Question

我有一个代码，可以得到数组格式的Json响应

{  "data": {
"findAllUsers": {
  "pendingUsers": [
    {
      "userId": "123",
      "firstName": "FirstNamemz8",
      "lastName": "LastName0Hz",
      "status": "Pending"
    },
    {
      "userId": "456",
      "firstName": "FirstNameEgl",
      "lastName": "LastNameVCC",
      "status": "Pending"
    }
  ]
}  }}

我需要以列表格式获取userID的列表。下面是我使用的代码不起作用。请通过有效的代码帮助我。

Answer 1

$query = "SELECT * FROM wp_posts WHERE post_status = 'publish' AND post_type LIKE 'product%'";

if ($result = $mysqli->query($query)) {
    printf("<br>Select returned %d rows.\n", $result->num_rows);
    echo "<table>";
    echo "<thead>";
    echo "<tr>";
    echo "<td>ID</td><td>Termék neve</td><td>Termék linkje</td><td>Termék kép</td>";
    echo "</tr>"; 
    echo "<thead>";
    while ($row = $result->fetch_assoc()) {
        $post_id = $row["ID"];

        echo "<tr>";
        echo "<td>".$post_id."</td>";
        echo "<td>".$row["post_title"]."</td>";
        echo "<td>".$row["guid"]."</td>";
        $query_img = "SELECT meta_value FROM wp_postmeta WHERE meta_key ='_thumbnail_id' AND post_id = $post_id";
        $result_img = $mysqli->query($query_img);
        $img = $result_img -> fetch_assoc();
        $query_img_2 = "SELECT meta_value FROM wp_postmeta WHERE meta_key ='_wp_attached_file' AND post_id = ".$img['meta_value']."";
        $result_img_2 = $mysqli->query($query_img_2);
        $img_2 = $result_img_2 -> fetch_assoc();
        echo "<td>".$img_2["meta_value"]."</td>";
        echo "</tr>";
    }       
    echo "</table>";

    /* free result set */
    $result->close();
}

需要从Java中的JSON响应中提取特定值

1 个答案: