[Solved] Single result from database using mysqli

When just a single result is needed, then no loop should be used. Just fetch the row right away. In case you need to fetch the entire row into associative array: $row = $result->fetch_assoc(); in case you need just a single value $row = $result->fetch_row(); $value = $row[0] ?? false; The last example will return … Read more

[Solved] How to Search value from input by mysqli in database

check this code .i think it will help you <html> <head> <meta http-equiv=”Content-Type” content=”text/html; charset=utf-8″> <title>PHP, jQuery search demo</title> <link rel=”stylesheet” type=”text/css” href=”https://stackoverflow.com/questions/42627922/my.css”> <script type=”text/javascript” src=”http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js”></script> <script type=”text/javascript”> $(document).ready(function () { $(“input”).keyup(function () { $(‘#results’).html(”); var searchString = $(“#search_box”).val(); var data=”search_text=” + searchString; if (searchString) { $.ajax({ type: “POST”, url: ‘search.php’, data: data, dataType: ‘text’, … Read more

[Solved] Can I use an if on foreach in such a case? [closed]

As was said, if you’re merely going to skip the record then you may as well not retrieve them in the first place (and thus incur the overhead for having to extract them into PHP memory, etc): SELECT * FROM `your_table` WHERE `gender` = ‘female’ AND `taken` = `available`; However, if you have a specific … Read more

[Solved] mysql_select_db() expects parameter 2 to be resource, object given PHP validation

Replace your code with this . <?php try{ $db = mysqli_connect (‘localhost’, ‘root’, ”, ‘car_rental’) or die (“SQL is Off”); } catch (Exception $e){ echo “SQL is Off”; exit; } echo “success”; $email = $_POST[“email”]; $pass = $_POST[“pass”]; mysqli_select_db($db,”car_rental”); $result = mysqli_query($db,”SELECT email, users FROM users WHERE email =$email”); //$row = mysql_fetch_array($result); $row = $result->fetch_array(MYSQLI_ASSOC); … Read more

[Solved] How to display data in MYSQLI using OOP in PHP .

To use PDO is apparently the best way to display data from MySQL database using OOP in PHP $stmt = $Connection->prepare(“SELECT * FROM countries WHERE name = ?”); $stmt->execute([“name_for_search”]); $data = $stmt->fetchAll(); solved How to display data in MYSQLI using OOP in PHP .

[Solved] How do i MD5 crypt password at registration?

You need to use like that for your INSERT Statement: $user_password= md5($_REQUEST[‘user_password’]); Now how can you select md5() password from database after insertion? Its very simple, you must need to follow same step: $user_password= md5($_REQUEST[‘user_password’]); // your input with md5 MD5() PHP Manual As per manual, it will return you the hash as a 32-character … Read more

[Solved] How do you show multiple markers in Google Maps using PHP from the database when searching?

The following is wholly untested! The Object Orientated style of using mySQLi is considerably less verbose than the procedural style and alas your code is vulnerable to SQL injection due to the use of POST data directly used within the SQL statement. You should consider using Prepared Statements to mitigate this threat. It appears that … Read more

[Solved] Turning a SQL row into an array

If you only want specific associative properties from all columns queried from the MySQL call, just set them in an array with their respective properties: $categoriesTest = array(); $sql = “SELECT * FROM `respondent_data` WHERE `respondent_firstname` = ‘John'”; $result = mysqli_query($conn, $sql); while($row = mysqli_fetch_assoc($result)) { $categoriesTest[] = array( ‘respondent_sdo’ => $row[‘respondent_sdo’], ‘respondent_dcto’ => $row[‘respondent_dcto’], … Read more

[Solved] How can I avoid SQL injection in my code below? [duplicate]

As suggested, prepared statements are the best way to achieve good protection from SQL injection. Shortened Example You will need to add entries to fill in all columns you wish to insert. $email = $_POST[‘e-mail’]; $fn = $_POST[‘firstname’]; $ln = $_POST[‘lastname’]; if ($stmt = $mysqli->prepare(“INSERT INTO `newcartdb`.`orders`(Email,Firstname,Lastname) values(?,?,?)”) { $stmt->bind_param(“sss”, $email, $fn, $ln); “sss” – … Read more

[Solved] mysqli_connect() [function.mysqli-connect]: (28000/1045): Access denied for user ‘uname’@’localhost’ (using password: YES) in /all_pts.php on line 16 [duplicate]

mysqli_connect() [function.mysqli-connect]: (28000/1045): Access denied for user ‘uname’@’localhost’ (using password: YES) in /all_pts.php on line 16 [duplicate] solved mysqli_connect() [function.mysqli-connect]: (28000/1045): Access denied for user ‘uname’@’localhost’ (using password: YES) in /all_pts.php on line 16 [duplicate]

[Solved] How to loop table in php [duplicate]

There are multiple issues with your current code. You fetch the first row with $row = mysqli_fetch_array($select); (line 3), but you don’t do anything with it. This means that the first result is discarded. Your while loop attempts to loop over an incorrect variable ($query is a string, not the result-object), and you’ve quoted it … Read more