mySQLi Prepared Statement Select with Escape Characters - php

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()

Related

Using PHP variable in SQL query

I'm having some trouble using a variable declared in PHP with an SQL query. I have used the resources at How to include a PHP variable inside a MySQL insert statement but have had no luck with them. I realize this is prone to SQL injection and if someone wants to show me how to protect against that, I will gladly implement that. (I think by using mysql_real_escape_string but that may be deprecated?)
<?php
$q = 'Hospital_Name';
$query = "SELECT * FROM database.table WHERE field_name = 'hospital_name' AND value = '$q'";
$query_result = mysqli_query($conn, $query);
while ($row = mysqli_fetch_assoc($query_result)) {
echo $row['value'];
}
?>
I have tried switching '$q' with $q and that doesn't work. If I substitute the hospital name directly into the query, the SQL query and PHP output code works so I know that's not the problem unless for some reason it uses different logic with a variable when connecting to the database and executing the query.
Thank you in advance.
Edit: I'll go ahead and post more of my actual code instead of just the problem areas since unfortunately none of the answers provided have worked. I am trying to print out a "Case ID" that is the primary key tied to a patient. I am using a REDCap clinical database and their table structure is a little different than normal relational databases. My code is as follows:
<?php
$q = 'Hospital_Name';
$query = "SELECT * FROM database.table WHERE field_name = 'case_id' AND record in (SELECT distinct record FROM database.table WHERE field_name = 'hospital_name' AND value = '$q')";
$query_result = mysqli_query($conn, $query);
while ($row = mysqli_fetch_assoc($query_result)) {
echo $row['value'];
}
?>
I have tried substituting $q with '$q' and '".$q."' and none of those print out the case_id that I need. I also tried using the mysqli_stmt_* functions but they printed nothing but blank as well. Our server uses PHP version 5.3.3 if that is helpful.
Thanks again.
Do it like so
<?php
$q = 'mercy_west';
$query = "SELECT col1,col2,col3,col4 FROM database.table WHERE field_name = 'hospital_name' AND value = ?";
if($stmt = $db->query($query)){
$stmt->bind_param("s",$q); // s is for string, i for integer, number of these must match your ? marks in query. Then variable you're binding is the $q, Must match number of ? as well
$stmt->execute();
$stmt->bind_result($col1,$col2,$col3,$col4); // Can initialize these above with $col1 = "", but these bind what you're selecting. If you select 5 times, must have 5 variables, and they go in in order. select id,name, bind_result($id,name)
$stmt->store_result();
while($stmt->fetch()){ // fetch the results
echo $col1;
}
$stmt->close();
}
?>
Yes mysql_real_escape_string() is deprecated.
One solution, as hinted by answers like this one in that post you included a link to, is to use prepared statements. MySQLi and PDO both support binding parameters with prepared statements.
To continue using the mysqli_* functions, use:
mysqli_prepare() to get a prepared statement
mysqli_stmt_bind_param() to bind the parameter (e.g. for the WHERE condition value='$q')
mysqli_stmt_execute() to execute the statement
mysqli_stmt_bind_result() to send the output to a variable.
<?php
$q = 'Hospital_Name';
$query = "SELECT value FROM database.table WHERE field_name = 'hospital_name' AND value = ?";
$statement = mysqli_prepare($conn, $query);
//Bind parameter for $q; substituted for first ? in $query
//first parameter: 's' -> string
mysqli_stmt_bind_param($statement, 's', $q);
//execute the statement
mysqli_stmt_execute($statement);
//bind an output variable
mysqli_stmt_bind_result($stmt, $value);
while ( mysqli_stmt_fetch($stmt)) {
echo $value; //print the value from each returned row
}
If you consider using PDO, look at bindparam(). You will need to determine the parameters for the PDO constructor but then can use it to get prepared statements with the prepare() method.

simple update and insert mysqli in php code can not be executed [duplicate]

