i'm trying make php script prints text file website on button click.
this code:
<?php mysqli_connect("host", "username", "password", "dbname"); if (isset($_post['button'])) { mysqli_query("select name,lastname users order rand()"); $name = null; $lastname = null; while ($row = mysqli_fetch_array()) { $name = $row['name']; $lastname = $row['lastname']; } echo '<p><small>output:</small></p><pre style="min-width:auto;display:table;">' . $name . '_' . $lastname . '</pre>format: <b>name_lastname</b> <span style="width:310px;">'; } ?> <form method="post"> <button type="submit" class="btn btn-default" name="button">button</button> <br> </form>
and error every time when button clicked:
anyone knows how fix this?
you need introduce database connection variable mysqli_* functions required, $con
in modified script:
<?php $con = mysqli_connect("host", "username", "password", "dbname") or die(mysqli_error($con)); if (isset($_post['button'])) { $result = mysqli_query($con, "select name,lastname users order rand()"); $name = null; $lastname = null; while ($row = mysqli_fetch_array($result)) { $name = $row['name']; $lastname = $row['lastname']; } echo '<p><small>output:</small></p><pre style="min-width:auto;display:table;">' . $name . '_' . $lastname . '</pre>format: <b>name_lastname</b> <span style="width:310px;">'; } ?> <form method="post"> <button type="submit" class="btn btn-default" name="button">button</button> <br> </form>
in addition need pass result set identifier returned mysqli_query(), $result
in case. can found @ w3schools.
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