Post and get at the same time in php

Posted by user225269 on Stack Overflow See other posts from Stack Overflow or by user225269
Published on 2010-05-01T09:20:06Z Indexed on 2010/05/01 9:27 UTC
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Do you have any suggestions with my problem. I need to use get and post at the same time. Get because I need to output what the user has typed. And post because I need to access the mysql database in relation to that input. It looks something like this:

<form name="x" method="get" action="x.php">
<input name="year" type="text">

<select name="general" id="general">
        <font size="3">
        <option value="YEAR">Year</option>

</form>

This will output the contents of mysql depending on what the user will check:

<form name="y" method="post" action"y.php">
<input name="fname" type="checkbox">
</form>

And the form action of those two combined will look something like this:

   <?php

               if($_POST['general'] == 'YEAR'){
                   ?>
                   <?php echo $_GET["year"]; ?>
                   <?php
            $result2 = mysql_query("SELECT * FROM student
    WHERE student.YEAR='$syear'");
    ?>
    <table border='1'>
            <tr>

                    <?php if ( $ShowLastName ) { ?><th>LASTNAME</th><?php } ?>
                    <?php if ( $ShowFirstName ) { ?><th>FIRSTNAME</th><?php } ?>
        </tr>

    <?php while ( $row = mysql_fetch_array($result2) ) {
        if (!$result2)  { 

    }
        ?>
            <tr> 
                    <td><?php echo $row['IDNO']?> </td>
                    <td><?php echo $row['YEAR'] ?> </td>
     <?php if ( $ShowLastName ) { echo('<td>'.$row['LASTNAME'].'</td>'); } ?></td>
                    <?php if ( $ShowFirstName ) { echo('<td>'.$row['FIRSTNAME'].'</td>'); } ?>

I really get lots of undefined errors when I do this. What can you recommend that I should do in order to get the value inputted by the user together with the mysql data.

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