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Selecting PHP Combo Box and Button

开发者 https://www.devze.com 2023-01-15 16:39 出处:网络
I am currently using PHP 5 with a MysSQL database with 2 tables. So far my PHP Combo Box is working however I need to access the values selected from the combo box. it goes like this:

I am currently using PHP 5 with a MysSQL database with 2 tables. So far my PHP Combo Box is working however I need to access the values selected from the combo box. it goes like this:

1) I select a value from the Combo Box. 2) I click on the Submit button 3) The Submit button brings me to another webpage.

The problem that my program is facing now is during step 3 when I click the submit button the开发者_高级运维re is no webpage generated. I think the problem is due to the sequencing of the Combo Box Codes and Button Codes.

My codes are as shown:

<?php
include "db_connect.php";
{
?>
<td valign=top><strong>Name:</strong></td>
<td>

<?php
echo '<select name="table_choice">'; 
echo "<option size =30 selected>Select</option>";

$result = mysql_query("show tables");

if(!$result) trigger_error("Query Failed: ".  mysql_error($db), E_USER_ERROR);

if(mysql_num_rows($result)) 
{ 
while($table_array = mysql_fetch_array($result)) 
{ 
    echo "<option>$table_array[0]</option>";
}  

$array_value = $_POST['table_choice'];

if(!$_POST['submit'])
{
?>

<input type="submit" name="submit" value="Submit">

<?php
 }
 else
{
  echo '<script type="text/javascript">
        alert("Redirecting you to the site main page");
        window.location="echo.php"</script>';
 }

 } 
else 
{
echo "<option>No Names Present</option>";  
} 
}

?>

Everything seems fine to me.


Where is the <form> tag?

Anyway, consider writing your web applications using some web framework or at least templates to separate the program logic (PHP code) from the presentation (HTML code). Else it will be a big unmaintainable mess soon (or maybe it already is).


<?php
include "db_connect.php";
{
?>

what's with the random bracket?


Never Mind got the answer. The answer is just to simply add an echo '</select>' above the $array_value = $_POST['table_choice'];. The answer is simply to end select with /select. Thanks for the extra tips guys.

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