依赖下拉PHP

Question

Hello I want to show a table that have all the values from a mysql table but depending on the values that I will select from a dropdown menu. For example the dropdown list menu have a value named open. I just want to see the rows from the table that have that status. I will need to use Ajax for this?

Here is my code?

$status = $_POST['TipoStatus'];
echo '<a href = "rnservices.php">  Create Service</a> ';
echo '</br>';
echo '</tr><tr><td><label for="TipoStatus"> Status:</label></td><td>';
$query = "SELECT TipoStatus FROM status"; // First Remar
         $result = queryMysql($query);

            if (!queryMysql($query)) {
    echo "Query fail: $query<br />" .
            mysql_error() . "<br /><br />";
                            }
    else
{
echo '<select name = "TipoStatus" size = "1">'; // or name="toinsert[]"
// echo '<option value="none" selected="selected">None</option>';
while ($row_1 = mysql_fetch_array($result)) {
  echo '<option value="' . htmlspecialchars($row_1['TipoStatus']) . ' selected="$row_1[9]" >' // Third remark

  . htmlspecialchars($row_1['TipoStatus'])
  . '</option>';

}
echo '</select>';
echo '</p>';
}    

echo '<table border="1" >';    
echo '<tr>';    
echo '</br>';
echo '<th> Service ID</th>';
echo '<th>Title</th>';
echo '<th>Description</th>';
echo '<th>Notes</th>';
echo '<th>Submit By</th>';
echo '<th>Assigned Employee</th>';
echo '<th>Assigned Group</th>';
echo '<th>Category</th>';
echo '<th>Status</th>';
echo '<th>Urgency</th>';
echo '<th>Customer</th>';
echo '<th>Day Created</th>';
echo '</tr>';

$query = ("SELECT ServiceID, Title, Description, Notes, SubmitBy, AssignedEmp, " .
"AssignedGroup, NameCategory, TipoStatus, TiposUrgencia, CustomerName, DayCreation FROM Service where TipoStatus = '$status'  ");
$result = queryMysql($query);
echo 'resultado' . mysql_num_rows($result);


while ($row = mysql_fetch_assoc($result)) {

    echo '<tr>';

    echo '<td><a href="rnservices1.php?ServiceID='.$row["ServiceID"].'"> '.$row['ServiceID'] .' </a></td>';
    echo '<td>' .$row['Title']. ' </td>';
    echo '<td>'.$row['Description'].'</td>';
    echo '<td>'.$row['Notes'].'</td>';
    echo '<td>'.$row['SubmitBy'].'</td>';
    echo '<td>'.$row['AssignedEmp'].'</td>';
    echo '<td>'.$row['AssignedGroup'].'</td>';
    echo '<td>'.$row['NameCategory'].'</td>';
    echo '<td>'.$row['TipoStatus'].'</td>';
    echo '<td>'.$row['TiposUrgencia'].'</td>';
    echo '<td>'.$row['CustomerName'].'</td>';
    echo '<td>'.$row['DayCreation'].'</td>';

    echo '</tr>';
}

mysqli_free_result($result);


echo $ticket_select;`enter code here`

echo '</table>';
echo '<form method = "post" action "rnseetickets.php">';
 ?>

Answer 1

Your code is a little meandering, and I can't quite tell if you're running this from the command line or as a web page, but it seems like you're wanting to filter a SQL result with a drop-down of options? That can be done with a simple form (if you don't mind a page refresh) or with AJAX (if you want it to filter without a page refresh):

// Run SQL
$sql = "SELECT TipoStatus FROM status";
if (!empty($_GET['status'])) {
    $sql .= "WHERE `value`=\"".addslashes($_GET['status'])."\"";
}
$result = queryMysql($sql);

// Filter form
echo "<form action=\"".$_SERVER['PHP_SELF']."\" method="get">";
echo "<select name=\"status\"><option>open</option><option>other option</option></select>";
echo "<input type=\"submit\" value=\"Filter\">";
echo "</form>";

// Table
// As you already have it