Question

Hello我有一个数组，其中包含其他两个数组，使用以下代码：

    foreach($relations as $rel){
        $data[$i]["relationTo"] = $rel["name"];
        $data[$i]["relation"] = $rel["relation"];
        $i = $i+1;
    }
    foreach($relations as $rel){
        $children[$i]["id"] = $rel["id2"];
        $children[$i]["name"] = $rel["sname"];
        $children[$i]["data"] = $data;
        $i = $i+1;
    }
    foreach($relations as $rel){
        $relationArray[$i]["id"] = $rel["id"];
        $relationArray[$i]["name"] = $rel["name"];
        $relationArray[$i]["children"] = $children;
        $i = $i+1;
    }

当我使用以下方式打印出来时

print_r($relationArray);

它打印以下内容：

Array ( [2] => Array ( [id] => 4 [name] => Albaraa [children] => 
Array ( [1] => Array (      [id] => 5 [name] => Sadi [data] => 
Array ( [0] => Array ( [relationTo] => Albaraa [relation] => Father ) ) ) ) ) )

我正在使用json_encode，我需要以某种方式在json中输出，不包括开头的数组索引计数...当我使用时json输出：

echo json_encode($relationArray);

目前是这样的：

 {"2":{"id":"4","name":"Albaraa","children":
 {"1":{"id":"5","name":"Sadi","data":  [{"relationTo":"Albaraa","relation":"Father"}]}}}}

在前两个数组的前面有“2”和“1”...这不是我想要实现的那样：

 {"id":"4","name":"Albaraa","children":
 {"id":"5","name":"Sadi","data":  [{"relationTo":"Albaraa","relation":"Father"}]}}}}

非常感谢任何帮助！

Answer 1

几种解决方案

1）不要通过[$i]输入值，准备新的完整内部数组并将其放入array_push中

2）如果你仍然想这样做，你只能提取值：

print_r(json_encode(array_values($array)));

如何摆脱PHP中数组中的索引数量

1 个答案: