Question

我的桌子上有10条记录。

id    value
------------
1      record-1
2      record-2
3      record-3
4      record-4
5      record-5
6      record-6
7      record-7
8      record-8
9      record-9
10     record-10

我想创建这个json：

[

{
"1":"record-1",
"2":"record-2"
},


{
"3":"record-3",
"4":"record-4"
},


{
"5":"record-5",
"6":"record-6"
},


{
"7":"record-7",
"8":"record-8"
},


{
"9":"record-9",
"10":"record-10"
}

]

假设所有记录都在$records变量中。

Answer 1

我们需要在这里使用array_chunk()功能。

试试这个：

$result = array_chunk($records, 2, true);
echo json_encode($result);

输出：

[{"1":"record-1","2":"record-2"},{"3":"record-3","4":"record-4"},{"5":"record-5","6":"record-6"},{"7":"record-7","8":"record-8"},{"9":"record-9","10":"record-10"}]

Answer 2

你可以试试这个：

$crecords = array_chunk($records , 2, true);

返回两个记录作为数组元素。现在使用for循环生成json：

foreach($crecords as $jsons){
    //print_r($jsons)...you got the json object by json_encode..
}

Answer 3

你可以试试这个：

<?php
$array = array(
    'record-1',
    'record-2',
    'record-3',
    'record-4',
    'record-5',
    'record-6',
    'record-7',
    'record-8',
    'record-9',
    'record-10',
    );




array_unshift($array, "phoney"); // changing array keys from starting 0 to 1 // dummy value
unset($array[0]); // unsetting or removing value from array key 0



$final_arr = array_chunk($array, 2,true);


$json = array();
foreach ($final_arr as $key => $value)
{
    $json[$key] = $value;
}
echo json_encode($json);
?>

这应该为您提供所需的json输出。

输出：

  [
    {
        "1": "record-1",
        "2": "record-2"
    },
    {
        "3": "record-3",
        "4": "record-4"
    },
    {
        "5": "record-5",
        "6": "record-6"
    },
    {
        "7": "record-7",
        "8": "record-8"
    },
    {
        "9": "record-9",
        "10": "record-10"
    }
  ]

如何从十个记录创建五个对象

3 个答案: