When working with the sqlsrv_query command I can request data from the MSSQL server.
This works
But!
When I want to add data it returns the error [error:array].
The code I use for this is:
$tsql= "INSERT INTO dbo.VERLOF_events (id,
username,
soort,
afdeling,
description,
evdate,
trdate)
VALUES
(?, ?, ?, ?, ?, ?, ?)";
$var = array('', $username, $soort, $afdeling, $description, $evdate, $trdate);
if (!sqlsrv_query($conn, $tsql, $var))
{
die('Error: ' . sqlsrv_errors());
}
echo "1 record added";
The array values are set in the POST statement.
$afdeling = $row['Afdeling'];
$submit = #$_POST['submit'];
$description = #$_POST["description"];
$evdate = #$_POST["evdate"];
$trdate = #$_POST["trdate"];
$username = #$_SESSION['username'];
$soort = #$_POST['Dagen'];
Why does it return the array error?
I looked it up but could not find the problem returning the error.
Any help is appreciated!
The problem is probably you're trying to add an empty value in the id field. If you set identity on it with auto-numbering, you don't need to include it in your query :
$tsql= "INSERT INTO dbo.VERLOF_events (
username,
soort,
afdeling,
description,
evdate,
trdate)
VALUES
(?, ?, ?, ?, ?, ?)";
$var = array($username, $soort, $afdeling, $description, $evdate, $trdate);
if (!sqlsrv_query($conn, $tsql, $var))
{
die('Error: ' . sqlsrv_errors());
}
echo "1 record added";
Related
Can't insert data into DB . When i remove user_id then data is inserted. Please check below my code and help me.
function adddata($data) {
global $db;
if (is_array($data)) {
$stmt = $db->prepare('INSERT INTO `pay` (id, payment, status, itemid, time, user_id) VALUES(?, ?, ?, ?, ?, ?');
$userid = 2;
$stmt->bind_param(
'sdssss',
$data['txn_id'],
$data['payment_amount'],
$data['payment_status'],
$data['item_number'],
date('Y-m-d H:i:s'),
$userid
);
$stmt->execute();
$stmt->close();
return $db->insert_id;
}
return false;
}
It's subtle, but your SQL string is missing a closing bracket:
$stmt = $db->prepare('INSERT INTO `pay` (...) VALUES (?, ?, ?, ?, ?, ?)');
Where the VALUES list was not properly closed.
A lot of problems can be detected and resolved by enabling exceptions in mysqli so mistakes aren't easily ignored. This should show up as a SQL error in your logs.
I am getting this error when trying to insert data into my database from my website:
Error: INSERT INTO newtask (new_category, new_department, new_required,
new_name, new_address, new_contact, new_email, new_logged, new_description)
VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?)
You have an error in your SQL syntax; check the manual that corresponds to
your
MySQL server version for the right syntax to use near '?, ?, ?, ?, ?, ?, ?,
?,
?)' at line 1
Here is my code:
<?php
$servername = 'localhost';
$username = 'root';
$password = '';
$dbname = 'tasks_db';
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO newtask (new_category, new_department, new_required,
new_name, new_address, new_contact, new_email, new_logged, new_description)
VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?)";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
Not sure what i am doing wrong here?
This is a prepared statement, so you need to prepare it, bind your values, and then execute it:
$stmt = $conn->prepare("INSERT INTO newtask (new_category, new_department, new_required,
new_name, new_address, new_contact, new_email, new_logged, new_description)
VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?)");
// One "s" per placeholder value, plus one value per "s" after
$stmt->bind_param("sssssssss", $new_category, $new_department, ...);
if ($stmt->execute()) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
Where $new_category is whatever value is going into that column and so on.
This is all covered in the documentation.
The mistake is you were trying to run a SQL query with placeholder values and no data. ? is not a valid value in MySQL. It's replaced at the driver level by the bind_param operation on statements produced with prepare. The query function executes code as-is with no changes.
I'm trying to use prepared statements to enter data in a database. The unprepared statement works but this prepared statement does not. I can't find out why.
Prepared version:
$stmt = $mysqli->prepare("INSERT INTO videos (file_name, upload_by, date, path)
VALUES (?, ?, ?, ?)");
$stmt->bind_param('ssss', $newstring, $id, $date->format('Y-m-d'), $location);
$stmt->execute();
Unprepared version:
$sql = "INSERT INTO videos (file_name, upload_by, date, path) VALUES ('$newstring', '$id', '
$date', 'Nominator/$location$newstring')";
mysqli_query($mysqli, $sql);
Replace $stmt-execute(); with $stmt->execute();
Also, don't use date and path in query. Rename them with some other name like date1 and path1.