This question already has an answer here:
Syntax error due to using a reserved word as a table or column name in MySQL
(1 answer)
Closed 5 years ago.
This little piece of code should be very easy basic coding, yet it doesn't work. The problem is within the INSERT / UPDATE code, because if I delete those and just echo simple text inside of the if/else code everything works just fine.
This is the code I have, whichs gives a HTTP ERROR 500.
$sql2 = mysqli_query($mysqli, "SELECT * FROM koppel WHERE userid = ".$_GET['userid']." AND msgid = ".$_GET['msgid']."");
$row = mysqli_fetch_assoc($sql2);
$check = $_GET['check'];
$msgid = $_GET['msgid'];
$userid = $_GET['userid'];
$ja = 'ja';
$nee = 'nee';
$tabel_content = $row['check'];
$tabel_id = $row['id'];
if ($tabel_content == $ja){
$stmt = $mysqli->prepare("UPDATE koppel SET check = ? WHERE id = ?");
$stmt->bind_param('si',
$nee,
$tabel_id);
$stmt->execute();
$stmt->close();
} elseif ($tabel_content == $nee){
$stmt = $mysqli->prepare("UPDATE koppel SET check = ? WHERE id = ?");
$stmt->bind_param('si',
$ja,
$tabel_id);
$stmt->execute();
$stmt->close();
} else {
$stmt = $mysqli->prepare("INSERT INTO koppel(userid,
msgid,check) VALUES (?, ?, ?)");
$stmt->bind_param('iis', $userid,
$msgid,
$check);
$stmt->execute();
$stmt->close();
}
What am I missing?
I don't see any error there, but make sure $mysqli is a valid mysqli connection to your database.
To debug your problem, try checking your server error logs (they will show the cause of your 500 error, and in which line) or try removing each part of your code until you understand exactly which line is failing.
You can also move all your "execute" and "close" calls to be below the if/elseif/else structure, as it always gets executed, to avoid repeating code.
Also "tabel" should be spelled "table".

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.

mysqli prepared statement without bind_param

I have this code for selecting fname from the latest record on the user table.
$mysqli = new mysqli(HOST, USER, PASSWORD, DATABASE);
$sdt=$mysqli->('SELECT fname FROM user ORDER BY id DESC LIMIT 1');
$sdt->bind_result($code);
$sdt->fetch();
echo $code ;
I used prepared statement with bind_param earlier, but for now in the above code for first time I want to use prepared statement without binding parameters and I do not know how to select from table without using bind_param(). How to do that?
If, like in your case, there is nothing to bind, then just use query()
$res = $mysqli->query('SELECT fname FROM user ORDER BY id DESC LIMIT 1');
$fname = $res->fetch_row()[0] ?? false;
But if even a single variable is going to be used in the query, then you must substitute it with a placeholder and therefore prepare your query.
However, in 2022 and beyond, (starting PHP 8.1) you can indeed skip bind_param even for a prepared query, sending variables directly to execute(), in the form of array:
$query = "SELECT * FROM `customers` WHERE `Customer_ID`=?";
$stmt = $db->prepare($query);
$stmt->execute([$_POST['ID']]);
$result = $stmt->get_result();
$row = $result->fetch_assoc();
The answer ticked is open to SQL injection. What is the point of using a prepared statement and not correctly preparing the data. You should never just put a string in the query line. The point of a prepared statement is that it is prepared. Here is one example
$query = "SELECT `Customer_ID`,`CompanyName` FROM `customers` WHERE `Customer_ID`=?";
$stmt = $db->prepare($query);
$stmt->bind_param('i',$_POST['ID']);
$stmt->execute();
$stmt->bind_result($id,$CompanyName);
In Raffi's code you should do this
$bla = $_POST['something'];
$mysqli = new mysqli(HOST, USER, PASSWORD, DATABASE);
$stmt = $mysqli->prepare("SELECT `fname` FROM `user` WHERE `bla` = ? ORDER BY `id` DESC LIMIT 1");
$stmt->bind_param('s',$_POST['something']);
$stmt->execute();
$stmt->bind_result($code);
$stmt->fetch();
echo $code;
Please be aware I don't know if your post data is a string or an integer. If it was an integer you would put
$stmt->bind_param('i',$_POST['something']);
instead. I know you were saying without bind param, but trust me that is really really bad if you are taking in input from a page, and not preparing it correctly first.

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

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.

Categories