使用选择选项值更改表格填充

Question

Im new to php and ajax and wanted to display a table that's filled with content of a database. I succeeded doing it but now I'm trying to change the tables content with a select . I know there are many sites explaining how to do it, but I somehow don't get it. The best solution for me is by changing the table without a reload of the page or reloading it with a separate button.

I read about doing it with ajax / javascript but, as I mentioned, Im not familiar with those things.

Below is my code thats already workin.

PHP:

<?php

$mysqlhost="localhost"; // 
$mysqluser="root"; //
$mysqlpwd=""; //
$mysqldb="wordpress"; //


$connection=mysql_connect($mysqlhost, $mysqluser, $mysqlpwd); 
mysql_select_db($mysqldb, $connection);


$sql = "SELECT id, user_email FROM wp_users";  
$db_query = mysql_query($sql);


?>
<table cellpadding="1" cellspacing="3" border="1">
    <tr>
        <td>ID</td>
        <td>Mail</td>
    </tr>
<?php

  while ($adr = mysql_fetch_array($db_query)){
?>

    <tr>
        <td><?=$adr['id']?></td>
        <td><?=$adr['user_email']?></td>
    </tr>
<?php
  }
?>
</table>

My selects:

<select name="Choose" title="chose">
<option value="one" id="One">One</option>
<option value="two" id="Two">Two</option>
<option value="three" id="Three">Three</option>
</select>

Id really appreciate some code or hints how to do it.

Answer 1

Here is the code:

<table id="tableid"> //mention id for a table
 ......
 ......
 </table>

// create an event for select
<select name="Choose" title="chose" onchange="getajax(this.value)">
<option value="one" id="One">One</option>
<option value="two" id="Two">Two</option>
<option value="three" id="Three">Three</option>
</select>

:

function getajax(value){
$.ajax({
type: "GET",
url: "Ajaxpage.php",
data: {text:value},
success: function(data) {
  $("#tableid").html(data);
}
});
}

<?php
$mysqlhost="localhost"; // 
$mysqluser="root"; //
$mysqlpwd=""; //
$mysqldb="wordpress"; //


$connection=mysqli_connect($mysqlhost, $mysqluser, $mysqlpwd); //use mysqli instead of mysql

mysqli_select_db($mysqldb, $connection);


$sql = "SELECT id, user_email FROM wp_users where someid='".$_GET['text']."'";  

$query = mysql_query($sql);
 while($row= mysql_fetch_array(query)){
 echo "<tr><td>".$row['id']."</td><td>".$row['user_email']."</td></tr>";

 }


?>

Answer 2

<script>
function reloadWithOptionValue(){
document.FilterFrom.submit();
}
</script>
<?php

$mysqlhost="localhost"; // 
$mysqluser="root"; //
$mysqlpwd=""; //
$mysqldb="wordpress"; //


$connection=mysql_connect($mysqlhost, $mysqluser, $mysqlpwd);

mysql_select_db($mysqldb, $connection);


$sql = "SELECT id, user_email FROM wp_users";  
if(isset($Choose) && !empty($Choose)){
      $sql.=" where id like '%$Choose%' or user_email like  '%$Choose%'";
}
$db_query = mysql_query($sql);


?>

<form method="post" name="FilterFrom">
<table cellpadding="1" cellspacing="3" border="1">

    <tr>

        <td>ID</td>

        <td>Mail</td>

    </tr>
<?php

  while ($adr = mysql_fetch_array($db_query)){

?>

    <tr>

        <td><?=$adr['id']?></td>

        <td><?=$adr['user_email']?></td>

    </tr>

<?php

  }

?>

</table>
<select name="Choose" title="chose" onchange="reloadWithOptionValue()">
<option value="one" id="One">One</option>
<option value="two" id="Two">Two</option>
<option value="three" id="Three">Three</option>
</select>
</form>

try this one