Firebase 数据库删除具体数据 php

Question

How can I remove firebase specific data? I use the php Kreait\Firebase library.

  $fg = $database->getReference('raw_check_out')->orderByChild('reciptno')->equalTo($recipt)->getSnapshot();

  $reb = $fg->getValue();
  $fg->remove();

but this is not working.

Answer 1

Based on your code example:

$fg = $database
    ->getReference('raw_check_out')
    ->orderByChild('reciptno')
    ->equalTo($recipt)
    ->getSnapshot();

Here $fg does not hold the reference, but the snapshot.

If you want to remove the reference after you have retrieved the data you need, you need the reference itself:

$fg = $database->getReference('raw_check_out');

$query = $fg->orderByChild('reciptno')->equalTo($recipt);
$reb = $query->getSnapshot()->getValue();

$fg->remove();

Answer 2

This function is for unsubscribing users if you want to remove "user-id-8776" and your structure is like:

public function unsubscribe($uid)
 {
    $ref = $this->database->getReference()->getChild('users')->getChild($uid)->orderByChild($uid)->getReference();
    $path = $ref->getUri()->getPath();
    $ref->remove();
 }

For any other data is more or less the same, you just have to find the reference of data you want to delete it and then you can remove it.

Answer 3

Check Image

You have another option to resolve this problem. This solution given below:

    $url = "https://<projectID>.firebaseio.com/chatList/".<jsonFileName>."json";
    $ch = curl_init();
    curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "DELETE");
    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
    curl_setopt($ch, CURLOPT_POST, 1);
    curl_setopt($ch, CURLOPT_HTTPHEADER, array('Content-Type: text/plain'));
    $result = curl_exec($ch);
    curl_close($ch);