I doing wrong? nothing I find seems to work.
The call to $stmt->get_result(); just gets the result from the mysqli_stmt object into a mysqli_result object, you still have to fetch result rows as a separate call
if(isset($_POST["id"])){
$stmt = $mysqli->prepare("SELECT * FROM account WHERE id = ?");
$stmt->bind_param("s", $_POST["id"]);
$stmt->execute();
$result = $stmt->get_result();
//You still need some version of a FETCH after $stmt->get_result();
$row = $result->fetch_assoc();
echo json_encode($row);
}
Related
This question already has answers here:
Single result from database using mysqli
(6 answers)
Closed 5 months ago.
I am trying to get a value from column "odznak" in "users" tab for user "user01" and store it in variable $odznak (for searching in another tab.
$stmt = $conn->prepare("SELECT odznak FROM users WHERE username = 'user01'");
$stmt->execute();
$result = $stmt;
$odznak;
You need to fetch the data (say into an associative array)
On the other hand, as a good practice, please use parameterized prepared statement in your select query
So, change to:
$stmt = $conn->prepare("SELECT odznak FROM users WHERE username = ?");
$stmt->bind_param("s", 'user01');
$stmt->execute();
$result = $stmt->get_result();
$row = $result->fetch_assoc();
$odznak=$row["odznak"];
Now, $odznak is the retrieved data
I've been doing SQL for over a year now, and have became completely stuck. For some reason, i'm not able to return any values from this table as I get the error
mysqli_fetch_array(): Argument #1 ($result) must be of type mysqli_result, mysqli_stmt given
I'm completely floored as to why this is happening, as i've used these kind of queries in the past
The code i'm using is
$user = "testuser";
$q = $conn->prepare("SELECT * FROM users WHERE username = ?");
$q->bind_param("s", $user);
$q->execute();
while($row = mysqli_fetch_array($q))
var_dump($row);
If I do var_dump($q), then I get an object object(mysqli_stmt)#3 (10) with no errors and the correct amount of fields. I'm just not able to read anything from this for some reason.
Thanks!
You need to call a get_result() before you can fetch your data
$user = "testuser";
$stmt = $conn->prepare("SELECT * FROM users WHERE username = ?");
$stmt->bind_param("s", $user);
$stmt->execute();
$result = $stmt->get_result();
while($row = $result->fetch_assoc()) {
var_dump($row);
}
PS. better to use fetch_assoc() instead of fetch_array()
This question already has answers here:
mysqli_stmt::bind_result(): Number of bind variables doesn't match number of fields in prepared statement
(2 answers)
Closed 1 year ago.
Iam getting this error
[07-Sep-2017 11:48:47 UTC] PHP Warning: mysqli_stmt::bind_result():
Number of bind variables doesn't match number of fields in prepared
statement
$stmt = $con->prepare("SELECT * FROM table where Id =?");
$stmt->bind_param("s", $_POST['Id']);
$stmt->execute();
$result = $stmt-> bind_result($Id);
$numRows = $result->num_rows;
if($numRows > 0) {
if($row = $result->fetch_assoc())
{
$Taxname=$row['TaxName'];
$Tid=$row['Id'];
}}
You need to alter your query. Instead of * you need to instead opt for picking out the data you actually want from the database.
For example, if the table table has columns Id,TaxName then you would execute like so:
<?php
$sqlData = array();
$stmt = $con->prepare("SELECT Id,TaxName FROM table where Id =?");
$stmt->bind_param("i", $_POST['Id']); //id is an integer? this should be i instead of s
$stmt->execute();
$stmt->store_result(); //store the result, you missed this
$stmt-> bind_result($Id,$TaxName); //bind_result grabs the results and stores in a variable
$numRows = $stmt->num_rows; //see the correction I made here
if($numRows >0){
while ($stmt->fetch()) { //propper way of looping thru the result set
$sqlData [] = array($Id,$TaxName);
//assoc array would look like:
//$sqlData [] = array("Id" =>$Id,"TaxName" => $TaxName);
}
}
$stmt->close(); //close the connection
?>
Then you would have an array of results that you can use after you've finished with mysqli queries.
Hope this helps you understand it a bit more.
$stmt = $con->prepare("SELECT Id,TaxName FROM table where Id =?");
$stmt->bind_param("s", $_POST['Id']);
$stmt->execute();
$result = $stmt-> bind_result($Id,$TaxName);
$stmt->store_result();
$numRows = $stmt->num_rows;
if($numRows > 0) {
while( $result->fetch_assoc())
{
$newid = $Id;
$newtaxname= $TaxName;
}
print_r($newid)."<br>";
print_r($newtaxname);
}
This code will give you the answer without any warnings.
Reference : http://php.net/manual/en/mysqli-stmt.bind-result.php
mysqli_stmt::bind_result — Binds variables to a prepared statement for
result storage
This question already has answers here:
SELECT statement with fetch_array in mysqli prepared statements
(3 answers)
Closed 1 year ago.
How do i get mysqli_fetch_array with prepared statements? i tried fetch_assoc() but i always get an error telling me that fetch_assoc() is not defined. How do i make it work?
$query = "SELECT * FROM users WHERE
username = ? AND
pass = ? LIMIT 1";
$stmt = $_SESSION['connessione']->prepare($query);
$stmt->bind_param('ss', $username, $password);
$stmt->execute();
$stmt->store_result();
if($stmt->num_rows == 1){
$row = $stmt->fetch_assoc();
// originally it was like this $row = mysqli_fetch_array($result);
// where $result was the query result
$this->login_iduser = $row['id'];
$this->login_profile_pic = $row['pic'];
$this->login_privileges = $row['admin'];
return TRUE;
}
return FALSE;
As you can see above i need to find the $row array.
You'll have to have MySQLND installed to do this. mysqli prepared statements don't result a mysqli_result object natively. You'll need to use get_result
$stmt->execute();
$result = $stmt->get_result();
if($result->num_rows == 1){
$row = $result->fetch_assoc();
I am receiving a fatal error in my php/mysqli code which states that on line 46:
Fatal error: Call to undefined method mysqli_stmt::fetch_assoc() in ...
I just want to know how can I remove this fatal error?
The line of code it is pointing at is here:
$row = $stmt->fetch_assoc();
ORIGINAL CODE:
$query = "SELECT Username, Email FROM User WHERE User = ?";
// prepare query
$stmt=$mysqli->prepare($query);
// You only need to call bind_param once
$stmt->bind_param("s",$user);
// execute query
$stmt->execute();
// get result and assign variables (prefix with db)
$stmt->bind_result($dbUser, $dbEmail);
//get number of rows
$stmt->store_result();
$numrows = $stmt->num_rows();
if ($numrows == 1){
$row = $stmt->fetch_assoc();
$dbemail = $row['Email'];
}
UPDATED CODE:
$query = "SELECT Username, Email FROM User WHERE User = ?";
// prepare query
$stmt=$mysqli->prepare($query);
// You only need to call bind_param once
$stmt->bind_param("s",$user);
// execute query
$stmt->execute();
// get result and assign variables (prefix with db)
$stmt->bind_result($dbUser, $dbEmail);
//get number of rows
$stmt->store_result();
$numrows = $stmt->num_rows();
if ($numrows == 1){
$row = $stmt->fetch_assoc();
$dbemail = $row['Email'];
}
The variable $stmt is of type mysqli_stmt, not mysqli_result. The mysqli_stmt class doesn't have a method "fetch_assoc()" defined for it.
You can get a mysqli_result object from your mysqli_stmt object by calling its get_result() method. For this you need the mysqlInd driver installed!
$result = $stmt->get_result();
row = $result->fetch_assoc();
If you don't have the driver installed you can fetch your results like this:
$stmt->bind_result($dbUser, $dbEmail);
while ($stmt->fetch()) {
printf("%s %s\n", $dbUser, $dbEmail);
}
So your code should become:
$query = "SELECT Username, Email FROM User WHERE User = ?";
// prepare query
$stmt=$mysqli->prepare($query);
// You only need to call bind_param once
$stmt->bind_param("s",$user);
// execute query
$stmt->execute();
// bind variables to result
$stmt->bind_result($dbUser, $dbEmail);
//fetch the first result row, this pumps the result values in the bound variables
if($stmt->fetch()){
echo 'result is ' . dbEmail;
}
Change,
$stmt->store_result();
to
$result = $stmt->store_result();
And
Change,
$row = $stmt->fetch_assoc();
to
$row = $result->fetch_assoc();
You have missed this step
$stmt = $mysqli->prepare("SELECT id, label FROM test WHERE id = 1");
$stmt->execute();
$res = $stmt->get_result(); // you have missed this step
$row = $res->fetch_assoc();
I realized that this code was provided as an answer somewhere on stackoverflow:
//get number of rows
$stmt->store_result();
$numrows = $stmt->num_rows();
I tried it to get the number of rows but realized that i didnt need the line $stmt->store_result();, and it didn't get me my number. I used this:
$result = $stmt->get_result();
$num_of_rows = $result->num_rows;
......
$row = $result->fetch_assoc();
$sample = $row['sample'];
It's best to use mysqlnd as Asciiom pointed out. But if you're in a weird situation where you are not allowed to install mysqlnd, it is still possible to get your data into an associative array without it. Try using the code in this answer
Mysqli - Bind results to an Array