Question

我需要从php获得一个看起来完全像这样的JSON：

{
    "nodes": [
                {"x": 250, "y": 250, "color": "grey", "name":"1", "r":"28", "fixed":true},
                {"x": 120, "y": 150, "name":"somename", "score": -12.2, "icon": "someicon"},
                {"x": 140, "y": 150, "name":"someothername", "score": -0.08, "icon": "someothericon"}
            ],
    "links": [
                {"source":  0, "target": 1, "distance": 180},
                {"source":  0, "target": 2, "distance": 180},
                {"source":  0, "target": 2, "distance": 180}
            ]
}

我从db获取的节点数组，所以我之后做的是计算有多少，因为我必须为每个节点做一个链接（从1到n）。我尝试过这样做：（count($res)会给我一些节点）

$count = count($res);

for($i=1;$i<$count;$i++)
{

    $strings['source'] = 0;
    $strings['target'] = $i;
    $strings['distance'] = 180;

}

在事先定义$strings = array();时，但这个计数器的作用是，它只会在数组中放置最后一个值，而实际上，我需要$count个，我需要保留这种特定格式，以便将其传递给JS进行进一步处理。

$res的内容如下。

（[0] =＆gt;数组（[ID] =＆gt; 67 [得分] => 0.05 [dscore] =＆gt; 0.24 [xcord] =＆GT; 91 [ycord] =＆gt; 391 [name] =＆gt; somename [ticker] =＆gt; tickname [icon] =＆gt; someicon [datafrom] =＆gt; 2017-01-01）[1] =＆gt;数组（[ID] =＆gt; 68 [得分] =＆GT; -0.32 [dscore] =＆gt; 0.55 [xcord] =＆gt; 120 [ycord] =＆gt; 428 [name] =＆gt; othername [ticker] =＆gt; tickname [icon] =＆gt; someothericon [datafrom] =＆gt; 2017-01-01）[2] =＆gt;数组（[ID] =＆gt; 69 [得分] =＆gt; -0.32 [dscore] =＆gt; 0.21 [xcord] =＆gt; 482 [ycord] =＆gt; 268 [name] =＆gt; mysanantonio [ticker] =＆gt; tickname [icon] =＆gt;其他一些图标[datafrom] =＆gt; 2017-01-01）

我需要

循环$i次，并为每个循环增加值 target中的links乘以
合并两个数组以获取JSON 输出与在beggining

Answer 1

你应该让$strings成为一个二维数组。

$k = 0;
for($i=1;$i<$count;$i++) {

    $strings[$k]['source'] = 0;
    $strings[$k]['target'] = $i;
    $strings[$k]['distance'] = 180;
    $k++;
}   

$strings = array_values( (array)$strings );
echo json_encode($strings);

php到json有两种特定的数据格式

1 个答案: