php代码不选择mysql表中的所有字段我做错了什么？

Question

我有一个PHP代码,必须从mysql表中选择所有字段三个字段,但问题,结果只是2个字段如何修复此错误？

mysql表"用户"

的Fileds:

• ID.
• 用户.
• 密码.
• 日期.

结果是:

{"userlist":[{"id":"2","user":"user2"},{"id":"3","user":"michel"},{"id":"4","user":"georges"},{"id":"5","user":"testtest1"}],"success":1}

这是代码

<?php

/*
 * Following code will list all the emp
 */

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


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

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

// get all emp from emp table
$result = mysql_query("SELECT *FROM users") or die(mysql_error());

// check for empty result
if (mysql_num_rows($result) > 0) {


    // looping through all results
    // emp node
    $response["userlist"] = array();

    while ($row = mysql_fetch_array($result)) {
        $response["success"] = 1;
        // temp user array
            $userlist = array();
            $userlist["id"] = $row["id"];
            $userlist["user"] = $row["user"];
            $userList["date"] = $row["date"];

            //$response["message"] = $userList["create_date"];


        // push single Employee into final response array
        array_push($response["userlist"], $userlist);
    }
    // success

    //$response["message"] = "DISPLAYED Success";

    // echoing JSON response
    echo json_encode($response);
} else {
    // no emp found
    $response["success"] = 0;
    $response["message"] = "No User found";

    // echo no users JSON
    echo json_encode($response);
}
?>

Answer 1

仔细看....

$userlist["id"] = $row["id"];
$userlist["user"] = $row["user"];
$userList["date"] = $row["date"];

注意的情况下,$userlist和$userList？将最后一个更改为小写,它将正常工作.