Question

我从fetch.php文件中以离子形式从HTTP源中获取数据：

$mysqli =mysqli_connect("test.com", "test", "test@123", "test");

mysqli_set_charset($mysqli,"utf8");

$query = "SELECT * FROM store";

$dbresult = $mysqli->query($query);

while($row = $dbresult->fetch_array(MYSQLI_ASSOC)){

    $data[] = array(
        'sto_gender_id' => $row['sto_gender_id'],
        'sto_category_id' => $row['sto_category_id'],
    );
}
if($dbresult){
    $result = json_encode($data);             
}
else {
    $result = "{'success':false}";
}
echo $result;

我通过此提供者category.ts

获得了这些数据

import { HttpClient } from '@angular/common/http';
import { Injectable } from '@angular/core';

@Injectable()
export class CategoryProvider {

  constructor(public http: HttpClient) {
    console.log('Hello CategoryProvider Provider');
  }
  getCategory(){
    return this.http.get<any[]>('https://www.test.com/test/fetch.php')
  }
}

我通过此html文件显示结果：

<button ion-button block *ngFor="let category of allCategory" (click)="goToGenderPage(category)">
  {{category.sto_category_id}}{{category.sto_gender_id}}
</button>

但是问题是它显示具有相同sto_category_id

的多个记录

[{"sto_gender_id":"1","sto_category_id":"1"},
{"sto_gender_id":"2","sto_category_id":"1"},
{"sto_gender_id":"1","sto_category_id":"4"},
{"sto_gender_id":"2","sto_category_id":"2"}]

，但我不希望这样做，因此有任何方法可以基于sto_category_id过滤结果，以仅显示多个中的一个sto_category_id。喜欢：

[{"sto_gender_id":"1","sto_category_id":"1"},
{"sto_gender_id":"1","sto_category_id":"4"},
{"sto_gender_id":"2","sto_category_id":"2"}]

那么最好的方法是什么。

Ionic：筛选Json数据以不显示具有相同ID的多个数据

0 个答案: