sunil_pandya

in CSS example of Grid is as under: .grid-container { display: grid; grid-template-columns: auto auto auto auto; gap: 10px; background-color: #2196F3; padding: 10px; } .grid-container > div { background-color: rgba(255, 255, 255, 0.8); border: 1px solid black; text-align: center; font-size: 30px; } The above example's HTML shows as under: <div class="grid-container"> <div>1</div> <div>2</div> <div>3</div> <div>4</div> <div>5</div> <div>6</div> <div>7</div> <div>8</div> </div> If you prepare your HTML div and then display: grid, gap, will not work in your practical example <div class="main"> <div class=1><p>Some Content</p> <p>Some Contet</p> </dv> <div class=2> <p>Some Content</p> <p>Some Cotent</p> </div> </div> Now if you use CSS Shown as per the example then any browsers will not catch the CSS shown in the example at all. If you the user want to display two div side by side it will not display side by side at all. that is it will not display as a grid, it will not catch the coloumn, nor your specified column width nor auto; Please if somebody can post real life example Sunil pandya

php+mysqli_real_escape_string do not work for email id, what could be reason? mysql table shows record for userid and password but do not show record of emailid. Following are my code, obtained from w3school php example <?php$link = mysqli_connect("hostname", "user", "password", "databasel");/* check connection */if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit();}mysqli_select_db($link,"spinall"); $username = mysqli_real_escape_string($link, $username); $password = mysqli_real_escape_string($link, $password); $emailid = mysqli_real_escape_string($link, $emailid);$sql="INSERT INTO job_newuser (username,password,emailid)VALUES('$_POST[username]','$_POST[password]','$_POST[emailid]')";if (!mysqli_query($link,$sql)) { die('Error: ' . mysqli_error($con)); }echo "1 record added";mysqli_close($link); ?>i am not able to understand what could be mistake? can any body tell me

<?php$link = mysqli_connect("hostname", "user", "password", "databasel");/* check connection */if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit();}mysqli_select_db($link,"spinall"); $username = mysqli_real_escape_string($link, $username); $password = mysqli_real_escape_string($link, $password); $emailid = mysqli_real_escape_string($link, $emailid);$sql="INSERT INTO job_newuser (username,password,emailid)VALUES('$_POST[username]','$_POST[password]','$_POST[emailid]')";if (!mysqli_query($link,$sql)) { die('Error: ' . mysqli_error($con)); }echo "1 record added";mysqli_close($link); ?>