PHP & SQL Insert data Into Table1 if input data exists in Table2 - php

I've been trying to figure out how to insert data received from a form into one table only if the received data exists in another table. If the data doesn't exist it moves onto another query and checks another table for the received data.
This is what I'm trying to do:
function addNewUser($username, $password, $email, $actcode){
$time = time();
$q = "UPDATE ".TBL_RELEASE_CODES." SET code = '$actcode' WHERE code = '$actcode'";
$result = mysql_query($q, $this->connection);
if (!$result || (mysql_numrows($result)) == 0){
$q = "INSERT INTO ".TBL_RELEASE_USERS." VALUES ('$username', '$password', '0', $ulevel, '$email', '$actcode', $time)";
return mysql_query($q, $this->connection);
}
The purpose of this is that when the user submits a special code the system will run checks to see if the code belongs to a certain table.
If it finds the submitted code in a table it will run the insert query associated with the check, if not then it breaks and returns an error saying no match was found.
I'm probably using the code incorrectly as I've been scrounging information from Google searches and testing them out. With no luck yet.
This code is being run off a website using PHP 5 and MySQL.

The first query doesn't do anything -- it sets code to $actcode only in the rows where code is already $actcode. You should use a SELECT, not UPDATE:
$q = "SELECT COUNT(*) FROM ".TBL_RELEASE_CODES." WHERE code = ?";
$stmt = mysqli_prepare($this->connection, $q);
mysqli_stmt_bind_param($stmt, 's', $actcode);
mysqli_stmt_bind_result($stmt, $count);
mysqli_execute($stmt) or die "Query failed: ".mysqli_stmt_error($stmt);
mysqli_stmt_fetch($stmt);
if ($count == 1) {
// Insert into TBL_RELEASE_USERS
} else {
// Return error saying no match found
}
You should also not use the mysql_XXX functions. They're deprecated and make it hard to avoid SQL-injection attacks. My code above uses mysqli_XXX, which supports prepared statements to protect against that. It also has an OO-style API if you like, but I didn't use that above.

Related

How to compare input from a user php post to a MySQL

I am teaching myself php and MySQL, and right now I have a problem with MySQL.
I want to compare the phone number that the user put in with the phone number in MYSQL, and if it is in MYSQL to not register it again.
My code:
<?php
require_once 'connection/connection.php';
// Variables from HTML to php
$worker_Name = $_POST['workerNameFromHtml']; // worker Name
$worker_City = $_POST['workerCityFromHtml']; // workerCity
$worker_career = $_POST['workerCareerFromHtml']; // worker career
$worker_PhoneNumber = $_POST['workerPhonNumberFromHtml']; // worker Phone Number
$worker_SecondPhoneNumber = $_POST['workerSecondPhoneNumberFromHtml']; // worker Second Phone Number
$submt=$_POST['submitFromHtml'];
if($submt){
$qry = ( "SELECT workrPhoneNumber FROM workersTable WHERE workrPhoneNumber = '$worker_PhoneNumber'") or die(mysql_error());
$result = $connect->query($qry);
$num = $result->num_rows;
if ($num == 1) {
$here = "INSERT INTO workersTable VALUES('','$worker_Name','$worker_City','$worker_career','$worker_PhoneNumber','$worker_SecondPhoneNumber')";
$query = $connect->query($here);
print "Successfully added!";
}
else {print "This number has already been entered Thank you for your cooperation!";}}
$connect->close();
So far I have not found a solution to this problem.
your biggest problem here is that you are trying to include variables inside of a string.
"SELECT workrPhoneNumber FROM workersTable WHERE workrPhoneNumber = '$worker_PhoneNumber'"
If you want to do it this way, you need to concatenate your variables with your string.
"SELECT workrPhoneNumber FROM workersTable WHERE workrPhoneNumber = '".$worker_PhoneNumber."'"
Keep in mind if you do this you will want to sanitize your variables first to prevent SQL injections. Also, when you INSERT variables, you will actually want to use a prepared statement like this:
"INSERT INTO table_name (column1, column2, column3,...) VALUES (value1, value2, value3,...)"
where the 1st set of values are the names of your columns in the database and the second set are your PHP variables you are putting into it.

How to keep temporary mysqli table available in php during statement execution?

I am busy trying to execute a set of statements that involve the use of a temporary table.
My goal is to create the temporary table, insert values to it and then do a like comparison of the temporary tables contents to another table.
These statements are working perfectly in phpmyadmin when executed from RAW SQL, but I'm assuming that the table is not available when I try to insert the data.
Below is the code for my php function + mysqli execution:
function SearchArticles($Tags){
global $DBConn, $StatusCode;
$count = 0;
$tagCount = count($Tags);
$selectText = "";
$result_array = array();
$article_array = array();
foreach($Tags as $tag){
if($count == 0){
$selectText .= "('%".$tag."%')";
}else {
$selectText .= ", ('%".$tag."%')";
}
$count++;
}
$query = "CREATE TEMPORARY TABLE tags (tag VARCHAR(20));";
$stmt = $DBConn->prepare($query);
if($stmt->execute()){
$query2 = "INSERT INTO tags VALUES ?;";
$stmt = $DBConn->prepare($query2);
$stmt->bind_param("s", $selectText);
if($stmt->execute()){
$query3 = "SELECT DISTINCT art.ArticleID FROM article as art JOIN tags as t ON (art.Tags LIKE t.tag);";
$stmt = $DBConn->prepare($query3);
if($stmt->execute()){
$stmt->store_result();
$stmt->bind_result($ArticleID);
if($stmt->num_rows() > 0){
while($stmt->fetch()){
array_push($article_array, array("ArticleID"=>$ArticelID));
}
array_push($result_array, array("Response"=>$article_array));
}else{
array_push($result_array, array("Response"=>$StatusCode->Empty));
}
}else{
array_push($result_array, array("Response"=>$StatusCode->SQLError));
}
}else{
array_push($result_array, array("Response"=>$StatusCode->SQLError));
}
}else{
array_push($result_array, array("Response"=>$StatusCode->SQLError));
}
$stmt->close();
return json_encode($result_array);
}
The first statement executes perfectly, however the second statement gives me the error of:
PHP Fatal error: Call to a member function bind_param() on a non-object
If this is an error to do with the Temp table not existing, how do i preserve this table long enough to run the rest of the statements?
I have tried to use:
$stmt = $DBConn->multi_query(query);
with all the queries in one, but i need to insert data to one query and get data from the SELECT query.
Any help will be appreciated, thank you!
You have a simple syntax error use the brackets around the parameters like this
INSERT INTO tags VALUES (?)
This is not an issue with the temporary table. It should remain throughout the same connection (unless it resets with timeout, not sure about this part).
The error is that $stmt is a non-object. This means that your query was invalid (syntax error), so mysqli refused to create an instance of mysqli_stmt and returned a boolean instead.
Use var_dump($DBConn->error) to see if there are any errors.
Edit: I just noticed that your query $query2 is INSERT INTO tags VALUES ? (the ; is redundant anyway). If this becomes a string "text", this would become INSERT INTO tags VALUES "text". This is a SQL syntax error. You should wrap the ? with (), so it becomes INSERT INTO tags VALUES (?).
In conclusion, change this line:
$query2 = "INSERT INTO tags VALUES ?;";
to:
$query2 = "INSERT INTO tags VALUES (?);";
also note that you don't need the ; to terminate SQL statements passed into mysqli::prepare.

