[ad_1] Here’s a basic structure of how an update query works: UPDATE tableName SET column = value WHERE condition = true; In your case, you’re updating the demo_table new table with a condition of new.trans_type = 9. The tricky thing here is you’re using your sub query as your value to set the columns as. … Read more
[ad_1] <?php if (isset($_GET[‘edit’])) { ?> <form action=”index.php” method=”post”> <input type=”text” name=”id” value=”<?=$id;?>”> <input type=”text” name=”nieuweprijs” placeholder=”vul nieuwe prijs in”> <input type=”submit” name=”submitnieuweprijs” value=”verzenden”><form> <?php } if (isset($_POST[‘submitnieuweprijs’])) { $nieuweprijs = Safesql($_POST[‘nieuweprijs’]); $id = Safesql($_GET[‘id’]); if(!$mysqli->query(“UPDATE prijzen SET prijs=””.$nieuweprijs.”” WHERE id='”.$id.”‘”)) { echo $mysqli->error;} Laden(0); } } ?> $id is the value obtained from the … Read more
[ad_1] Like @B-and-P describes in his comment. You can do this using LPAD. UPDATE customer_db SET number = LPAD(FLOOR(number + rand()*1000000),10,0) LPAD uses 3 parameters; string, total amount of characters and last but not least which character should be used for padding. 1 [ad_2] solved How to random a zero leading string field mysql [duplicate]
[ad_1] I’ll give a very simple example that should get you started. The values are accessible via (most likely) $_POST[‘input_name’]. Without using Post/Redirect/Get, you can just get the input values like: $input_name = isset($_POST[‘input_name’]) ? $_POST[‘input_name’] : ”; Then later you’ll display it in the form like: echo ‘<input name=”input_name” value=”‘ . htmlspecialchars($input_name, ENT_QUOTES) . … Read more
[ad_1] Your sql is wrong. Apart from the gaping wide open SQL injection attack vulnerability, you’re generating bad sql. e.g. consider submitting “Fred” as the first name: $First_Name2 = “Fred”; $query = “UPDATE people SET Fred = First_name WHERE ….”; now you’re telling the db to update a field name “Fred” to the value in … Read more
[ad_1] Add value to select box options echo “<option value=”$id”>$id</option>”; 7 [ad_2] solved Failed syntax UPDATE id in MariaDB
[ad_1] Is this some kind of trick question? update yourtable set columna=upper(columna), columnb=upper(columnb) 2 [ad_2] solved The most efficient way to UPPER CASE two columns in a single query [closed]
[ad_1] you are not connecting to database. you variables are strings. change this $conn = mysql_connect(“$DB_HostName”, “$DB_User”, “$DB_Pass”) to $conn = mysql_connect($DB_HostName, $DB_User, $DB_Pass) and your update is wrong . you have to use math part outside the query, try use this $RH = $RANK * $HEALTH ; $SP = $Skills + $POWER ; $SPRH … Read more