The below script inputs data to a database this takes some information from a form then stores them in to the database. And I'm also using uplodify to upload a image file and store the file name in the database but my issue is this data processing script keeps updating the row ID one never jumps to the second line I tried every thing can some one help me with this or show me what I'm doing wrong.
Also this checks the ID and if it's not equal to 1 then does an insertion if it's equal then update it but this not happening.
The ID is auto incrementing.
My script
<?php
/**
* #author SiNUX
* #copyright 2013
*/
include ('connect.php');
$getId = mysql_query("SELECT ID FROM poiinfo ORDER BY ID DESC LIMIT 1");
$row = mysql_fetch_array($getId);
$poiName = $_REQUEST['Name'];
$poiDes = $_REQUEST['Descrip'];
$poiCon = $_REQUEST['ConInfo'];
//$poiId = $_REQUEST['pID'];
if($row['ID'] != "1"){
$dbData = "INSERT INTO poiinfo(`Name`, `Des.`, `Contact`) VALUES ('$poiName','$poiDes','$poiCon')";
$putData = mysql_query($dbData);
if ($putData){
echo "Data inserted";
}else {
echo "Not Done";
}
}else {
$updLn = "UPDATE `poiinfo` SET `Name`='$poiName',`Des.`='$poiDes',`Contact`='$poiCon'";
$updDone = mysql_query($updLn);
if ($updDone){
echo "Data inserted";
}else {
echo "Not Done";
}
}
?>
I tried u r suggestions but it's still the same now my code for the update is looks like this.
$updLn = "UPDATE `poiinfo` SET `Name`='$poiName',`Des.`='$poiDes',`Contact`='$poiCon' WHERE `ID`='".$row['ID']."'";
But still it keeps up dating the ID 1 not moving on to the next one.
Your update query is missing a WHERE clause. Try this:
$updLn = "UPDATE `poiinfo` SET `Name`='$poiName',`Des.`='$poiDes',`Contact`='$poiCon' WHERE ID = '".$row['ID']."'";
Also be beware of MySQL Injections: http://en.wikipedia.org/wiki/SQL_injection
To check why your update failed, you should call mysql_error in your last else clause :
} else {
echo mysql_error();
}
As for the first problem : if you never insert a new record (I don't see how that could happen, provided your code), you will never have a record whose ID is 2.
$updLn = "UPDATE `poiinfo` SET `Name`='$poiName',`Des.`='$poiDes',`Contact`='$poiCon'";
You need a where clause in this sql to specify a record to update. Currently it is updating all records.
$updLn = "UPDATE `poiinfo` SET `Name`='$poiName',`Des.`='$poiDes',`Contact`='$poiCon' WHERE `ID` = ".$row['id']";";
You will need to set an $id variable for this to work.
Related
Hello I have a table in my database called files and I want to delete a row by its id but whenever I run the code I get the response "Successfully Deleted" but it does not delete the row.
This is my code.
I would appreciate any help.
<?php
require "conn.php";
$id = $_POST["id"];
if($conn){
$sql = "SELECT * FROM files WHERE id LIKE '$id'";
$query = mysqli_query($conn,$sql);
if (mysqli_num_rows($query)>0) {
$sqlDelete = "DELETE FROM files WHERE id LIKE '$id'";
echo "Successfully Deleted";
} else {
echo "Failed to delete";
}
} else {
echo "Connection Error";
}
?>
I guess the issue is because id in database is Int but I'm passing Strings to it so how should I solve this problem?
After you assign
$sqlDelete = "DELETE FROM files WHERE id LIKE '$id'";
you need to
if (mysqli_query($conn, $sqlDelete))
{
echo "Successfully Deleted";
}
in order to execute the delete query. BTW the select statement is redundant.
Please note that your code is very unsafe and SQLi prone. Use prepared statements instead of text substitution.
Try this
$sqlDelete = "DELETE FROM files WHERE id = ".$id;
mysqli_query($conn,$sqlDelete );
What I want is that php check if the client IP address is the same one which in the DB if it already exists, if not to insert new data.
well, it works if the client isn't already inserted in the database, but if he already exists php is skipping the update and trying to insert it again in the database............
I don't know whats wrong with it and couldn't figure out.
Here is my code:
<?php
$corruser = $_SESSION['user_name'];
$client_ip = $_SERVER['REMOTE_ADDR'];
require_once 'connections/dbc.php';
if (!$conn) {
echo "Error connecting the database";
exit();
} else{
$GUI = "SELECT * FROM `customers` WHERE user_name='$corruser'";
$GUI_response = mysqli_query($conn, $GUI);
if (!$row = mysqli_fetch_assoc($GUI_response)) {
echo "Error while query the database";
exit();
} else{
$customer_id = $row['customer_id'];
$check = "SELECT * FROM `users-ipdb` WHERE customer_id='$customer_id' AND user_name='$user_name' ";
$check_response = mysqli_query($conn,$check);
$check_result = mysqli_fetch_array($check_response, MYSQLI_NUM);
if ($check_result[0] > 1) {
$update_ip = "UPDATE `users-ipdb` SET `client_ip`='$client_ip' WHERE customer_id='$customer_id' AND user_name='$corruser' ";
$update_ip_result = mysqli_query($conn, $update_ip);
if (!$update_ip_result) {
echo "ERROR UPDATING DATA BASE";
exit();
}
} else{
$insert_new = "INSERT INTO `users-ipdb`(`customer_id`, `user_name`,`client_ip`) VALUES ('$customer_id','$corruser','$client_ip')";
$insert_new_result= mysqli_query($conn, $insert_new);
if (!$insert_new_result) {
echo "Error inserting new data in the database";
exit();
}
}
}
}
?>
I think you made an error with this code :
$check = "SELECT * FROM `users-ipdb` WHERE customer_id='$customer_id' AND user_name='$user_name' ";
$user_name variable doesn't exist, you should replace it with $corruser
That's why the code never goes into the UPDATE
First, make sure that your condition does work as expected. If customer_id is not a number the following line:
if ($check_result[0] > 1) {
can be possibly evaluated as if(0 > 1) let you read this:
Comparing String to Integer gives strange results.
The other comments mention "UPSERTS" which are explained here https://mariadb.com/kb/en/library/insert-on-duplicate-key-update/
The basic idea is that you can do
INSERT INTO `users-ipdb`(`customer_id`, `user_name`,`client_ip`)
VALUES ('$customer_id','$corruser','$client_ip')"
ON DUPLICATE KEY UPDATE client_ip='$client_ip';
and you get rid of the all the php logic. For this to work properly customer_id and user_name must be both part of the PRIMARY KEY.
If you need to query multiple tables, you can use joins - if you use ON DUPLICATE KEY UPDATE you don't need them, but still a good thing to know - https://mariadb.com/kb/en/library/join-syntax/
Last, but not least - it is a good habit to escape any value which may come from other sources. Maybe it is not your case, but some people tend to create usernames like Joe';DROP TABLE mysql.user;SELECT ' and it will destroy your database, because your query will become
SELECT * FROM `users-ipdb` WHERE customer_id='$customer_id' AND user_name='Joe';DROP TABLE mysql.user;SELECT ''
So be careful.
On a table that displays data from a database, I have a form that has a text area on which a user can type a receipt number and submit to save in a database for a specif row. The PHP code below is what updates the database after the form is submitted.
I want to pick the rest of the details for the specific row so I used the $_POST['id'] on which the receipt has been submitted. The id is the primary key. I'm however having a challenge since I can't fetch data from the database using $id = $_POST['id'];I created before outside the function The update statement works perfectly but the SELECT STATEMENTdoesn't . How do I go about it? Any one?
if(isset($_POST['submit'])) {
$rec = $_POST['receipt'];
$id = $_POST['id'];
//reate connection
$sql = "UPDATE customer SET `receipt` = '".$_POST['receipt']."', `date_entered` = NOW(), `receipt_lock` = 1 WHERE `id` = '".$_POST['tid']."' AND receipt_lock = 0";
if ($conn->query($sql) === TRUE) {
// echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
exit();
$conn->close();
}
function wall(){
global $recp;
global $id;
// Create con
$sql = "SELECT * FROM customer WHERE id ='$id'";
$result1 = mysqli_query($conn, $sql);
$resultarr = mysqli_fetch_assoc($result1); // fetch data
$name = $resultarr['name'];
echo "$name"; //Does not display
$amount = $resultarr['amount'];
$transaction_id = $resultarr['trans_id'];
$date = $resultarr['time_paid'];
}
else {
echo "this is not right!;
}
wall();
Ignoring all the (valid) questions about SQL security and just addressing your problem - how about passing the $id variable as a parameter to your wall function.?
wall($id);
function wall($id){
$sql = "SELECT * FROM customer WHERE id ='$id'";
// ... use prepared statements for security...
...
}
Looks like you are using $_POST['tid'] instead of $_POST['id'] or $id in your SQL-query.
What you are doing there is a big nono, in terms of security. Make sure you escape your POST parameters before adding them inside your query.
$id = $_POST['id'];
$id = mysqli_real_escape_string($conn, $id);
http://php.net/manual/ro/mysqli.real-escape-string.php
Think about sending data as parameters of a function function wall($id) instead of a global parameter.
Continuing with my simple CRUD, I'm stuck again...
So I have a table created called "usuaris" and a column called "id" which is my auto-increment and then another column called "usuari_nom". Now, I want to add "delete function", so when I am displaying the records of my table I've added a to delete it:
<div id="main">
<?php
global $conn;
$query = "SELECT * FROM usuaris";
if($grup_usuaris = mysqli_query($conn, $query)) {
echo "<table>";
echo "<tr><th>Usuaris</th><th>Accions</th></tr>";
while($row = mysqli_fetch_assoc($grup_usuaris)) {
echo "<tr><td>" . $row['usuari_nom'] . "</td><td>Eliminar usuari</td></tr>";
}
echo "</table>";
echo "+ Afegeix Usuari";
mysqli_free_result($grup_usuaris);
} else {
echo "query failed";
echo("Error description: " . mysqli_error($conn));
}
?>
</div>
So now, If I click on "eliminar usuari" it goes to the file where I am adding the query to delete, plus the id of that user; for example: "http://localhost/calendario/elimina_usuari.php?subject=6". But then, in the file elimina_usuari.php, how do I select the id to know what record to delete?
I've thought with $_GET but it doesn't seems to work, either with $_POST:
elimina_usuari.php
<?php
global $conn;
$usuari_id = $_GET['id'];
$query = "DELETE FROM subjects WHERE id = {$usuari_id} LIMIT 1";
$result = mysqli_query($conn, $query);
if ($result && mysqli_affected_rows($conn) == 1) {
redirect_to("calendari.php");
} else {
echo "no eliminat";
}
?>
Any clue how can I get its id? Should I take it from the url somehow?
Thanks
you're doing fine.
just need to change this
$usuari_id = $_GET['id'];
to
$usuari_id = $_GET['subject'];
as you're setting subject instead of id in your url
http://localhost/calendario/elimina_usuari.php?subject=6
^
and if you want to process id, like $_GET['id'], you need to change URL.
"http://localhost/calendario/elimina_usuari.php?id=6"
^ change here
EDIT
as per your comment,
you can use any $variable to $_POST or $_GET, it has nothing to do with the database column name.
Like you can use following.
"http://localhost/calendario/elimina_usuari.php?eve_mf=6"
and on elimina_usuari.php page,
$id = $_GET['eve_mf'];
and second part, why can I do that and I don't need to call it id as it is called in my db table?
Again, it's not the issue what you call variables in you local environment, all you to do(and should take care of) is to put right parameters in your sql query.
$query = "DELETE FROM subjects WHERE id = {$usuari_id} LIMIT 1";
Here id is the name of your column name in your database. You can't change it here if you even want it to.
however, $usuari_id is your local variable, and you can change it whatever you want.
Hope I've explained what you're looking for :)
You can get the id with $_GET['subject'].
Please be aware about SQL injection as you are wrongly get the id of the user to be deleted:
$usuari_id = mysqli_real_escape_string($conn, $_GET['subject']);
<?php
global $conn;
$usuari_id = $_GET['subject'];
$query = "DELETE FROM subjects WHERE id = {$usuari_id} LIMIT 1";
$result = mysqli_query($conn, $query);
if ($result && mysqli_affected_rows($conn) == 1) {
redirect_to("calendari.php");
} else {
echo "no eliminat";
}
?>
You just need to Get the exact variable name or parameter name which you have sent with your url
I mean see your url contains subject=6
that means you have to get subject instead of id;
please replace this code
$usuari_id = $_GET['id'];
to
$usuari_id = $_GET['subject'];
try this in elimina_usurai.php
<?php
global $conn;
$usuari_id = $_GET['subject'];
$query = "DELETE FROM subjects WHERE id = {$usuari_id} LIMIT 1";
$result = mysqli_query($conn, $query);
if ($result && mysqli_affected_rows($conn) == 1) {
redirect_to("calendari.php");
} else {
echo "no eliminat";
}
?>
I'm updating some MySQL data for specific entries selected through checkboxes, my code for that looks like this:
<?php
if($_POST['code_approve']){
for($i=0;$i<count($_POST['checkbox']);$i++){
$approval_id = $checkbox[$i];
$sql = "UPDATE table SET status='approved', used='processed' WHERE id='$approval_id'";
$result = mysql_query($sql);
}
if($result){
echo "<meta http-equiv=\"refresh\" content=\"0;URL=table.php\">";
}
}
?>
Now I want to know how many checkboxes have been selected, and for every checkbox selected add 25 to table "members" row "balance".
I tried like this:
<?php
if($_POST['code_approve']){
for($i=0;$i<count($_POST['checkbox']);$i++){
$approval_id = $checkbox[$i];
$approvedamount = count($_POST['checkbox'])
$approvedx25 = $approvedamount * 25;
$sql = "UPDATE table SET status='approved', used='processed' WHERE id='$approval_id'";
$sql2 = "UPDATE members SET balance = + '$approvedx25'";
$result = mysql_query($sql);
}
if($result){
echo "<meta http-equiv=\"refresh\" content=\"0;URL=table.php\">";
}
}
?>
But that didn't work at all, what's the proper way to do this ?
//EDIT
Please ignore this question, it worked, i just forgot to add the mysql_query.
Your update query should be something like this:
$sql2 = "UPDATE members SET balance = balance + ".$approvedx25;
The way you wrote it, the update in withing the for clause, thus executed once for every checkbox. I'm not sure you want that, since you already multiply by the number of checkboxes.
Also, mind, your second query ($sql2) is not executed in your code.