Insert result into multiple tables

EDIT:
Im trying to submit a form with a title and body but i want the title to go to one table and body to go to another table, this in itself i can do but i need the ID generated from the title being inserted into its table to then be inserted into a field in the table the body is inserted so as to keep them linked.
What i have so far: I know its not pretty and its not safe, i will be reworking them once i learn how to do it properly.
if (#$_POST['post'])
{
$body = #$_POST['body'];
$title = #$_POST['title'];
$BoardID = #$_POST['BoardID'];
$MemberID = #$_POST['MemberID'];
$date = date("Y-m-d H:i:s");
include ('connect.php');
$insert = mysql_query("INSERT INTO threads VALUES ('','$BoardID','$title','$date','$MemberID','','')");
if($insert) {
header("location: ?p=posts&thread=$Thread_ID");
exit();
}
}
I need to somehow get $Thread_ID which has been generated in the insert and add that to a second insert for adding body to the post table, if that makes sense.
I tried getting the latest $Thread_ID and adding +1 but if multiple threads are posted at once they might get crossed over.
How would i go about fixing this?
The PHP manual tell us:
This extension Mysql is deprecated as of PHP 5.5.0, and is not recommended for writing new code as it will be removed in the future. Instead, either the mysqli or PDO_MySQL extension should be used.
(see ref.)
You must use mysqli or PDO, to make a connection between PHP and a MySQL database.
mysqli
If you want the id of the inserted row, you can use $mysqli->insert_id (ref)
Example:
$query = "INSERT INTO myCity VALUES (NULL, 'Stuttgart', 'DEU', 'Stuttgart', 617000)";
$mysqli->query($query);
printf ("New Record has id %d.\n", $mysqli->insert_id);
PDO
If you want the id of the inserted row, you can use $dbh->lastInsertId(); (ref)
And don't forget to sanatize all your inputs.
You need to execute both insert queries separately.
$insert = "INSERT INTO threads VALUES ('','$BoardID','$title','$date','$MemberID','','')";
$result = #mysql_query($insert);
$Thread_ID=#mysql_insert_id();
$insert = "INSERT INTO posts VALUES ('','$BoardID',$Thread_ID','$body','$date','$MemberID')";
$result = #mysql_query($insert);
Thanks,

Checking if row exists under criteria (PDO, prepare???)

The code below indicates my attempts to try and find out whether a row exists with the criteria gave in the code. It defaults to the else statement, correctly, but doesn't work with the 'if' statement if the if statement appears to be true (there are no emails down as ashfjks#sdhja.com), and instead the code proceeds. The latter part of this code is mostly to expand on the situation. the row can only exist or not exist so I don't understand why it's not strictly doing one or the other. I am converting into PDO for site secuirty, thats why not all is in PDO, yet. I am sorry if this question is too localised?
$stmt = $pdo->prepare("SELECT * FROM table WHERE email = ?");
$stmt->execute(array("$email"));
$row3 = $stmt->fetch(PDO::FETCH_ASSOC);
while($row = $stmt->fetch()) {
if ( ! $row3) {
// Row3 doesn't exist -- this means no one in the database has this email, allow the person to join
$query = "INSERT INTO table (username, email, password, join_date) VALUES ('$username', '$email', SHA('$password1'), NOW())";
mysqli_query($dbc, $query);
$query = "SELECT * FROM table WHERE username = '$username'";
$data2 = mysqli_query($dbc, $query);
while ($row = mysqli_fetch_array($data2)) {
$recipent = '' . $row['user_id'] . '';
$query = "INSERT INTO messages (recipent, MsgTit, MsgR, MsgA, sender, time, readb, reada, MsgCon) VALUES ('$recipent', '$MsgTit', '$MsgR', '$MsgA', '$sender', NOW(), '$readb', '$reada', '$MsgCon')";
mysqli_query($dbc, $query);
// Aftermath.
echo '<p>Your new account has been successfully created. You\'re now ready to log in. After this you should implement basic character-details on your users profile to begin the game.</p>';
mysqli_close($dbc);
exit();
} }
else {
// An account already exists for this email, so display an error message
echo '<p class="error">An account already exists for this e-mail.</p>';
$email = "";
}
}
Your if statement will never be executed. You need to check the number of rows returned. This is what you want:
Note: I originally used $stmt->rowCount(), but the OP said that didn't work for him. But I'm pretty sure the cause of that error was coming from somewhere else.
if (!($stmt = $pdo->prepare("SELECT * FROM table WHERE email = ?"))) {
//error
}
if (!$stmt->execute(array("$email"))) {
//error
}
//The $row3 var you had was useless. Deleted that.
$count = 0;
while ($row = $stmt->fetch()) {
$count++;
}
//The query returned 0 rows, so you know the email doesn't exist in the DB
if ($count== 0) {
$query = "INSERT INTO table (username, email, password, join_date) VALUES ('$username', '$email', SHA('$password1'), NOW())";
if (!mysqli_query($dbc, $query)) {
//error
}
$query = "SELECT * FROM table WHERE username = '$username'";
if (!($data2 = mysqli_query($dbc, $query))) {
//error
}
while ($row = mysqli_fetch_array($data2)) {
$recipent = '' . $row['user_id'] . '';
$query = "INSERT INTO messages (recipent, MsgTit, MsgR, MsgA, sender, time, readb, reada, MsgCon) VALUES ('$recipent', '$MsgTit', '$MsgR', '$MsgA', '$sender', NOW(), '$readb', '$reada', '$MsgCon')";
if (!mysqli_query($dbc, $query)) {
//error
}
// Aftermath.
echo '<p>Your new account has been successfully created. You\'re now ready to log in. After this you should implement basic character-details on your users profile to begin the game.</p>';
mysqli_close($dbc);
exit();
}
}
//The query did not return 0 rows, so it does exist in the DB
else {
// An account already exists for this email, so display an error message
echo '<p class="error">An account already exists for this e-mail.</p>';
$email = "";
}
And you should totally convert the rest of those queries to use PDO.
+1 to answer from #Geoff_Montee, but here are a few more tips:
Make sure you check for errors after every prepare() or execute(). Report the error (but don't expose your SQL to the user), and fail gracefully.
Note that even though you checked for existence of a row matching $email, such a row could be created in the brief moment of time since your check and before you INSERT. This is a race condition. Even if you SELECT for a row matching $email, you should also use a UNIQUE constraint in the database, and catch errors when you execute the INSERT in case the UNIQUE constraint blocks the insert due to conflict.
SELECT email instead of SELECT *. If you have an index on email, then the query runs more efficiently because it can just check the index for the given value, instead of having to read all the columns of the table when you don't need them. This optimization is called an index-only query.
Likewise use SELECT user_id instead of SELECT *. Use SELECT * only when you really need to fetch all the columns.
Bcrypt is more secure than SHA for hashing passwords.

