Question

因此,我正在从PHP的一个数据库获取数据。我打算比较从询问中获得的数据中得出的每个物体的数值。迄今情况良好。

当我试图在物体创建后从物体中提取数据时,这个问题就产生了。

我不是PHP dev,因此我不知道我是否遵循适当的PHP逻辑。我用于联合材料、AS3和Java,因此,物体和价值物体与我所了解的情况相比,在购买力平价中是很小的。

任何人都知道如何检索我的数据?

<?php
    include("../config.php");

    class userVO
    {
        public  $uid;
        public  $name;
        public  $email;
        public  $list;
        public  $num_list_items;
        public  $matches;
        public  $num_matches;

        public function __construct()
        {
            $this->matches = array();
        }
    }

    mysql_connect(DB_HOST, DB_USER, DB_PASSWORD) or die(mysql_error());
    mysql_select_db(DB_NAME) or die(mysql_error());

    $json_array = array();
    $result = mysql_query("...");

    $num_results = mysql_numrows($result);
    mysql_close();

    $users = array();
    $i = 0;
    while($i < $num_results)
    {
        $match = new userVO;
        $match->uid = mysql_result($result, $i, "uid");
        $match->name = mysql_result($result, $i, "name");
        $match->email = mysql_result($result, $i, "email");
        $users[] = userVO;
        $i++;
    }

    $num_users = count($users);
    echo "num users: " . $num_users . "<br>";

    $i = 0;
    while($i < $num_users)
    {
        echo "--- i: " . $i . " ---<br>";
        $current_user = $users[$i];

        echo "users[" . $i . "]: " . $users[$i] . "<br>";
        echo "users[" . $i . "]->name: " . $users[$i]->name . "<br>";
        echo "current user: " . $current_user . "<br>";
        echo "current user name: " . $current_user->name . "<br>";

        $i++;
    }
?>

Answer 1

你

<代码> 用户[] = 用户;

这应当做到:

<代码>用户[] = 美元;

这是你将新的用户表示反对的变量。

如果你制造物体,你也应当有母子。

http://www.ohchr.org。

Answer 2

One thing that I see that doesn t look right is this line here in the while loop: $users[] = userVO;, I think that should be $users[] = $match;

然后,你也可以使用<代码>foreach(>,在用户阵列上,而不是在下方使用。

我想写一下法典的最后部分,如:

$result = mysql_query("...");

$users = array();

while($row = mysql_fetch_assoc($result))
{
    $match = new userVO();
    $match->uid = $row["uid"];
    $match->name = $row["name"];
    $match->email = $row["email"];
    $users[] = $match;
}
mysql_close();

$num_users = count($users);
echo "num users: " . $num_users . "<br>";

foreach($users as $key => $user)
{
    echo "--- key: " . $key . " ---<br>";
    echo "users[" . $key . "]: " . $user . "<br>";
    echo "users[" . $key . "]->name: " . $user->name . "<br>";
    echo "current user: " . $user . "<br>";
    echo "current user name: " . $user->name . "<br>";
}


?>

Oh, also, with a bit of a change to the __construct function you can do this:

public function __construct($uid, $name, $email)
{
    $this->matches = array();
    $this->uid = $uid;
    $this->name = $name;
    $this->email = $email;
}

那么,你就能够做到:

while($row = mysql_fetch_assoc($result))
{
    $users[] = new userVO($row["uid"],$row["name"],$row["email"]);
}

Answer 3

Hmm,我可以看到几个问题:

$users[] = userVO; You re not adding $match to the array, you re adding userVO, which is the name of the class, not the variable holding the instance of the class.
你们是否需要为此设立一个类别? 除非你再做大量繁重的工作,否则你很可能会放弃多维阵列......我sql_fetch_assoc()对此非常重要,因为这会创造出一系列相关的表格。我通常做的是:


$sql = "SELECT * FROM table WHERE foo= bar ";
$result = mysql_query($sql);

if (!$result) {
   echo  Could not run query:   . mysql_error();
   echo  Query was:   . $sql;
   exit;
}

$stuff = array();
if (mysql_num_rows($result) > 0) {
   while ($row = mysql_fetch_assoc($result)) {
      $stuff[] = $row;
   }
}

之后,我做了一些整理,然后用了以下文字:


foreach($stuff as $row){
   $uid = $row[ uid ];
   $name = $row[ name ];
   $email = $row[ email ];

   echo "$uid $name $email"; 
}

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