missingcolor123

thanks, i rewrote it without doing the if empty check and it worked <option value="Option1" <?php if ($row1['instruction'] == 'Option1') echo 'selected = "selected"'; ?>>Option1</option> thanks for your help

Hi, I am trying to create a drop down menu that will select a value that is stored in the database - right now the code creats a dropdown (with nothing selected) - hope someone can help. in the database, the values are stored as --null- Option1 Option2 Option3 my code is $instruction = $_GET['instruction']; <?php <select id="instruction" name="instruction"> <option value="" <?php if (!empty($instruction) && $instruction == '' ) echo 'selected = "selected"'; ?>></option> <option value="Option1" <?php if (!empty($instruction) && $instruction == 'Option1') echo 'selected = "selected"'; ?>>Option1</option> <option value="Option2" <?php if (!empty($instruction) && $instruction == 'Option2') echo 'selected = "selected"'; ?>>Option2</option> <option value="Option3" <?php if (!empty($instruction) && $instruction == 'Option3') echo 'selected = "selected"'; ?>>Option3</option> </select> <?

the data that is stored in table1 - will be changing on a regular basis when i create the record in table2 - i want it to store what table1 was showing at the time ... not sure if that makes sense, but just playing around. what would be a better/proper way of doing it? i am very new to php

I am using the following code to generate a dynamic drop down menu from a column in table1(name) - and then inserting the selected option into table2. What I am trying to do is, select data from 2 columns from table1(name,age) and then store and insert both of those values into table2 I am hoping someone can help me out or at least point me in the rigth direction. ..I would like to insert into table 2 the age that matches the person selected by the drop down menu from table1. For example - if from the drop down menu I choose John - when the form is sumbitted i want to insert into table2 John's name and age (based on what is stored in table1, in this case John,22) ...Any ideas??? table1 id|name|age|score 01-john-22-1547 02-jane-22-1245 table2 id|county|name|age the dynamic drop down is generated from table1 (name) with following code <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <table width="100%" border="0" cellpadding="5" cellspacing="0"> <?php require_once('sql.php'); $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); $query="SELECT name FROM table1 ORDER BY name ASC"; $result = mysqli_query ($dbc,$query) or die(mysqli_error()); $dropdown = "<select name='name'>"; while($row = mysqli_fetch_array($result)) { $dropdown .= "\r\n<option value='{$row['name']}'>{$row['name']}</option>"; } $dropdown .= "\r\n</select>"; echo $dropdown; ?> <tr> <td height="50" align="right"> <label for="scheduled_time">Country</label><br/> </td> <td align="left"> <input type="text" id="country" name="country" class="input" maxlength="30"" /> </td> </table> <p></p> <input type="submit" value="Add Info" name="submit" /> </form> I am inserting into table2 with the following code <?php $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); if (isset($_POST['submit'])) { $county = mysqli_real_escape_string($dbc, trim($_POST['country'])); $name = mysqli_real_escape_string($dbc, trim($_POST['name'])); $query = "INSERT INTO table2 (country, name) VALUES ('$country', '$name'')"; $result = mysqli_query($dbc, $query); if (!$result) { printf("query error : <br/> %s\n", mysqli_error($dbc)); } if ($result) { echo 'Success'; } // close dbc mysqli_close($dbc); exit(); } ?>