mySQLi Prepared Statement Select with Escape Characters

I am trying to select from a mySQL table using prepared statements. The select critera is user form input, so I am binding this variable and using prepared statements. Below is the code:
$sql_query = "SELECT first_name_id from first_names WHERE first_name = ?";
$stmt = $_SESSION['mysqli']->prepare($sql_query);
$stmt->bind_param('s', $_SESSION['first_name']);
$stmt->execute();
$stmt->store_result();
if ($stmt->num_rows == '1') {
$stmt->bind_result($_SESSION['first_name_id']);
$stmt->fetch();
} else {
$stmt->close();
$sql_query = "INSERT INTO first_names (first_name) VALUES (?)";
$stmt = $_SESSION['mysqli']->prepare($sql_query);
$stmt->bind_param('s', $_SESSION['first_name']);
$stmt->execute();
$_SESSION['first_name_id'] = $_SESSION['mysqli']->insert_id;
}
$stmt->close();
Obviously my code is just determining whether or not the first_name already exists in the first_names table. If it does, it returns the corresponding ID (first_name_id). Otherwise, the code inserts the new first_name into the first_names table and gets the insert_id.
The problem is when a user enters a name with an escape character ('Henry's). Not really likely with first names but certainly employers. When this occurs, the code does not execute (no select or insert activity in the log files). So it seems like mySQL is ignoring the code due to an escape character in the variable.
How can I fix this issue? Is my code above efficient and correct for the task?
Issue #2. The code then continues with another insert or update, as shown in the code below:
if (empty($_SESSION['personal_id'])) {
$sql_query = "INSERT INTO personal_info (first_name_id, start_timestamp) VALUES (?, NOW())";
} else {
$sql_query = "UPDATE personal_info SET first_name_id = ? WHERE personal_info = '$_SESSION[personal_id]'";
}
$stmt = $_SESSION['mysqli']->prepare($sql_query);
$stmt->bind_param('i', $_SESSION['first_name_id']);
$stmt->execute();
if (empty($_SESSION['personal_id'])) {
$_SESSION['personal_id'] = $_SESSION['mysqli']->insert_id;
}
$stmt->close();
The issue with the code above is that I cannot get it to work at all. I am not sure if there is some conflict with the first part of the script, but I have tried everything to get it to work. There are no PHP errors and there are no inserts or updates showing in the mySQL log files from this code. It appears that the bind_param line in the code may be where the script is dying...
Any help would be very much appreciated.
you should validate/escape user input before sending it to the db.
checkout this mysql-real-escape-string()

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