php - php mysql如何解决此错误:Undefined variable :id in C:\

Question

如何解决错误:
undefined variable :38 中c:\ wamp \ www \ android_connect \ status.php中的ID
我在做什么错???

status.php

<?php
ob_start();

session_start();
//array for JSON response
$response = array();

if(isset($_SESSION['id']))
{

   $id= $_SESSION['id'];


} 

//var_dump ($id);

// include db connect class
    require_once '/db_connect.php';

    // connecting to db
    $db = new DB_CONNECT();

// check for required fields
if(empty($_POST['status'])){
       $response["success"] = 0;
       $response["message"] = "Your Text Field is empty";
        // echoing JSON response
        die (json_encode($response));


}
else if (isset($_POST['status'])) {

    $status = $_POST['status'];


    $sql = mysql_query("INSERT INTO status (status_text, uid)VALUES('$status', '$id')")or die(mysql_error());


      if ($sql) {
        // successfully inserted into database

        $response["success"] = 1;
        $response["message"] = "Status Saved.";

        // echoing JSON response
        die (json_encode($response));
    } else {
        // failed to insert row
        $response["success"] = 0;
        $response["message"] = "Error in saving the status.";

        // echoing JSON response
        die (json_encode($response));
    }

  }

ob_end_flush();
?>

该错误是在插入查询，但我不知道如何解决它。
$ _SESSION有东西可以帮助我吗？

Answer 1

您得到的错误意味着未设置$ id。在您的代码中，没有检查$ id变量。

将else if (isset($_POST['status'])) {更改为else if (isset($_POST['status']) && isset($id)) {
并添加一个else语句，如下所示:

else {
   // failed to insert row because of missing $id
    $response["success"] = 0;
    $response["message"] = "Error in saving the status, session variable $id missing";

    // echoing JSON response
    die (json_encode($response));
}