php 中的多维数组？

发布于 2024-11-28 13:27:50 字数 592 浏览 1 评论 0原文

我在 PHP 中遇到数组问题。那么，让我们看看：我在数据库中有一个表“用户”，其中包含以下字段：“姓名、年龄、评级”。用户数量接近 100。我需要从数据库中获取所有用户，将其发布到数组，并使用 JSON end show 对其进行编码。所以我想，我需要执行以下操作：

从数据库获取一行。
将所有数据添加到某些数组字段，例如关联数组。
将该数组推送到某个数组容器。
将数组容器编码为 JSON。

但是当我尝试编码时，我只得到数组容器中的最后一个元素。我写了类似的东西：

$arrContainer = array();
$arr = array();
$i = 0;
while($row = getDataFromDB)
{
  arr[$i] = $i;
  arr["name"] = row["name"];
  arr["age"] = row["age"];

  array_push($arrContainer, $arr);
  $i++;
}

JSON.encode($arrContainer);

问题：如何用一些数据制作数组数组？

原文

I have a problem with arrays in PHP. So, lets see: I have a table "Users" in database with such fields: "name, surname, age, rating". The number of users is nearly 100. I need to get all of them from database, post it to array, encode them with JSON end show. So I suppose, I need to do the follows:

Get one row from DB.
Add all data to some array fields like associative array.
Push that array to some array container.
Encode array container to JSON.

But when I try to encode, I get only last element in array container. I writted something like that:

$arrContainer = array();
$arr = array();
$i = 0;
while($row = getDataFromDB)
{
  arr[$i] = $i;
  arr["name"] = row["name"];
  arr["age"] = row["age"];

  array_push($arrContainer, $arr);
  $i++;
}

JSON.encode($arrContainer);

QUESTION: How can I make array of arrays with some data?

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原来分手还会想你 2024-12-05 13:27:50

while($row = getDataFromDB)
{
   $arr[]=$row; // add $row as element of $arr array
}

现在你得到 multiDimension-array 和可以Json编码吗

另外，一些数据库扩展可以自动将所有数据返回到多维数组。
请参阅 PDOStatement::fetchAll 手册

while($row = getDataFromDB)
{
   $arr[]=$row; // add $row as element of $arr array
}

Now you get multidimensional-array and can Json encode it

Besides, some database extensions can returns all data to multidimensional array automatically.
See PDOStatement::fetchAll manual

回复收藏 0 原文

月野兔 2024-12-05 13:27:50

试试这个：

$arr = array();
$i = 0;
while($row = getDataFromDB){
  $arr[$i] = $row;
  $i++;
}
JSON.encode($arr);

try this:

$arr = array();
$i = 0;
while($row = getDataFromDB){
  $arr[$i] = $row;
  $i++;
}
JSON.encode($arr);

回复收藏 0 原文

半窗疏影 2024-12-05 13:27:50

您可以像 @RiaD 的答案一样创建数组
然后输出json为..

while($row = getDataFromDB) // @RiaD's code
{
   $arr[]=$row; // add $row as element of $arr array
}

//json output
header("Expires: Mon, 26 Jul 1997 05:00:00 GMT" ); 
header("Last-Modified: " . gmdate( "D, d M Y H:i:s" ) . "GMT" ); 
header("Cache-Control: no-cache, must-revalidate" ); 
header("Pragma: no-cache" );
header('Content-type: application/json');
echo  json_encode($your_array);

You can create array as in @RiaD's answer
and then output json as ..

while($row = getDataFromDB) // @RiaD's code
{
   $arr[]=$row; // add $row as element of $arr array
}

//json output
header("Expires: Mon, 26 Jul 1997 05:00:00 GMT" ); 
header("Last-Modified: " . gmdate( "D, d M Y H:i:s" ) . "GMT" ); 
header("Cache-Control: no-cache, must-revalidate" ); 
header("Pragma: no-cache" );
header('Content-type: application/json');
echo  json_encode($your_array);

回复收藏 0 原文

分開簡單 2024-12-05 13:27:50

您发布的代码是正确的，但是在以下几行中，您忘记将 $ 放在 $arr 变量之前：

arr[$i] = $i;
arr["name"] = row["name"];
arr["age"] = row["age"];

还有行 arr[$i] = $i; 可能应该类似于 $arr['rowNum'] = $i; 让键与值相同似乎是多余的。

应该说，按照 @RiaD 建议使用 $arrContainer[]=$row; 是获取行数组的更快方法，array_push() 更快捷旨在向数组添加多个值或者要将其视为堆栈。

The code you have posted is correct, however in the following lines you forgot to put $ before the $arr variable:

arr[$i] = $i;
arr["name"] = row["name"];
arr["age"] = row["age"];

Also the line arr[$i] = $i; should probably be something like $arr['rowNum'] = $i; having the key be the same as the value seems redundant.

And it should be said that using $arrContainer[]=$row; as @RiaD suggested is a quicker way to get an array of the row, array_push() is more intended to add multiple values to an array or if you want to treat it as a stack.

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心如荒岛 2024-12-05 13:27:50

$allValuesArr = array();
$arr= array();
while($myRow = mysql_fetch_array($result))
{
    $arr["uid"] = $myRow["uid"];
    $arr["name"] = $myRow["name"];
    $arr["surname"] = $myRow["surname"];
    $arr["age"] = $myRow["age"];
    $arr["telephone"] = $myRow["telephone"];
    $arr["prof_id"] = $myRow["prof_id"];
    //$allValuesArr[] = $arr;
    array_push($allValuesArr, $arr);
}

$strJSON = json_encode($allValuesArr);
echo $strJSON;

我得到的回应：

[{"uid":"1","name":"Sergii","surname":"Surname1","age":"26","telephone":"0976543135","prof_id":"1225423"},{"uid":"2","name":"Slava","surname":"Surname2","age":"24","telephone":"06353524","prof_id":"13584384"},{"uid":"3","name":"Desame","surname":"Surname3","age":"28","telephone":"0973153584","prof_id":"35843815"}]

$allValuesArr = array();
$arr= array();
while($myRow = mysql_fetch_array($result))
{
    $arr["uid"] = $myRow["uid"];
    $arr["name"] = $myRow["name"];
    $arr["surname"] = $myRow["surname"];
    $arr["age"] = $myRow["age"];
    $arr["telephone"] = $myRow["telephone"];
    $arr["prof_id"] = $myRow["prof_id"];
    //$allValuesArr[] = $arr;
    array_push($allValuesArr, $arr);
}

$strJSON = json_encode($allValuesArr);
echo $strJSON;

response I get:

[{"uid":"1","name":"Sergii","surname":"Surname1","age":"26","telephone":"0976543135","prof_id":"1225423"},{"uid":"2","name":"Slava","surname":"Surname2","age":"24","telephone":"06353524","prof_id":"13584384"},{"uid":"3","name":"Desame","surname":"Surname3","age":"28","telephone":"0973153584","prof_id":"35843815"}]

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