arjunking Posted October 26, 2012 Share Posted October 26, 2012 (edited) Hi i have set up a blog based on wordpress. I want to add data which is read-only for the users. So i created two fields : user_a and user_b Now i want the users to be able to read the data . I dont know how to use session function so i created a page which asks for the user_id and user_pass and returns the field user_survey and user_survey_complete Instead of returning the data in user_survey and user_survey_complete , it just returns the heading user_survey and user_survey_complete the code which asks for data is : <form id="FormName" action="surveycheck.php" method="post" name="FormName"> <table width="448" border="0" cellspacing="2" cellpadding="0"> <tr> <td width="150" align="right"><label for="user_id">Username</label></td> <td><input name="user_id" maxlength="32" type="text" value="<?php echo stripslashes($user_id) ?>"></td> </tr> <tr> <td width="150" align="right"><label for="user_pass">Password</label></td> <td><input name="user_pass" maxlength="32" type="text" value="<?php echo stripslashes($user_pass) ?>"></td> </tr> <tr> <td colspan="2" align="center"><input name="" type="submit" ></td> </tr> </table> </form> The code which I used to display the data <html><head><title>MySQL Table Viewer</title></head><body> <?php $db_host = 'localhost'; $db_user = 'userid'; $db_pwd = 'password'; $database = 'wordpress_management'; $table = 'wp_users'; $user_id = trim(mysql_real_escape_string($_POST["user_id"])); $user_pass = trim(mysql_real_escape_string($_POST["user_pass"])); if (!mysql_connect($db_host, $db_user, $db_pwd)) die("Can't connect to database"); if (!mysql_select_db($database)) die("Can't select database"); // sending query $result = mysql_query("SELECT user_survey_link, user_survey_compelete FROM {$table} WHERE {$user_id}=user_login AND {$user_pass}=user_pass "); if (!$result) { die("Query to show fields from table failed"); } $fields_num = mysql_num_fields($result); echo "<h1>Table: {$table}</h1>"; echo "<table border='1'><tr>"; // printing table headers for($i=0; $i<$fields_num; $i++) { $field = mysql_fetch_field($result); echo "<td>{$field->name}</td>"; } echo "</tr>\n"; // printing table rows while($row = mysql_fetch_row($result)) { echo "<tr>"; // $row is array... foreach( .. ) puts every element // of $row to $cell variable foreach($row as $cell) echo "<td>$cell</td>"; echo "</tr>\n"; } mysql_free_result($result); ?> </body></html> Please help me in displaying the data under the fields. If someone know wordpress then please help me use session instead of asking for userid and pass I am a noob and not good at php so please provide the solution in detail Thanks a lot Edited October 26, 2012 by arjunking Quote Link to comment Share on other sites More sharing options...
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