如何在php mysqli中生成json对象
我可以用PHP中的MYSQLi查询生成这个JSON吗如何在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
{
"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[]