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如何在php mysqli中生成json对象_Php_Json_String_Mysqli - Fatal编程技术网

如何在php mysqli中生成json对象

如何在php mysqli中生成json对象,php,json,string,mysqli,Php,Json,String,Mysqli,我可以用PHP中的MYSQLi查询生成这个JSON吗 { "name": "Top Node", "children": [ { "name": "Bob: Child of Top Node", "parent": "Top Node", "children": [ { "name": "Son of Bob", "parent": "Bob: Chi

我可以用PHP中的MYSQLi查询生成这个JSON吗

{
    "name": "Top Node",
    "children": [
      {
        "name": "Bob: Child of Top Node",
        "parent": "Top Node",
        "children": [
          {
            "name": "Son of Bob",
            "parent": "Bob: Child of Top Node"
          },
          {
            "name": "Daughter of Bob",
            "parent": "Bob: Child of Top Node"
          }
        ]
      },
      {
        "name": "Sally: Child of Top Node",
        "parent": "Top Node"
      }
    ]
  }
您可以使用对sql查询的结果进行编码

例如:

    $rows = array();
    while ($r = mysqli_fetch_array("SELECT * FROM TABLE_NAME", MYSQLI_ASSOC)) {
        $rows[] = $r;
    }
    echo json_encode($rows);  //print result in json format

检查这个可能重复的感谢您的回答,但我想问一下在父子树中使用的嵌套json。要嵌套json,您需要嵌套查询输出。我的意思是,将嵌套的sql结果传递给
行[]