[Solved] Fetch Json from MySQL in Jquery calender

Depends on the programming language you are using, if you’re using Java/php, create a service method which would get the JSON from the database and return the JSONobject through ajax, and introduce it into the calendar through jquery javascript. i would also recommend you to use full calendar as its very easy to grasp. $(‘#calendar’).fullCalendar({ … Read more

[Solved] MySql query don’t work [closed]

No need to have ” for the table column , it’s needed for the variable that need to be inserted into the table column. Between, remove the ; after ). The correct code should be looked like this: $query = “INSERT INTO Zakazes( id, dateDelivery, timeCok, nameClient, phoneClient, metro, adress, comments, product, summ, skidka, result, … Read more

[Solved] PHP inserting data into database results in an error [closed]

Your immediate problem is that long is a reserved word. You need to wrap it in backticks, eg INSERT INTO … pribank, `long`, avbal, … See http://dev.mysql.com/doc/refman/5.0/en/identifiers.html This says nothing for the security of your application. I strongly suggest you read up on PDO, PDO::prepare() and PDOStatement::bindParam() 1 solved PHP inserting data into database results … Read more

[Solved] PDO login.php needs help finishing [closed]

The prepare statement returns an object, you can execute that object. Thus $query->execute() – the first parameter of execute can be an array. Alternatively, you can use the bindParam function first on that same $query object instead of passing an array though. Anyways, here you are: <?php if(isset($_POST[‘username’], $_POST[‘password’])){ try{ $username=”root”; $password = ”; $conn … Read more

[Solved] Trying to convert mysql to mysqli, not working

Just for a quick testing purpose, try this below which is a bare bones method. You can then slowly build up sanitizing and troubleshoot from thereon. <?php $username = $_POST[‘username’]; $password = sha1($_POST[‘password’]); $link = mysqli_connect(‘xxx’, ‘xxx’, ‘xxx’, ‘xxx’); $query = “SELECT password, id FROM users WHERE username=”$username” AND password=’$password'”; $result = mysqli_query($link, $query); if(mysqli_num_rows($result) … Read more

[Solved] PHP sql database not updating after update query, indexes are not defined? [closed]

At first you have to change: <input type=”date” id=”updateltext” name=”listdate” value=””/> to: <input type=”date” id=”updateltext” name=”list_date” value=””/> To match what you have in your php code and apply it to the rest of the html inputs then change: WHERE product_id = ‘ .$id. ‘ to: WHERE product_id = ‘$id’ Notice Don’t forget to format for=”” … Read more

[Solved] AUTO_INCREMENT Uncontrolled Increase

This means that you have a process that is trying to insert records but fails – such a operation increase AUTO INCREMENT number but does not add records. You can always pull back AUTO INCREMENT number – you can do it in PHPMyAdmin in Operations tab. 2 solved AUTO_INCREMENT Uncontrolled Increase

[Solved] Php function doesn’t run [closed]

$book=mysql_real_escape_string($_POST[‘books’]); $sql=”INSERT INTO books (bID, book) VALUES (”,’$book’)”; if (!mysql_query($sql,$con)) { die(‘Error: ‘ . mysql_error($con)); }` <— Here is a rogue quote.. The syntax highlighter if StackOverflow spotted this problem. Do you use an editor with a highlighter too? Since the code is now invalid, PHP cannot parse the file and will fail. It doesn’t … Read more

[Solved] what does this sql query do? SELECT column_1 FROM table_1,table_2;

In more layman’s terms, it means that for each record in Table A, you get every record from Table B (all possible combinations). TableA with 3 records and Table B with 3 records gives 9 total records in the result: TableA-1/B-1 TableA-1/B-2 TableA-1/B-3 TableA-2/B-1 TableA-2/B-2 TableA-2/B-3 TableA-3/B-1 TableA-3/B-2 TableA-3/B-3 Often used as a basis for … Read more

[Solved] Error in an update mysql query execution with php and mysqli [duplicate]

You need to escape the single quotes using php’s str_replace, e.g.: $exp_title = str_replace(“‘”, “\'”, $_REQUEST[‘exp_title’]); $exp_description = str_replace(“‘”, “\'”, $_REQUEST[‘exp_description’]); $exp_time = $_REQUEST[‘exp_time’]; $update=”UPDATE experience SET exp_title=””.$exp_title.”” , exp_description='”.$exp_description.”‘ , exp_time=””.$exp_time.”” WHERE expid='”.$id.”‘”; However, you should really really use preparedstatements instead of concatenating strings and escaping characters, e.g.: $exp_title = $_REQUEST[‘exp_title’]; $exp_description = $_REQUEST[‘exp_description’]; … Read more