4 Topics
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| | It has been a few years since I actually did real coding... well a few years since i had a job for that. Just dumping about 1000 photos from a directory and want to display them. No magic of fancy stuff. Thanks  |
| | I am new for PHP. I tried to display text box value based on drop down selection. it seems code working. no error display here. But when i change the drop down, text box value shows empty. Even print the data it will display only null. HTML:- <select name="cat" id="cat"> … |
| | a) Write MARIE code to: 1. Input a number and store it in X . 2. Y will be equal to “  |
| | Warning: mysqli_query() expects parameter 2 to be string, object given in E:\xampp\htdocs\cadastro\cadastrar.php on line 46 Alguem pode me ajudar? <!DOCTYPE html> <html> <head> <meta charset="UTF-8" /> <title>Cadastrar</title> </head> <body> <?php include "config.php"; $nome = isset( $_POST['nome'] ) ? $_POST['nome'] : 'valor_padrao'; $sobrenome = isset( $_POST['sobrenome'] ) ? $_POST['sobrenome'] : 'valor_padrao'; …  |