使用mysql数据显示表单下拉列表&插入记录

时间:2014-05-28 23:26:00

标签: php mysql

我试图让一个下拉列表显示来自表单上的mysql数据库的数据以添加记录。我找不到让它工作的方法,它没有显示任何东西,所以我甚至无法尝试。我正在学习php和mysql,所以我不知道代码中有什么问题,虽然连接很好。建议?在此先感谢,这是代码。

<?php
 include("includes/connect.php");
 doDB();


if (!$_POST) {
$display_block = "
<form method=\"post\" action=\"".$_SERVER["PHP_SELF"]."\">        

   <div class=\"form-group\">  
        <div class=\"row\">      
            <div class=\"col-md-8\">
      <label>Dominio</label>
     <input  class=\"form-control input-sm\" name=\"dominio\" type=\"text\"  placeholder=\"dominio\">
            </div>
        </div>
    </div>";


//get plan list
$get_planlist_sql = "SELECT * FROM PLANES ORDER BY plan ASC";
$get_planlist_res = mysqli_query($mysqli, $get_planlist_sql)
    or die(mysqli_error($mysqli));

if (mysqli_num_rows($get_planlist_res) > 0) {

    $display_block .="<div class=\"form-group\">  
<div class=\"row\">      
    <div class=\"col-md-8\">
        <label>Plan</label>
           <select class=\"form-control input-sm\">
            <option value=\"\>Selecciona</option>";

       while ($getplans = mysqli_fetch_array($get_planlist_res)) {
        $plan = $getplans{'plan'];
        $display_block .= "<option value=\"".$plan"\">".$plan."</option>";                   
    }

        $display_block .= "
        </select>
        <div class=\"form-actions\">
            <button type=\"submit\" class=\"btn btn-primary btn-lg\">Ingresar usuario</button>
        </div>
        </form>";
}
//free result
    mysqli_free_result($get_planlist_res);


} else if ($_POST) {
//Revisar los campos requeridos
if (($_POST["dominio"] == "") || ($_POST["plan"] == "")) {
    header("Location: cliente.php");
    exit;
}

//Add to table datos_cliente
$add_datosclientes_sql = "INSERT INTO datos_cliente (dominio, plan) VALUES ('".$_POST["dominio"]."', '".$_POST["plan"]."')";
$add_datosclientes_res = mysqli_query($mysqli, $add_datosclientes_sql)
or die(mysqli_error($mysqli));
mysqli_close($mysqli);
     $display_block = "<p>La incidencia ha sido enviada. Deseas <a href=\"nuevo-cliente.php\">agregar otra</a>?</p>";
}

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>

<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<link rel="stylesheet" href="css/bootstrap.css" />


<title>Untitled Document</title>



</head>

<body>

<div class="container">
<div class="row">

<?php echo $display_block; ?>
</div>

</div><!--end container-->

</body>
</html>

0 个答案:

没有答案