Update your Query like below that will surely work (Tested Offline):
<?php
$mysqli = new mysqli('localhost', 'root', '', 'test2');
if ($mysqli->errno) {
printf("Connect failed: %s\n", $mysqli->error);
exit();
}
$stmt = $mysqli->prepare("INSERT INTO videos (file_name, upload_by, date1, path1) VALUES (?, ?, ?, ?)");
$stmt->bind_param('ssss', $file_name, $upload_by, $date1, $path1);
$date1 = date("Y-m-d");
$file_name = "test.jpg";
$upload_by = "amit";
$path1 = "test";
if ($result = $stmt->execute()){
echo "success";
$stmt->free_result();
} else {
echo "error";
}
$stmt->close();
?>
You are binding your parameter twice, if you are using only ?, don't bind parameter again just execute directly.
//Prepare your query first
$stmt = $mysqli->prepare("INSERT INTO videos (file_name, upload_by, date, path)
VALUES (?, ?, ?, ?)");
//Just pass your argument and execute directly without binding the parameter (The parameter is binded already)
$stmt->execute('ssss', $newstring, $id, $date->format('Y-m-d'), $location);
I'm trying to insert some value into my database, but I got no result, but the code got no error, and the result label said it is succeed. My database connection working. How to check the issue here, I confused.
My Code Here
// insert new data to menu table
$sql_query = "INSERT INTO tbl_jadwal (Nama_Lokasi, Category_ID, Longitude, Latitude, Phone, Email, Menu_image, Description)
VALUES(?, ?, ?, ?, ?, ?, ?, ?)";
$upload_image = 'upload/images/' . $menu_image;
$stmt = $connect->stmt_init();
if ($stmt->prepare($sql_query))
{
// Bind your variables to replace the ?s
$stmt->bind_param('sssssss',
$nama_lokasi,
$category_ID,
$longitude,
$latitude,
$phone,
$email,
$upload_image,
$description
);
// Execute query
$stmt->execute();
// store result
$result = $stmt->store_result();
$stmt->close();
}
This should do, you were missing one s in the param string
$stmt->bind_param('ssssssss',
$nama_lokasi,
$category_ID,
$longitude,
$latitude,
$phone,
$email,
$upload_image,
$description
And you have way too much code. Only a very little part of it is relevant
I can conect to my database just fine, but whenever I try to insert data I get a "Call to a member function bind_param() on a non-object" error, which I know means that I must have mistyped the name of a slot, or something, but I just can't see it, I'm connecting locally, do I have to set up something in MyPhP to allow data to be added from a php file? Thanks in advance.
<?php
include 'connect.php';
if (isset($_POST['Slot1'])) {
$Slot1 = $_POST['Slot1'];
}
if (isset($_POST['Slot2'])) {
$Slot2 = $_POST['Slot2'];
}
if (isset($_POST['Slot3'])) {
$Slot3 = $_POST['Slot3'];
}
if (isset($_POST['Slot4'])) {
$Slot4 = $_POST['Slot4'];
}
if (isset($_POST['Slot5'])) {
$Slot5 = $_POST['Slot5'];
}
if (isset($_POST['Slot6'])) {
$Slot6 = $_POST['Slot6'];
}
if (isset($_POST['Slot7'])) {
$Slot7 = $_POST['Slot7'];
}
$stmt = $db->prepare("INSERT INTO `tabel` (Slot1, Slot 2, Slot3, Slot4,Slot5,
Slot6, Slot7)
VALUES (?, ?, ?, ?, ?, ?,?");
$stmt->bind_param('sssssss',$Slot1,$Slot2,$Slot3,$Slot4,$Slot5,$Slot6,$Slot7);
$stmt->execute();
$stmt->close();
header("Location: Display.php")
?>
You have missed one end parenthese
Replace
$stmt = $db->prepare("INSERT INTO `tabel` (Slot1, Slot 2, Slot3, Slot4,Slot5,
Slot6, Slot7)
VALUES (?, ?, ?, ?, ?, ?,?");
To
$stmt = $db->prepare("INSERT INTO `tabel` (Slot1, Slot 2, Slot3, Slot4,Slot5,
Slot6, Slot7)
VALUES (?, ?, ?, ?, ?, ?,?)");
Try to change your query to
$stmt = $db->prepare("INSERT INTO `tabel` (Slot1, Slot 2, Slot3, Slot4,Slot5,Slot6, Slot7)VALUES (?, ?, ?, ?, ?, ?,?)");